Working with Polynomials:A How-To Manual

Kent M. Willis

RaptorMath.com

Copyright © 2015 by Kent M. Willis. All rights reserved.

One of the most troublesome parts of algebra for many people is dealing with polynomials,especially factoring them. It has so many rules, and “if this, then that” situations that a lotof people are tempted to give up on it. Instead, you should take the time to work on theseskills until you feel fairly confident that you can handle one kind of factoring — then moveon to the next and practice until you conquer that one too.

Then comes the ultimate challenge: factoring any kind of polynomial. Don’t worry aboutthat one until you are ready for it — first you need to learn factoring one step at a time.

So, first we will cover the basics of how to add, subtract, multiply, divide, and evaluatepolynomials; these skills must be mastered before you begin factoring. When you are goodat them, master one kind of factoring at a time. Get plenty of practice — this stuff is notsomething that you can merely read if you want to retain it.

Some places in this book have optional topics that appear in shaded boxes. Study these onlyif you are interested in learning a little more. Often this is material on why something worksinstead of merely how to use a procedure. Sometimes it is a topic that helps to prepare astudent for courses beyond Algebra, such as Calculus. You may always skip over the shadedboxes without fear that you are missing out on something you need to learn for Algebra.

Sample of a Shaded Box

The material in a shaded box is optional.

1

Contents

1 Basics of Polynomials 4

1.1 Prime Factorization of Integers . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.1.1 Best Method for Prime Factorization . . . . . . . . . . . . . . . . . . . . 4

1.1.2 Using a Cheat Sheet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.2 Greatest Common Factor (GCF) . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.2.1 Finding the GCF of Natural Numbers . . . . . . . . . . . . . . . . . . . 6

1.2.2 Finding the GCF of Terms with Variables . . . . . . . . . . . . . . . . . 7

1.3 Evaluating Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.3.1 Plug-in Method of Evaluation . . . . . . . . . . . . . . . . . . . . . . . . 8

1.3.2 Synthetic evaluation: a nifty alternative . . . . . . . . . . . . . . . . . . 9

1.3.3 (Optional) Why synthetic evaluation is possible . . . . . . . . . . . . . 11

1.4 Adding and Subtracting Polynomials . . . . . . . . . . . . . . . . . . . . . . . 11

1.4.1 Adding Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.4.2 Subtracting Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

1.4.3 Working with more than one variable . . . . . . . . . . . . . . . . . . . 12

1.5 Multiplying Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

1.5.1 Multiplying One Term times One Term (One-by-One) . . . . . . . . . . 13

1.5.2 Multiplying One Term times Two or More Terms (One-by-More) . . . 14

1.5.3 Multiplying Two or More Terms times Two or More Terms (More-by-More) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

1.5.4 Procedure: Multiplying on the Grid . . . . . . . . . . . . . . . . . . . . 15

1.5.5 Procedure: Clarence the Clam . . . . . . . . . . . . . . . . . . . . . . . . 16

1.5.6 Multiplying More Things Together . . . . . . . . . . . . . . . . . . . . . 17

1.6 Special Products (Shortcuts) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

1.6.1 Squaring a binomial (Double-Stuff Shortcut) . . . . . . . . . . . . . . . 17

1.6.2 Multiplying a Sum and Difference (No-Stuff Shortcut) . . . . . . . . . 19

1.6.3 Finding Greater Powers of a Polynomial . . . . . . . . . . . . . . . . . 19

1.7 Dividing Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

1.7.1 Dividing one term by one term (one-by-one) . . . . . . . . . . . . . . . 21

1.7.2 Dividing two or more terms by one term (more-by-one) . . . . . . . . 21

1.7.3 Dividing two or more terms by two or more terms (more-by-more) . . 22

1.7.4 Alternate Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

2

2 Introduction to Factoring 26

2.1 Earning a Black Belt in Factoring . . . . . . . . . . . . . . . . . . . . . . . . . . 26

2.2 How to know when you are done factoring . . . . . . . . . . . . . . . . . . . . 27

3 Fundamental Factoring Methods 28

3.1 White Belt: Factoring out the Greatest Common Factor (GCF) . . . . . . . . . 28

3.2 Yellow Belt: Factoring by Grouping . . . . . . . . . . . . . . . . . . . . . . . . . 29

3.2.1 Unfactorable polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . 32

3.2.2 Some tips for oddball problems: . . . . . . . . . . . . . . . . . . . . . . 32

3.3 Orange Belt: Factoring Bare Trinomials x2 + bx + c . . . . . . . . . . . . . . . . 33

3.3.1 The Key Number Method . . . . . . . . . . . . . . . . . . . . . . . . . . 33

3.3.2 Special notes regarding negatives . . . . . . . . . . . . . . . . . . . . . 34

3.4 Purple Belt: Factoring General Trinomials ax2 + bx + c . . . . . . . . . . . . . 35

4 Special Pattern Factorizations 38

4.1 Green Belt: Factoring the Difference of Squares . . . . . . . . . . . . . . . . . . 38

4.2 Blue Belt: Factoring the Sum or Difference of Cubes . . . . . . . . . . . . . . . 39

4.3 Red Belt: Factoring Perfect Square Trinomials . . . . . . . . . . . . . . . . . . . 41

5 Mixed Factoring (Conclusion of Red Belt) 42

5.1 How to Choose the Correct Factoring Method . . . . . . . . . . . . . . . . . . 42

5.2 Things to Watch Out For . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

6 Brown Belt: Advanced Factoring 43

6.1 Basic Synthetic Division . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

6.2 Advanced Synthetic Division . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

6.3 Using Verified Roots to Factor Polynomials . . . . . . . . . . . . . . . . . . . . 48

6.3.1 Listing all possible rational roots . . . . . . . . . . . . . . . . . . . . . . 49

6.3.2 Using a Graphing Calculator to Narrow the Choices . . . . . . . . . . . 50

6.4 Analyzing graphs to find roots . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

6.4.1 Factors associated with roots . . . . . . . . . . . . . . . . . . . . . . . . 51

6.4.2 Number of “zero roots” (roots equal to zero) . . . . . . . . . . . . . . . 51

6.4.3 Using a graph to find non-zero real roots . . . . . . . . . . . . . . . . . 52

6.4.4 Adjusting for integer coefficients . . . . . . . . . . . . . . . . . . . . . . 53

6.4.5 Looking for multiplicity . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

6.4.6 Irrational and Irreal roots always come in pairs . . . . . . . . . . . . . . 53

A Appendix of Reference Materials 54

3

1 Basics of Polynomials

1.1 Prime Factorization of Integers

This is a process of breaking down a number into its prime factors. There are several waysto do this. Textbooks usually show the tree method of factoring (which probably everyonehas seen at least once), but there are several shortcomings to this method:

• Suppose you are given a number that you can’t think of how to get there with multi-plying two other numbers? Do you just give up and say it’s prime after 5 minutes, 20minutes, or 4 weeks?

• The ends of the sticks with the prime factors are jagged and all over the place and notin any order. It’s too easy to miss seeing one when you are collecting them for the finalanswer.

• You have to sort your answer out at the end unless you are happy with something like17 · 7 · 2 · 5 · 2 · 7 · 2. I would much rather see this: 23 · 5 · 72 · 17.

Fortunately, there is a method that fixes all these shortcomings.

1.1.1 Best Method for Prime Factorization

Keep a list of prime numbers at least up through 20 handy! 2, 3, 5, 7, 11, 13, 17, 19

1. We always start with 2 as the Number To Check (NTC) because it’s the smallest primenumber. We will repeat the following substeps, working our way up through the primenumbers, small to large.

(a) See if the number is divisible by the NTC.

(b) If it is, do the division.

(c) See if the answer is also divisible by the NTC. Repeat until the answer is “no, it isnot divisible by the NTC.”

(d) Choose the next prime number from the list as the new NTC. Don’t skip to a largerone or you won’t know when to quit. Repeat from step 1a for the new NTC.

2. Keep repeating for bigger and bigger NTCs until one of these things happens:

(a) the answer to a division problem is a prime number, or

(b) the answer to an attempted division is smaller than the NTC.

(c) If there were no successful divisions before the answer to an attempted divisionwas smaller than the NTC, then the number you were working with was prime.

3. The prime factors are the collection of numbers you divided by together with the primeanswer to the last division.

4

There is only one correct answer, and everyone should get the same one correct answer byany valid method.

Example: Find the prime factorization of 60.

1. Try dividing by 2 repeatedly until it doesn’t work any more:60 ÷ 230 ÷ 215

Notice the prime factors are collecting on the right side after the ÷ signs.

2. Now we’re done with the 2s, so try dividing by 3, since it is the next prime number:60 ÷ 230 ÷ 215 ÷ 35

The answer to the last division was a prime number, so we are done. The answer is acollection of the numbers on the right end of each line (including the lonely one on thebottom line), all strung together with multiplication signs between them:2 · 2 · 3 · 5 or 22 · 3 · 5

Always check your answer to see that each number in the list is really a prime number, andthat they multiply together to get back the number you started with.

Example Find the prime factorization of 24.24 ÷ 212 ÷ 26 ÷ 23

Answer 2 · 2 · 2 · 3 or 23 · 3

Example Find the prime factorization of 35.35 ÷ 2 doesn’t work35 ÷ 3 doesn’t work35 ÷ 57

Answer 5 · 7

Example Find the prime factorization of 194.194 ÷ 297 ÷ 3 doesn’t work97 ÷ 5 doesn’t work97 ÷ 7 doesn’t work97 ÷ 11 doesn’t work, but the answer to 97 ÷ 11 is less than 11.

So we stop. 97 must be prime.Answer 2 · 97

5

1.1.2 Using a Cheat Sheet

Sometimes students are allowed to use a page of notes that includes the prime factorizationsof a multiptude of prime factorizations. Such a page is included in the appendix at the endof the book.

1.2 Greatest Common Factor (GCF)

1.2.1 Finding the GCF of Natural Numbers

Procedure

1. Find the prime factors for each of the numbers (see page 4), organizing them intolists.

2. Loop through the first number’s list of factors, checking to see if that factor is ineach and every list of factors. If it is not in every number’s list of factors, it is notcommon, so skip it and continue on with the next factor in the first list.

3. If a factor is common to every list of factors, write down the factor in the commonfactors list, joining factors together with multiplication dot, and put the exponentto be the lowest exponent on any of the lists. Repeat step 2 with the next factor inthe first list.

4. If there is nothing in the common factors list, the answer is 1 (since 1 is a factorof every number). If there is only one number in the common factors list, it is theGCF. Otherwise, multiply all the common factors together.

Example Find the GCF for 12, 18, and 96

1. Each number gets its own list of prime factors:12 = 22 · 318 = 2 · 32

96 = 25 · 3CF =

2. Begin with the first factor of the first list (2 in this case). Since all three lists have a2 to some power in them, then 2 is common.

3. Write down the lowest power of 2 that appears in any list. One way of thinking ofthis is to ask, "Every term has at least how many factors of 2?" The list for 18 hasa plain 2 with no exponent. That is the lowest power of 2 in any of the three lists.So write down a 2 in the common factors list.Repeat these two steps for the next factor in the first list, namely 3. All three listshave a 3 to some power, so it is common also. The lowest power of 3 that appearsin any list of factors is a plain 3 in the list for 12. So write a multiplication dot afterthe 2 you found before and then write the 3 that you found in common.The first list has no more factors, so we are done repeating steps 2 and 3. Ourcommon factor list looks like: 2 • 3

6

4. Multiplying the common factors together, we get the GCF: 6

Example: Find the GCF for 14 and 46

1. Each number gets its own list of prime factors:14 = 2 • 746 = 2 • 23CF =

2. Begin with the first factor of the first list (2 in this case). Since both lists have a 2 tosome power in them, then 2 is common.

3. Write down the lowest power of 2 that appears in any list. The list for 14 has aplain 2 with no exponent. That is the lowest power of 2 in both lists. So writedown a 2 in the common factors list.Repeat for the next factor in the first list. So check the 7. The list for 46 has nofactor of 7, so we skip 7.The first list has no more factors, so we are done repeating steps 2 and 3. Ourcommon factor list looks like: 2

4. There is nothing to multiply in the common factors list, so the GCF is: 2

1.2.2 Finding the GCF of Terms with Variables

Procedure Just the same as for plain numbers, except it’s actually easier. We just say that thevariables are already in prime factored form. So find the lowest exponent for any letterthat is common to all the terms.

Example Find the GCF of 4x2y and 3xyz

1. Each term gets its own list of prime factors:4x2y = 22x2y3xyz = 3xyzCF =

2. Begin with the first factor of the first list (2 in this case). Since the second list doesnot have a 2, it is not common so we skip it.Next factor is an x. Since both terms have an x, it is common.

3. The lowest exponent of x is the plain x with no exponent. So write it down as aCommon Factor.The next factor in the first list is y. The second term also has a y, and the lowestexponent is the plain y. It becomes the second factor in the Common Factors list.The first list has no more factors, so we are done repeating steps 2 and 3. Ourcommon factor list looks like: x · y

4. Multiplying the common factors list together, we find the GCF is: xy

7

1.3 Evaluating Polynomials

1.3.1 Plug-in Method of Evaluation

When you are given a polynomial and a value for the variable in it, you can simply plug inthe value in place of each spot the letter appears. Then use the order of operations to simplifythe result, and you should end up with a plain number value.

Example Find the value of 2x3 − 3x2 + 8x− 6 if x = 2.

2x3 − 3x2 + 8x− 6

plugin 2 · 23 − 3 · 22 + 8 · 2− 6

expos 2 · 8− 3 · 4 + 8 · 2− 6

mult 16− 12 + 16− 6

add 14

There are a few things to be careful about:

• The exponent on the x will transfer to the value of x, but not to the coefficient that ismultiplied times the x.

• Always take care of simplifying exponents before multiplying the coefficients.

• Be careful with signs!

• We can never solve a polynomial to find what x is equal to — we have to be told whatx is before we can do this evaluation.

We can also evaluate a polynomial with other letters and using negative values.

Watch out! Be sure to use parentheses around negative values that you plug in for thevariable.

Example Find the value of 2y3 − 3y2 + 8y− 6 if y = −1.

2y3 − 3y2 + 8y− 6

plugin 2(−1)3 − 3(−1)2 + 8(−1)− 6

expos 2(−1)− 3(1) + 8(−1)− 6

mult −2− 3− 8− 6

add -19

Remember that the negative is also affected by the exponent. So, for each term in the polyno-mial, if the exponent is odd, you get a negative value; but if the exponent is even, the valueis positive.

8

1.3.2 Synthetic evaluation: a nifty alternative

Synthetic evaluation is a time-saving and mildly fun method that condenses evaluating apolynomial down to a simple alternation between multiplying and adding. You won’t needto find what (−2)5 is, or worry about exponents at all. It can be a little confusing at first, butonce you get used to it, you can zip through one of these evaluation problems in less thanhalf the time needed for the conventional standard method.

Here is the generic setup for a problem. We will evaluate things like ax3 + bx2 + cx + d whenx = n.

space for “carrying” numbers up

+ a b c d addition area above this line

n multiplication row

The top rows (the “addition area” above the top horizontal line) are for numbers that areadded together in columns. The multiplication row is on the bottom. After adding thenumbers in the column above, the sum is written on the bottom row and then multipliedby the number on the left edge (n as shown here), and the result of that is carried up to thetop of the next column. We rope off the right-most column with a second vertical line. Theanswer will appear in the bottom right spot. Now we’ll step through the process in detail.Some of the question marks in your head should start to disappear when you see the processin action.

Procedure Synthetic Evaluation

For ease in explanation, we will be working an example as we go through the procedure. We willevaluate 2x2 + 3x− 14 when x = 2.

1. Arrange the terms in order of descending powers (from highest power of x down tolowest). Our example is already good to go.

2. Put in a placeholder for any “missing terms.” For example, x2 + x + 1 has no missingterms, because from highest to lowest, the exponent on the x doesn’t skip anything.But x3 + x + 1 has a missing term: there is no x2 term in it. Placeholders can be insertedwith a zero times the missing term. So x3 + x + 1 should be written as x3 + 0x2 + x + 1.Our example has no missing terms.

3. Write the coefficents in a row with a little space between each one. Put a horizontal lineunder the coefficients, and leave a blank row above the coefficient. Draw two verticallines, one to the left of all the coefficients and one singling out the rightmost coefficient.The setup looks like this:

+ 2 3 −14

9

4. Write the number you are plugging in for x in the bottom left corner. This is called the“multiplication row.” Your setup should look like this:

+ 2 3 −14

2

5. Now we begin a pattern. It goes like this: beginning on the left, add a column in theaddition area (above the horizontal line), put the answer in the same column in themultiplication row (under the line), multiply by the number at the left of the multipli-cation row, and put the answer at the top of the next column (in the addition area). Thispattern repeats until all the columns are full. Here is what your work should look likeat each step of the way:

+ 2 3 −14

2 2

Nothing to add to the 2, so it drops down

4

+ 2 3 −14

2 2

Multiplied (2)(2), answer goes in the next column.

4

+ 2 3 −14

2 2 7

Added 4+3

4 14

+ 2 3 −14

2 2 7

Multiplied 2 times 7

4 14

+ 2 3 −14

2 2 7 0

Added -14+14.

6. The bottom number in the last column is the answer to the evaluation. In this case, it is0.

Example Find the value of 2y3 − 3y2 + 8y− 6 if y = −1.

Here is the setup: + 2 −3 8 −6

−1

10

Here it is completely filled out:

−2 5 −13

+ 2 −3 8 −6

−1 2 −5 13 −19

Answer: −19

1.3.3 (Optional) Why synthetic evaluation is possible

If we look at a polynomial such as 2x3 + 3x2− 5x + 4, one of the problems we encounter withevaluation is those exponents. We have to get the signs correct and not mess up anywhereif we want the correct answer. But there is a way, using a little bit of factoring out a com-mon factor (a method covered elsewhere), to avoid having any exponents at all. The onlyoperations will be addition and multiplication. Let’s take a look.

If we group all but the last term together, we can factor out a common x, and we have have:x(2x2 + 3x − 5) + 4. For the sake of making things neater for later, let’s put the x after theparentheses. It doesn’t change anything except make it easier to see a pattern. So we get(2x2 + 3x− 5)x + 4. Now, inside the parentheses, we can do that same thing again, and weget: ((2x + 3)x− 5)x + 4. If we had a higher order polynomial, we could keep repeating thatprocess as many times as needed until there are no more exponents higher than 1.

Now, if we are careful and thorough, we can see that there is a nice alternating pattern ofmultiplication and addition. For any given input value (that gets plugged in for x), wemultiply 2 by the input, then add 3, then multiply by the input, subtract 5, multiply bythe input, then add 4. The result will be the same as taking 2 times the cube of the inputplus 3 times the square of the input minus 5 times the input plus 4 (as we would do by theconventional method).

That is exactly what synthetic evaluation does for us, but we can use it without factoring anddoing so much writing. We condense it down to a very simple state and alternate betweenmultiplying and adding. Once you get used to it, you can zip through one of these evaluationproblems in less than half the time needed for the plug-in method.

1.4 Adding and Subtracting Polynomials

1.4.1 Adding Polynomials

There is only one thing we can do to add polynomials. That is to combine like terms. Anyonewho does more than that is trying to do the impossible, and will get a wrong answer.

Some people like to put one polynomial per line, lining up like terms under like terms above.This is sometimes called “adding polynomials vertically.” Or you can do it horizontally, likethe examples below in “Subtracting Polynomials.”

Example Add 4x3 − 3x2 + 2x and 6x3 + 2x2 − 3x

11

4x3 −3x2 +2x

6x3 +2x2 −3x

10x3 −x2 −x

Answer: 10x3 − x2 − x

Example Add x2 − 2x + 5 and 4x2 − 2

Watch out for missing terms! This has a missing x term in the second polynomial. You caneither add in a 0x or just leave a gap. Either way gets the same answer.

x2 −2x +5

4x2 +0x −2

5x2 −2x +3

(Using a 0x placeholder)

x2 −2x +5

4x2 −2

5x2 −2x +3

(Leaving a gap instead)

1.4.2 Subtracting Polynomials

Instead of directly subtracting polynomials, we should distribute the minus sign and thenadd. Why? It’s just about guaranteed that you’ll miss changing a sign if you are trying tokeep too many things in your head at once. So just distribute and then add. It works.

Note: You can work these in “vertical format” just as well. These will be worked horizontallyjust as an example.

Example Subtract (14y3 − 6y2 + 2y)− (2y3 − 7y2 + 6)

(14y3 − 6y2 + 2y)− (2y3 − 7y2 + 6)

dist 14y3 − 6y2 + 2y− 2y3 + 7y2 − 6

like 12y3 + y2 + 2y− 6

1.4.3 Working with more than one variable

Watch out! Be sure that you only combine like terms. If they don’t match exactly on theletters, don’t force the issue.

Example Simplify (5m3n + 3m2n2 − 4mn)− (7m3n−m2n2 + 6mn)

12

(5m3n + 3m2n2 − 4mn)− (7m3n−m2n2 + 6mn)

dist 5m3n + 3m2n2 − 4mn− 7m3n + m2n2 − 6mn

like −2m3n + 4m2n2 − 10mn

�� ��Practice Time!

Add or subtract the polynomials as indicated:

1.(3x2 − 2x + 12

)+

(x2 − 4

)

2.(4x3 − 3x

)−

(3x2 − 5

)

3.(5x2 − 3x + 2

)+

(x2 + 8x− 7

)−

(3x2 − 6x + 5

)

1.5 Multiplying Polynomials

One trick that we use a lot in Algebra is to break complicated problems down into a groupof simpler ones. This is certainly true for multiplying polynomials. We begin with the basicbuilding block of one term times one term. All the other kinds of multiplication are accom-plished by doing a bunch of these basic ones and adding the results together. So learn wellhow to do the simplest ones, because sometimes you may have 30 of them to do for onecomplicated problem.

1.5.1 Multiplying One Term times One Term (One-by-One)

This is the basic building block for this entire section. All the other kinds of multiplicationare broken down to this level, so make sure you really get this.

1. Regroup the numbers and letters, putting similar things together.

2. Multiply them using exponent-combining techniques learned in previous sections.

3. If some element has nothing similar to it, just put it in as it is.

4. Normally, we write the answer with the number first (including a minus sign if needed),then all the letters in alphabetical order.

5. We always get one single term for the answer of a one-by-one problem.

13

Example Multiply 3x4 · 4x

3x4 · 4x

regroup 3 · 4 · x4 · x

simplify 12x5

Example Multiply −3x4 · y2

−3x4 · y2

nothing to regroup −3x4 · y2

put it together −3x4y2

Most people (after grasping the concept well) will skip writing down the steps in this process,since they are easy enough to do in your head most of the time.

1.5.2 Multiplying One Term times Two or More Terms (One-by-More)

This turns out to be the usual kind of distributing. So the terms together in parentheses getseparated, and each one is multiplied by the term outside the parentheses. Each one of thosemultiplications is just 1 term times 1 term, as described above.

• Treat the + and − signs as part of the term immediately after the sign, and then all theterms are added together.

• We will get the same number of terms in the answer as we started with inside theparentheses.

Example Find the product: 2x4(3x2 + 2x− 5)

2x4(3x2 + 2x− 5)

dist 2x4 · 3x2 + 2x4 · 2x + 2x4(−5)

simp 6x6 + 4x5 − 10x4

Example Find the product: −8m3(4m3 + 3m2 + 2m− 1)

−8m3(4m3 + 3m2 + 2m− 1)

dist −8m3 · 4m3 + (−8m3)3m2 + (−8m3)2m + (−8m3)(−1)

simp −32m6 − 24m5 − 16m4 + 8m3

14

1.5.3 Multiplying Two or More Terms times Two or More Terms (More-by-More)

This is the most complicated kind of polynomial multiplication and can be called “multitermdistribution;” it has to be broken down into a bunch of one-by-one problems, each term fromthe first parentheses times each term from the second parentheses. The answers for each ofthese one-by-one may have some like terms, so we look to see if we can combine them. Thenall the terms are added together.

There are three systems that I know about for multiterm distribution: FOIL, Clarence theClam, and the Grid.

FOIL method: Many people have been taught to use the FOIL method for this, and it’s OKas far as it goes. The main point of it is to remember all the combinations of what to multiplytogether. But FOIL only works for two-by-two multiplication.

Clarence the Clam: Clarence the Clam works for two-by-two or more-by-more, so you canskip the FOIL method and just have one method to learn. If you want to keep using FOIL, goahead, but you will need to use Clarence or another system of multiterm distribution whenyou have more than 2 terms in a set of parentheses.

The Grid: The grid provides three advantages:

• it has a neat rectangular placement that helps make sure you don’t leave any term out,

• it helps you locate like terms that you can combine,

• and it also works in reverse to help you with factoring later on.

I recommend the grid for students who have never been good at multiplying polynomialsbefore.

The main point of all multiterm distribution methods is that each and every term from oneset of parentheses must be multiplied by each and every term in the other set of parentheses.As long as you keep careful track and don’t skip any terms, you will be OK.

1.5.4 Procedure: Multiplying on the Grid

1. Write the first polynomial above the grid, leaving good spacing so the terms are sep-arate with each one above a cell of the grid. Keep any minus signs with the termfollowing it.

2. Write the second polynomial on the left of the grid, with similar separation and spacing.

3. In each cell, write the product of the term at the top of its column times the term left ofits row.

4. Like terms will line up along diagonal lines — you can add them together to simplify.

15

Example Multiply(3x2 + 2x− 1

)(2x− 7)

After writing both polynomials (steps 1 and 2), we get a grid like this:

3x2 2x −1

2x

−7

Now multiply the term above by the term on the left by column and row and we get(being careful with the signs):

3x2 2x −1

2x 6x3 4x2 −2x

−7 −21x2 −14x 7

Notice that like terms are on the diagonals, and we can combine them:

Answer: 6x3 − 17x2 − 16x + 7

1.5.5 Procedure: Clarence the Clam

This is named for the shape of the arcs that help keep track of what is finished.

1. Rearrange the parentheses (if needed) to put the parentheses containing the fewestterms first.

(a) Multiply the first term from the first parentheses times each term from the secondparentheses. This is like a one-by-n problem above.

(b) Repeat step 2 with the second term from the first parentheses times each term fromthe second parentheses. Write the answer for each one-by-one problem below alike term from step 2 (if there is a like term).

(c) Repeat again with the third term from the first parentheses and so on until yourun out of terms in the first parentheses.

(d) Combine all like terms.

Example Find the product: (x2 + 3x + 5)(x− 4)

1. Rearrange to (x− 4)(x2 + 3x + 5) to make things easier and neater.2. Multiplying the x from the first parentheses times each of the terms in the second

parentheses, we get: x3 + 3x2 + 5x3. Multiplying the−4 from the first parentheses times each of the terms in the second

parentheses, we get: −4x2− 12x− 20 and when we line those terms up under liketerms from what we got on step 2, we have:

x3 +3x2 +5x

−4x2 −12x −20

4. We are out of terms in the first parentheses, so stop repeating.5. Combining like terms, we get the answer: x3 − x2 − 7x− 20

16

1.5.6 Multiplying More Things Together

So far we only talked about multiplying two things together. But what if we stacked upmore, like 3(x + 1)(x− 4). In these cases, we can only multiply two at a time to get a partialanswer, which is enclosed in parentheses; then multiply the partial answer by the remainingpart(s). The best advice here is to save the simplest and easiest multiplication for last. Somepeople refuse to believe me, and they end up with working something that is harder than ithad to be. So really and truly, get the hardest stuff done first, and save the easiest for last.OK?

Example Simplify 3(x + 1)(x− 4)

We will save the multiplying by 3 for last, since it is the easiest!

3(x + 1)(x− 4)

clam 3(x2 − 4x + 1x− 4)

like 3(x2 − 3x− 4)

dist 3x2 − 9x− 12

Notice that the answer from multiplying the two sets of parentheses remains insideparentheses so that the 3 is distributed. Always put these partial answers in parenthe-ses. Then check to see if the parentheses are doing a job. If not, they can be deleted.�� ��Practice Time!

Multiply the polynomials as indicated:

1.(3x2 − 2x + 12

) (x2 − 4

)

2.(4x3 − 3x

) (3x2 − 5

)

3. 3x (8x− 7) (−6x + 5)

1.6 Special Products (Shortcuts)

1.6.1 Squaring a binomial (Double-Stuff Shortcut)

One thing that comes along quite often is working with a polynomial times itself. If it hap-pens to have two terms, we have a shortcut that makes your homework go quicker. Let’slook at an example worked out the regular way from last section, and then dissect it a littleand come up with a general shortcut.

17

For example, let’s look at (x + 4)2. The first thing to do is to talk about what that meansand what it does not mean. Remember that an exponent of 2 is a shortcut way of writingsomething multiplied by itself. So what is that something here? It is (x + 4) taken as a unit.

Watch out! We absolutely CANNOT ever say that (x + 4)2 means x2 + 42. IT DOES NOT!

Sharing the power ONLY works with multiplication or division inside the parentheses. Sowhat is the correct way to write it out the long way? That would be (x + 4)(x + 4). And if wemultiply that out using multiterm distribution, we get x2 + 4x + 4x + 16, and then combinelike terms, right?

But before we combine the like terms, let’s look at what we have. We do have an x2 and a 42,like the warning above talked about. But we also have other stuff. We have 4x added in. Twotimes, as a matter of fact. Where did they come from? From the first x times the last 4, andthe last x times the first 4. Since we’re talking about the first and last paren groups alwaysbeing the same, we can see that we will ALWAYS get a deal like that: multiply the two termstogether and then double it because it gets added in twice. That is why this Double-Stuffshortcut works.

Procedure Double-Stuff Shortcut

1. Square the first term to get the first cookie.

2. Multiply the first and last terms together and then double it to get the double-stuff.Be careful with the sign! If one of the terms is negative, this double-stuff term willbe negative and you would put − double-stuff. Otherwise, put + double-stuff.

3. Square the last term to get the last piece to get the second cookie. Put + (ALWAYSplus) and this cookie at the end.

Example Simplify (y− 3)2

1. Squaring the first term (y) gives us y2 for the first cookie.

2. The double-stuff is 2(y)(−3) = −6y.

3. Squaring the last term (−3) gives us positive 9 for the second cookie.

Answer: y2 − 6y + 9

Example Simplify (10m + 7)2

1. The first term is 10m. So the first cookie is 100m2.

2. The double-stuff is 2(10m)(7) = 140m.

3. The last term is 7. So the last cookie is 49.

Answer: 100m2 + 140m + 49

18

1.6.2 Multiplying a Sum and Difference (No-Stuff Shortcut)

This kind looks sort of like squaring a binomial except that one time we have a plus and onetime a minus, like (x + 2)(x − 2). They may try to hide it in tricky ways, like changing theorder as in (x + 2)(−2 + x). But if can be put into the pattern of the same two terms, oncewith a plus and once with a minus, then we have a good shortcut for it.

If we work (x + 2)(x − 2) out the long way, we get x2 − 2x + 2x − 4, and the two middleterms add to 0, so we end up with x2 − 4. It turns out that those middle terms will ALWAYSdrop out in this kind of multiplication. So we have another shortcut: the No-Stuff Shortcut.

Procedure No-Stuff Shortcut

1. Square the first term to get the first cookie.

2. Square the last term to get the last cookie.

3. Put a minus sign in between the cookies (there is no stuff!)

Example Simplify (3 + y)(3− y)

1. The first term is 3. Don’t rearrange them when the sum/difference pattern is clear.So the first cookie is 32 = 9.

2. The last term is y. We don’t worry about the sign, since it’s + in one case and − inthe other. So the last cookie is y2.

3. Put together with a minus sign, we get the answer: 9− y2

Example Simplify (10m + 7)(10m− 7)

1. The first term is 10m. So the first cookie is 100m2.

2. The last term is 7. So the last cookie is 49.

3. Put together with a minus sign, we get the answer: 100m2 − 49

1.6.3 Finding Greater Powers of a Polynomial

There isn’t really much of a shortcut here, except to realize that we can break down higherpowers into groups of squares and use the double-stuff shortcut. After that, it gets messywhen we multiply the parital answers together.

Please note that there is a method for finding greater powers of a binomial, called “binomialexpansion.” However, it is fairly complicated for beginners and is usually considered worth-while for quite high powers, such as (x + y)6 or higher. By then, it is worth the time to learna new technique. For now, we will not worry about binomial expansion, and just carry onwith the procedures we are more familiar with.

Here is a glimpse of what binomial expansion looks like, but it is beyond the scope of thisbook to explain what all this fancy stuff means.

19

Binomial Expansion

(a + b)n =n

∑k=0

(nk

)an−kbk

where (nk) =

n!k! (n− k)!

Example Simplify (3k− 2)4

Let’s split this into two pieces: (3k− 2)2(3k− 2)2.

Then we can use the double-stuff shortcut to change this to (9k2− 12k + 4)(9k2− 12k +4).

Unfortunately, there is no shortcut for squaring a polynomial with more than twoterms, so we have to use normal multiterm distribution.

Answer: 81k4 − 216k3 + 216k2 − 96k + 16

Example Simplify (4x + 1)3

Let’s split this into two pieces: (4x + 1)(4x + 1)2

Then we can use the double-stuff shortcut to change this to (4x + 1)(16x2 + 8x + 1)

Unfortunately, there is no shortcut for this either, so we have to use normal multitermdistribution.

Answer: 64x3 + 48x2 + 12x + 1�� ��Practice Time!

Find the special products:

1. (5x− 2)2

2. (2x + 5y) (2x− 5y)

3. (x + 5)3

1.7 Dividing Polynomials

There are three levels of polynomial division, going from pretty easy to pretty difficult. Thekey to success with them is to get really good at the easy ones, and then see how the moredifficult ones use the easy kind. In effect, we break a complicated division problem downinto a series of easy ones!

20

1.7.1 Dividing one term by one term (one-by-one)

This is the simplest type of dividing with polynomials, and is a basic block for all the othertypes. All we really do with this is to set up the division as a fraction and reduce it. Youshould reduce it in steps, first the sign, then the numbers, then each of the letters in turn.

Example Divide12x2

6xEverything is positive so the answer will be positive.

Working with the plain numbers, we see that 126 reduces to 2

1 , or just plain 2 upstairs.

Working with the x terms, we see thatx2

xreduces to

x1

, or just x upstairs.

Since the whole downstairs become a 1, just write the upstairs on the ground floor.

Answer: 2x

Example Divide147x

The number parts reduce to 2 upstairs.

The letter parts don’t reduce, leaving x downstairs.

Answer:2x

1.7.2 Dividing two or more terms by one term (more-by-one)

This type is recognized by having only one term downstairs, with two or more terms up-stairs. We break this into a series of one-by-one divisions by reversing what we do when weadd fractions. If we had to add the fractions x

7 + 37 , since they have the same downstairs,

we would just add the tops. But x and 3 are not like terms, so our answer would be x+37 .

Now let’s look at x+37 as a division problem (x + 3) ÷ 7. We can break it into two simpler

division problems: x7 + 3

7 , which is the reverse of what we did with adding the fractions. Inthis illustration, it doesn’t help because neither of the fractions we broke it into reduce. Let’stry an example that does do something.

Example Divide (12m6 + 18m5 + 30m4)÷ 6m2

We break this into three fractions, one for each term upstairs. Notice the downstairs

part is repeated without change:12m6

6m2 +18m5

6m2 +30m4

6m2

Then we reduce each fraction, independent of the others: 2m4 + 3m3 + 5m2

Sometimes we have subtractions to worry about. Just keep the sign with the term followingit and add all the fractions together. And sometimes we get messy fractional results. That isOK. Just be sure that you don’t push everything upstairs just because it looks more finishedthat way.

21

Example Divide50m4 − 30m3 + 20m

10m3

Broken into three separate fractions, we have50m4

10m3 +−30m3

10m3 +20m10m3

Be careful! Note that the minus sign for the middle term followed along to the up-stairs of the middle fraction.

Each of these is reduced for the final answer: 5m− 3 +2

m2

1.7.3 Dividing two or more terms by two or more terms (more-by-more)

This method is modeled after long division from elementary school, only it looks about tentimes worse with all the alphabet soup mixed in. There is no substitute for watching a prob-lem in action, and I can’t everything you need into notes that don’t move and don’t talk. Youhave to see it to believe it.

Procedure Long Division with Polynomials

1. Put the terms in order of descending powers and fill in any “missing” terms. Put thedividend inside the division bracket and the divisor outside it.

2. Repeat this pattern:

(a) Divide the first term of the dividend by the first term of the divisor.

(b) Put the answer above the bracket.

(c) Multiply the divisor by the answer part you just wrote (not the whole thing everytime), and put that underneath the dividend, lining up like terms.

(d) Subtract by changing the signs of what you wrote below the divisor and thenadding the like terms. The highest order term will drop out if you are doing itright. The remaining terms become the new dividend.

(e) Stop repeating if the power of x in the new dividend is lower than the power of xin the divisor. At that point, the “new dividend” is called the remainder.

3. Write the answer as it appears above the bracket, and if there is a remainder, write it asa fraction added on at the end. Remainder over divisor.

Watch out! Some important things that folks forget to do on these division problems:

• Always put polynomials in order of descending powers before dividing.

• Always check for “missing” terms. Fill them in with zero times the missing powerbefore you begin.

22

• Always add the remainder fraction, if the remainder is not zero. If the remainder isnegative, still put a plus sign and then the remainder fraction, with the minus signupstairs. The downstairs will always be the divisor you are dividing BY. The upstairsis the remainder.

• Stop when the remainder has a lower power of x than the divisor you are dividing BY.

Example Dividex2 + 5x + 6

x + 3First, we set up the problem. The number we are dividing up (the upstairs) is calledthe dividend, and we put it inside a bracket. The number we are dividing by (thedownstairs) is called the divisor, and we put it outside the bracket. You should have asetup like this now:

x + 3 x2 +5x +6

Next, we pay attention to the first term of the dividend and the first term of the divisor.Don’t worry about the rest of the terms. They will get worked out in their own goodtime. We divide the first term inside the bracket by the first term outside the bracket,

and reduce:x2

x= x. This means the first part of the answer is x. Put it above the

bracket. You don’t need to worry about aligning it with anything. Just put it up on topsomewhere. You now have a setup like this:

x

x + 3 x2 +5x +6

Next, we multiply x(x + 3) and put that below the dividend. You then have:

x

x + 3 x2 +5x +6

x2 +3x

Then, in order to subtract easily and accurately, we change the sign of each term thatwe just wrote, then ADD the like terms. We get:

x

x + 3 x2 +5x +6

−x2 −3x

2x +6

We are ready to repeat the cycle, this time with2xx

= 2. Since that is positive, we put+2 up above the bracket. Now we have:

23

x +2

x + 3 x2 +5x +6

−x2 −3x

2x +6

Multiply the +2(x + 3) and put it below the dividend:

x +2

x + 3 x2 +5x +6

−x2 −3x

2x +6

2x +6

Change signs and then add:

x +2

x + 3 x2 +5x +6

−x2 −3x

2x +6

−2x −6

0

The zero remainder says we can stop and the answer is above the bracket.

Answer: x + 2

The remaining examples will not be shown in step-by-step fashion, but you should be ableto duplicate them by following the same pattern.

Example Divide2x3 + 5x + x2 + 13

2x + 3(Note the terms are out of order.)

x2 −x +4

2x+3 2x3 +x2 +5x +13

2x3 +3x2

−2x2 +5x

−2x2 −3x

8x +13

8x +12

1

24

Answer: x2 − x + 4 +1

2x + 3

Example Dividex3 − 8x− 2

(Note the missing x2 and x terms.)

x2 +2x +4

x− 2 x3 +0x2 +0x −8

x3 −2x2

2x2 +0x

2x2 −4x

4x −8

4x −8

0

Answer: x2 + 2x + 4

Example Divide2m5 + m4 + 6m3 − 3m2 − 18

m2 + 3(Note the strange alignment to keep like terms

together)

2m3 +m2 −6

m2 + 3 2m5 +m4 +6m3 −3m2 +0m −18

2m5 +6m3

m4 +0m3 −3m2

m4 +3m2

−6m2 +0m −18

−6m2 −18

0

Answer: 2m3 + m2 − 6

Example Divide3x3 + 7x2 + 7x + 8

3x + 6(Sometimes we have fractions to deal with)

25

x2 +13 x +5

3

3x + 6 3x3 +7x2 +7x +8

3x3 +6x2

x2 +7x

x2 +2x

5x +8

5x +10

−2

Answer: x2 + 13 x + 5

3 +−2

3x + 6

1.7.4 Alternate Method

There is another method that can be used to divide polynomials, but it is more abstractand can be confusing to beginners. It is called “synthetic division,” and it is covered in theadvanced section near the end of the book. If you are anxious to try it, start with the basicsynthetic division on page 43. If you get good at synthetic division now, you will be aheadof the game later on.

2 Introduction to Factoring

2.1 Earning a Black Belt in Factoring

Similar to the martial arts, you will earn “belts” of different colors as your skills in factoringdevelop. Your instructor has a Black Belt in factoring, and will help you along your ownpath. The different colors of the belts, from the most basic to most advanced are listed be-low. Generally, students only need to go through Red Belt until they take College Algebra.College Algebra will require some more advanced methods detailed in the Brown Belt, andteachers will need a Black Belt.

26

White Factor out the GCF: Find factors that all the terms have in common, and pullthem out front.

YellowFactor by grouping: If four or more terms have no common factors,sometimes we can group the terms by some commonality, find a commonfactor for each little group, then find a common factor for the groups.

Orange Key Number method: A step-by-step method for factoring trinomials likex2 + 7x + 6 (ones with a bare x2 term).

PurpleGeneral Trinomial method: A two step-by-step method for factoringtrinomials like 2x2 + 7x + 10 (with some number other than 1 times the x2

term).

Green Difference of squares: A pattern-based method for factoring binomials likex2 − 4.

Blue Sum/difference of cubes: A pattern-based method for factoring binomials likex3 + 8 or x3 − 8.

Red Mixed factoring: Factoring a wide variety of the above types without beingtold which method to use, as well as multi-step factorizations.

Brown Advanced factoring: Limited-guess and check procedures for polynomialsnot covered previously, such as x4 − 3x3 +−5x2 + 7x− 12.

Black Mastery: Expertise gained through helping others learn how to factor.

When factoring, there is a definite order of steps to take. Taking things out of order can leadto using the wrong method and generally leads to the wrong answer. Beginners usually feela little overwhelmed at first with all the different methods. But these methods are givenin a good order, and if you keep referring back to this section as you progress through thedifferent belts, it will all come together in the end. Sometimes the simple and easy stuff tripspeople up more than the harder stuff, so always follow these steps and you should be OK.

2.2 How to know when you are done factoring

1. There are no addition or subtraction signs outside of parentheses.

2. Each and every factor inside parentheses cannot be factored any further.

3. The answer has been checked by multiplying out the answer to see if we get the originalproblem back again.

27

3 Fundamental Factoring Methods

3.1 White Belt: Factoring out the Greatest Common Factor (GCF)

There are several ways to do this factoring. Usually people who have done this quite a fewtimes develop their own style. You begin to see common factors by inspection, and can pullout more common factors later if you did not find all the common factors the first time. Justbe sure that when you are done, you really have found all the common factors.

For students who are not used to factoring, here is one sure-fire method for factoring out theGCF.

Procedure: Factoring out the GCF

1. Find the GCF for all of the terms(see 1.2.2 on page 7).

2. Write the GCF first, followed by emptyparentheses for the “leftovers.”

3. Compute the leftovers term by term,putting them inside the parentheses.Here is how to do this, one term at atime:

(a) Make a fraction with the originalterm upstairs, and the GCF down-stairs.

(b) Reduce the fraction. The down-stairs should all cancel out.

(c) Put each term’s leftovers inside theparentheses, making sure to keepthe plus or minus sign the originalterm has.

Note: If the GCF is exactly identical toone of the terms, the fraction will re-duce to 1. Put the 1 in its place in theleftovers.

4. Check the leftovers again to see if thereis another common factor.

5. Check your answer by distributing.You should get the original problemback.

Example: Factor out the GCF:

24x2y4z + 42x3y2

1. Find the GCF of all numerical val-ues, then find the GCF for each let-ter: 6x2y2.

2. Put the GCF first,with parenthesesafter it for the “leftovers”, like this:6x2y2 ( ).

3. The first term’s leftovers are:24x2y4z6x2y2 which reduces to 4y2z and

the second term’s are: +42x3y2

6x2y2

which reduces to +7x so the finalresult is 6x2y2 (4y2z + 7x

). That

will be the final answer, as long aswe don’t find any trouble.

4. A quick check of the leftoversinside the parentheses shows nocommon factors for the leftovers.That’s good.

5. Check: Distributing your answeryields the original problem back.That’s good, but don’t put theproblem down as the answer. Youranswer should have parentheses,with no addition or subtractionoutside.

Example: Factor out the GCF: 2x + 2

28

1. Find the GCF of all numerical values. GCF = 2. In this case, there is nothing toadditional for the GCF for each letter, since the second term has no letters at all.

2. Put the GCF first, with parentheses after it for the “leftovers”, like this: 2 ( )

3. The first term’s leftovers are:2x2

which reduces to x and the second term’s are:+22

which reduces to +1 so the final result is 2 (x + 1)

4. A quick check of the leftovers inside the parentheses shows no common factorsfor the leftovers. That’s good.

5. Distributing your answer yields the original problem. Check!

3.2 Yellow Belt: Factoring by Grouping

Usually “factor by grouping” problems have exactly four terms, and we usually group theterms in pairs. Later on, you might see some more peculiar “factor by grouping” problems,and there are a few tips later on in this section for those.

There are two different approaches to factoring by grouping. The first one is the way thatalmost everyone uses and you may have seen in a previous class. The second way uses agrid, and it helps keep the plus and minus parts straight with very little effort. So, if you arealready used to factoring by grouping the usual way, feel free to keep using it. However, ifyou have never had success at this kind of factoring, give the grid method a try. You mightlike it!

Procedure Factoring by grouping (traditional method)

1. Pick the pairs by choosing the biggest ugliest part (BUP) and grouping it with theterm that resembles it the most. The choice is a judgment call, and it won’t stopthe show if you pick a different one from what I would.

2. Rearrange the terms if needed, and put parentheses around the pairs. Thereshould be a plus sign between the two sets of parentheses. If there is a minussign, hold on. You have a little extra work:

(a) Factor out a –1 from the second set of parentheses.(b) Check that when you distribute the minus sign, everything would end up like

when you started.

3. Factor out the GCF (see page 28) for each set of parentheses. Then the leftoversfrom the first set should be the same as the leftovers from the second set. If theydon’t match, try again, grouping the terms in a different way.

4. Treat the leftovers in parentheses as a single unit. It becomes the GCF for the twosets. So factor it out, making a new set of parenthesized leftovers.

5. Check your answer by multiplying it out!

Example Factor by grouping: 2ab + 2b + 7a + 7

1. I would say the BUP is the first term, 2ab. The term that most resembles it is 2b.

29

2. There is no need to rearrange the terms this time. So putting the groups in paren-theses, we get: (2ab + 2b) + (7a + 7). It’s good that there was a plus sign therebetween the parentheses. No extra work!

3. Now to factor out the GCF for each group. Our example becomes: 2b (a + 1) +7 (a + 1). And look! The stuff inside the parentheses matches! Yay!

4. Factoring out the (a + 1) that the two parts have in common, we get: (a + 1) (2b + 7).Check your answer!

Example: Factor by grouping: 8x2 + 6x− 12xy− 9y

1. I would say the BUP is the first term, 8x2. The term that most resembles it is 6x.You might disagree and work it out another way. That’s OK.

2. There is no need to rearrange the terms this time. But if we just merrily put thegroups in parentheses, we get something bad:

(8x2 + 6x

)− (12xy− 9y), which

has a problem. What’s wrong? There is a minus sign there between the parenthe-ses. That means extra work! Backing up, first we have to factor out –1 from the lasttwo terms: − (12xy + 9y). Then putting it together right, we have:

(8x2 + 6x

)−

(12xy + 9y)

3. Now to factor out the GCF for each group. Our example becomes: 2x (4x + 3)−3y (4x + 3). And look! The stuff inside the parentheses matches! Yay!

4. Factoring out the (4x + 3) that the two parts have in common, we get: (4x + 3) (2x− 3y).

5. Check your answer!

Procedure Factoring by grouping with the grid

1. Assuming you have a usual problem with four terms, make a 2-by-2 grid, and fill it inwith the four terms like this:

BUP (if it is positive)

term least like the BUP

or

BUP (if it is negative)

term least like the BUP

The remaining two terms go in any order into the last two cells, but the top left cellmust be positive for everything to work out correctly.

2. Factor out the GCF (see page 28) above each column and on the left of each row, makingthe sign match the nearest cell.

3. Do a quick check to see if each cell is equal to the product of the GCF in its column andthe GCF on its row. This will catch over 90% of all errors. Do not continue if even oneof the four does not check!

4. Wrap the GCF parts in parentheses, and the answer is:(the sum of the top terms)(the sum of the left terms)

30

5. Check your answer by multiplying it out!

The following examples are the same as we did using the usual method, so we should getthe same answer here.

Example Factor by grouping: 2ab + 2b + 7a + 7

1. I would say the BUP is the first term, 2ab. The term that least resembles it is 7.Since the BUP is positive, it goes in the top left cell. The grid gets filled in like this:

GCFs

2ab 2b

7a 7

Note that the 7a and the 2b terms could trade places, and it would not change theanswer.

2. Now fill in the GCF for each column and each row:GCFs a 1

2b 2ab 2b

7 7a 7

3. The quick check gives a go, since 2ab = a · 2b, 2b = 1 · 2b, 7a = 7 · a, and 7 = 7 · 1.

4. Wrapping the top GCFs in parentheses (with a plus sign since they are positive)and the side GCFs in like manner, we get: (a + 1) (2b + 7).

5. Check your answer!

Example: Factor by grouping: 8x2 + 6x− 12xy− 9y

1. I would say the BUP is the first term, 8x2. The term that least resembles it is −9y.You might disagree and work it out another way. That’s OK. So we fill in the gridlike this:GCFs

8x2 −12xy

6x −9y

Again, the 6x and the −12xy could be switched, and you will still get the sameanswer.

2. Filling in the GCF for each column and for each row, we get:

GCFs 2x −3y

4x 8x2 −12y

3 6x −9y

Note that the GCF for the right column is negative because the nearest cell wasnegative.

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3. The quick check gives a go: 8x2 = 4x · 2x, −12y = 4x (−3y), 6x = 3 · 2x, and−9y = 3 (−3y).

4. Wrapping our GCFs in parentheses, we get: (2x− 3y) (4x + 3).

5. Check your answer!

3.2.1 Unfactorable polynomials

Sometimes there are polynomials that don’t factor out right. You can try grouping themdifferently and see if it comes out that way. But some of them never can be factored bygrouping, and we would like to know how to spot them before we waste our time on them.

How to spot trouble before you fry your brain:

• Exactly one term is negative. With the sole exception of the second oddball problembelow, you can’t factor these by grouping. Because we put terms in pairs, you have tohave negatives in pairs also.

• None of the terms have any common factors. If you can’t match them up with anotherterm, it doesn’t do any good to group them. These can’t be factored.

3.2.2 Some tips for oddball problems:

• Usually six terms are grouped into two groups of three terms.

• Sometimes four terms are grouped three and one. Then the three are hopefully a perfecttrinomial square and the one lonely term is subtracted and is a perfect square. Thenyou can factor those groups as a difference of squares. (You should not see any of theseuntil Red Belt or beyond.)

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3.3 Orange Belt: Factoring Bare Trinomials x2 + bx + c

Sometimes instructors and textbooks only have one method for these factorizations, com-monly called the “guess and check” method. While this works (with practice), it also hasdrawbacks. If you find a polynomial that is unfactorable, you will always wonder if youtried every guess that you could.

There is a clear, step-by-step method that will let you check every possibility quickly, leavingno stone unturned. It is highly recommended, especially for beginners. It is called the “KeyNumber” method. If you want help with the “guess and check” method, look in a textbook.

3.3.1 The Key Number Method

The Key Number method is used only for trinomials (three terms) with no number (or thenumber 1) as the coefficient of the squared term. You can think of it as having a “bare x2

term.” This kind of factoring is quite common.

Examples of factoring problems that this method works for:

• x2 + 5x + 4

• x2 − 8x− 9

Note these have three terms, and the x2 term has no number in front of it. Examples offactoring problems that this method will not work for:

Polynomial Why the Key Number method won’t work

2x2 + 5x− 4 The x2 has a 2 in front of it.

x2 − 16 There are only two terms.

x2 − 4x + y2 − 4y There are four terms.

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Procedure: Key Number Method

1. Make sure the terms are in descendingorder of powers. Usually they are, butsome need to be rearranged.

2. When it is arranged in descending or-der of powers, the last term is the “keynumber.”

3. Make a list of pairs of numbers thatmultiply together to the key number.Watch your signs; see the special notesregarding negatives below.

4. Pick the pair that adds together to getthe middle term’s coefficient. If there isno pair that works, the trinomial is un-factorable, or prime.

5. Set up two sets of parentheses with thevariable plus one of the pair in each.

6. Check your answer by multiplying itout and combining like terms. Youshould get the original trinomial back.

Example: Factor 7x + x2 + 12

1. We need to rearrange the terms:x2 + 7x + 12

2. The last term is the key number: 12

3. Starting with 1, then 2, 3, and soon, list all the pairs of numbersthat multiply together to the keynumber:

1 122 63 4

4. Pick 3 and 4, because 3+ 4 = 7, thenon-x part of the middle term.

5. Set up two sets of parentheses withx plus one of the pair in each:(x + 3) (x + 4)

6. Check your answer!

3.3.2 Special notes regarding negatives

If the key number is positive, then both the numbers in the key pair have the same sign (+or –). The sign of the middle term tells you what sign they both have.

Example A x2 + 5x + 4 has key number +, so key pair have same sign. middle term is +, soboth +

Example B x2–5x + 4 has key number +, so key pair have same sign. middle term is −, soboth −

If the key number is negative, then one of the pair is positive and the other one is negative.The sign of the middle term tells you the sign of the heavyweight.

Example C x2 − 3x − 4 has key number –, so key pair mixed signs. middle term is −, soheavyweight −

Example D x2 + 3x − 4 has key number −, so key pair mixed signs. middle term is +, soheavyweight +

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Matching exercise

Match the following factored polynomials to Examples A through D above, based on thesigns of the numbers:

1. (x− 4) (x + 1)

2. (x + 4) (x + 1)

3. (x− 4) (x− 1)

4. (x + 4) (x− 1)

Example: Factor x2 − 7x + 12

1. Terms are in proper order.

2. The last term is the key number: 12

3. List all the pairs of numbers that multiply together to the key number. They mustmatch in sign because the key number is positive. And they must both be negativebecause the middle term is negative.

−1 −12−2 −6−3 −4

4. Pick −3 and −4, because −3 + (−4) = −7, the non-x part of the middle term.

5. Set up two sets of parentheses with x plus one of the pair in each: (x− 3)(x− 4)

6. Check your answer!

3.4 Purple Belt: Factoring General Trinomials ax2 + bx + c

There are several methods that I have seen to factor general trinomials, which are the kindwith a coefficient on the squared term, such as 3x2 + 5x − 12. These are different from thetrinomials that we used the Key Number method for; here we have trinomials (three terms)with a number greater than 1 as the coefficient of the squared term.

The first method I learned is called the “guess and check” method. It has several drawbacks,but it can be very quick if you are good at guessing the answer. It entails guessing theanswer and then using multiterm distribution to multiply out your guess, so you can checkto see if it multiplies back to the original problem. It does not, however, alert the student ifa polynomial is unfactorable. You are left feeling stupid if you can’t find the right answer toguess. I give this method a big Mr. Yuck sitcker.

Later I learned a method called the “Texas Box” method and then the “A-C” method. Theyboth have merit, and are perfectly fine to use. But when I analyzed them in detail, I foundthe two methods are actually identical except for the detail of how to finish up with factoringby grouping. We already have two methods for factoring by grouping, so there is nothing

35

new to learn here about that. So I merged them together in this book, and I will call this onething the General Trinomial method.

Once this method has done its work of splitting the middle term into two, you can finish upwith factoring the four terms by grouping — either by the traditional method or by using thegrid. Although the grid looks a little scarier, students who use it tend to make fewer errors,so I recommend it. You are always free to use the traditional method if you choose to.

Procedure: General Trinomial method

1. Make sure the terms are in descending order of powers. Usually they are, butsome need to be rearranged. Also, make sure the squared term is positive. If it isnot, factor out a –1 to force it positive. The method is difficult to use if the first term isnegative.

2. Take only the number part of the first term and multiply it by the last term to getthe “key number.” This is where the A-C method got its name — for Ax2 + Bx+C,you multiply A and C together.

3. Make a list of pairs of numbers that multiply together to the key number. Watchyour signs just like in the Key Number method (Blue Belt).

4. Pick the key pair that add together to get the middle term’s coefficient (non-x part).If there is no pair that works, the trinomial is unfactorable, or prime.

5. Take the middle term of the original trinomial, and tack the variable part of it ontoeach one of the pair you just found. Replace the middle term by the two terms(which add up to the original middle term).

6. Finish up by factoring by grouping (using either the grid or the traditional method).

Example Factor 6x2 − 17x + 10 using the General Trinomial method.

1. Terms are in proper order by powers.

2. Key number is 6 · 10 = 60.

3. The list of all pairs of numbers that multiply together to the key number. But thesigns must both be negative, because the key number is positive (meaning thesigns match) and the middle term is negative (meaning that both are negative).Refer to the Blue Belt’s Special Notes Regarding Negatives on page 34.

-1 -60-2 -30-3 -20-4 -15-5 -12-6 -10

4. The key pair that add together to get the middle term’s coefficient: –5 and –12,because they add to –17.

5. The middle term as the variable x, so we take our pair and put the same vari-able on them: −5x and −12x. We then put them in place of the middle term. (Ifeither of the pair were positive, we would need to put a + sign in with it.) Ourpolynomial is now: 6x2 − 5x− 12x + 10.

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6. We finish up by factoring by grouping.(a) If you use the traditional method, rearrange the terms to get the more simi-

lar parts together: 6x2− 12x− 5x + 10. Then add parentheses, being careful tochange the signs for the second pair because of the negative sign:

(6x2 − 12x

)−

(5x− 10).This becomes 6x (x− 2)− 5 (x− 2) when we factor the GCF from each pair,and so the final answer is (x− 2) (6x− 5).

(b) If you use the grid, the terms can be arranged like this:

GCFs

6x2 −12x

−5x 10Then find the GCFs for each column and each row:GCFs x −2

6x 6x2 −12x

−5 −5x 10So the final answer is (x− 2) (6x− 5).

Note: You do not need to use both the grid and the traditional method for the finishing upusing grouping. Choose your favorite, and then excel with it.

Example Factor using the General Trimomial method: 20x2 + 3x− 2

Note: You could use guess-and-check if you like it.

1. The terms are in proper order.2. The key number is (20) (−2) = −40.3. The list of factor pairs is:

–1 40

–2 20

–4 10

–5 8

4. The key pair is −5 and 8, since they add to positive 3 (the middle term’s coeffi-cient).

5. Replacing the middle term with the new pair, we get:

20x2 + 3x− 2 = 20x2 − 5x + 8x− 2

6. Factoring by grouping yields the final answer:

GCFs 4x −1

5x 20x2 −5x

2 8x −2Answer: (5x + 2) (4x− 1)

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4 Special Pattern Factorizations

4.1 Green Belt: Factoring the Difference of Squares

Use this table to help you determine perfect squares for factoring the difference of squares.

Square Root 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Square 1 4 9 16 25 36 49 64 81 100 121 144 169 196 225

Important: You can not factor the sum of squares. x2 + y2 is not factorable.1 Period. It isprime. But the difference of squares, like x2 − y2, is factorable. There is a pattern to it. Firstwe find the “square roots” (which we call A and B) and then we plug them into the pattern(A + B)(A – B) as the answer.

Procedure: Factoring the difference of squares

1. Find the square root of the first term and call it A. Then find the square root of thesecond term and call it B.

(a) To find the square root of a plain number, locate the number on the line labeled“square” in the table above. The square root is immediately above the numberyou found, in the line labeled “square root”. Or you can use the square rootfunction of your calculator.

(b) To find the square root of the letter parts (each letter individually), take halfthe exponent, or if you prefer, divide the exponent by 2.

2. Plug the square roots into the pattern (A + B)(A – B).

3. Check your answer.

1In an advanced method, we find a way to do this using complex numbers. But there is still no way to factorit using real numbers. So, unless you are working on advanced stuff in College Algebra, just remember that itis not factorable.

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Example: Factor x2 − 9y2

1. The first term’s square root is found by dividing the exponent by 2. This firstsquare root we call A:A = x1 = xThe second term’s square root is found in two steps. First, locate 9 on the squaresline of the table. The square root of 9 is 3. Next, the square root of y2 is y (dividingthe exponent by 2). So the second square root we call B:B = 3y

2. Plugging the square roots into the pattern (A + B)(A – B) we get:(x + 3y)(x− 3y)

3. Check your answer!

Example: Factor 4x2 − 25y2z4

1. The first term’s square root is found in two steps. First, locate 4 on the squares lineof the table. The square root is just above it. So the square root of 4 is 2. Next, welook at the letter part, and divide the exponent by 2. This first square root we callA:A = 2xThe second term’s square root is found in three steps. First, locate 25 on thesquares line of the table. The square root of 25 is 5. Next, the square root of y2

is y, and lastly, the square root of z4 is z2. So the second square root we call B:B = 5yz2

2. Plug the square roots into the pattern (A + B)(A – B) to get:(2x + 5yz2)(2x− 5yz2)

3. Check your answer!

4.2 Blue Belt: Factoring the Sum or Difference of Cubes

Use this table to help you determine perfect cubes for factoring the sum or difference ofcubes.

Cube Root 1 2 3 4 5 6 7 8 9 10 11 12

Cube 1 8 27 64 125 216 343 512 729 1000 1331 1728

This type of factoring requires using a pattern, kind of like the factoring the difference ofsquares. But there are some differences. Namely, the pattern is different. And we use cuberoots instead of square roots. Also you can factor with either a + or – (sum or difference). Sowe have two similar patterns: one for + and one for –.

Procedure: Factoring the sum or difference of cubes

1. Find the cube root of the first term and call it A. Then find the cube root of thesecond term and call it B.

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(a) To find the cube root of the number part, locate the number on one of the lineslabeled “cube” in the table above. The cube root is immediately above thenumber you found, in the line labeled “cube root”.

(b) To find the cube root of the letter parts (each letter individually), take one-third the exponent, or if you prefer, divide the exponent by 3.

2. Make a list to refer to while plugging things into the pattern. The list should con-tain five things: A, B, A2, AB, and B2. To find A2, multiply the A cube root timesitself. To get AB, multiply the two cube roots together. And to get B2, multiply theB cube root times itself.

3. Select the pattern that fits the original problem. Here are the two patterns. Theyare very similar.A3 − B3 = (A− B)

(A2 + AB + B2)

A3 + B3 = (A + B)(

A2 − AB + B2)4. Plug the five items from your list into the correct pattern.5. Check your answer.

Special Notes Regarding the Patterns

When selecting the correct pattern, there is some additional information that may help youmake the right choice, and maybe even memorize the patterns. Memorization helps yourspeed greatly.

In the first set of parentheses, put A + B or A − B, whichever sign matches the originalproblem. For x3 − y3, the first parentheses contain x − y; for x3 + y3, the first parenthesescontain x + y.Remember: same sign.

The second set of parentheses contains 3 terms. The first is A2. Square each number anddouble the exponent on each letter. The second term is AB (the two cube roots multipliedtogether), with the opposite sign as was used in first set of parentheses. And the last is B2

(always positive). So for x3 − y3, the last set of parentheses would contain x2 + xy + y2. Forx3 + y3, the last set of parentheses would contain x2 − xy + y2. The first and last terms areALWAYS plus, and the middle terms is the opposite of the sign used in the first parentheses.Remember: positive, opposite sign, positive.

Example: Factor 27x3 + 64y6

1. Find A. The cube root of 27 is 3, and the cube root of x3 is x1 or just plain x. SoA = 3xFind B. The cube root of 64 is 4, and the cube root of y6 is y2. So B = 4y2

2. Find A2: A2 = (3x)(3x) = 9x2

Find AB: AB = (3x)(4y2) = 12xy2

Find B2: B2 = (4y2)(4y2) = 16y4

3. Choose the pattern with + like the original problem: A3 + B3 = (A + B)(

A2 − AB + B2)4. Plug our As and Bs into the pattern, to get: (3x + 4y2)(9x2 − 12xy2 + 16y4)

5. Check your answer.

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4.3 Red Belt: Factoring Perfect Square Trinomials

There is one more factoring method to cover that is optional for beginners. Without it, youcan always use either the Key Number method or the General Trinomial method, whicheveris applicable, and get the same result. The advantage to learning this method is that it is alittle faster and easier if you are good at recognizing and dealing with patterns. So if youhate the Green and Blue Belts’ problems with patterns, you can skip this part at first. But besure to come back and learn it when you are more advanced — it can save you a ton of time.

This kind of factoring is very specialized, but if it applies to a given problem, it saves a lotof time. Most of the procedure below has to be completed before you know whether themethod works or not, so you will want to do a quick check first to save time.

Put the terms in order of descending powers (for one variable) and then check whether thefirst and last terms are both perfect squares. If so, this method is worth a shot.

Procedure: Perfect trinomial square shortcut

1. Rearrange the terms (if needed) into descending order of powers.

2. Find the square roots of the first and last terms (square roots, just as in Yellow Belt).Call them A and B.

3. Multiply the two square roots together, and double the answer. Call this 2AB.

4. Check to see if this 2AB matches the middle term of the original problem. Or if−2AB matches, that is good, too. Either way is OK. However, if it does not match,this method will not work. Stop and try another way.

5. Select the pattern that matches the original problem. The patterns are:A2 + 2AB + B2 = (A + B)2

A2 − 2AB + B2 = (A− B)2

6. Plug in your A and B. The answer is (A + B)2 or (A− B)2, whichever matches thepattern.

7. Check your answer.

Example: (This one works!) Factor: 4x2 − 12x + 9

1. Terms are in descending order of powers. Great!

2. The square roots A and B are: A = 2x, B = 3

3. Now find 2AB: (2)(2x)(3) = 12x

4. Is that equal to the middle term, or else the negative of the middle term? Yes, 12xis the negative of −12x

5. Choose the pattern that matches: A2 − 2AB + B2 = (A− B)2

6. Plug in your A and B. The answer is (2x− 3)2

7. Check your answer.

Example: (This one fails!) Factor: x2 + 3x + 9

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1. Terms are in descending order of powers. Great!

2. The square roots A and B are: A = x, B = 3

3. Now find 2AB: (2)(x)(3) = 6x

4. Is that equal to the middle term, or else the negative of the middle term? No,6x does not match 3x. Since it does not match, we can’t continue this method.The next step would be to try the Key Number method, since the squared termdoes not have a number part. The General Trinomial method would be used if thesquared term did have a number in front of it.

5 Mixed Factoring (Conclusion of Red Belt)

Now that you are good at all these different kinds of factoring, it is time to learn how tofactor a polynomial without being told which method to use. There are definite clues aboutwhich method to use, and also several pitfalls to learn about so that you don’t fall into them.First, study this section to learn what you can from a book, then practice determining whichmethod(s) to use for a random problem.

5.1 How to Choose the Correct Factoring Method

1. Always and always factor out all common factors first thing. (White Belt) If there aren’tany common factors, that’s OK. But you’ve got to check. Some of the methods belowwill not work if you don’t.

2. Count the terms. Then choose the correct method as follows:

• 4 or more terms: factor by grouping (Yellow Belt)

• 3 terms:

– perfect square trinomial (Red Belt, if possible)– Key number (if it has a bare x2 term) (Orange Belt)– General trinomial (otherwise, such as: 3x2 + 5x− 4) (Purple Belt)

• 2 terms:

– difference of squares (if exponents are multiples of 2) (Green Belt)– difference of cubes or sum of cubes (if exponents are multiples of 3) (Blue Belt)

3. Check each factor to see if it can be factored further.

4. Check your answer. If you multiply the factors together, you should get the originalproblem back again.

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5.2 Things to Watch Out For

There are three things that people commonly forget to do, any of which can leave them withthe wrong answer:

• forgetting to first and always check for common factors,

• forgetting to see if a factor can be factored further, and

• forgetting to check the answer by multiplying it out.

Every factoring problem answer should be checked!

6 Brown Belt: Advanced Factoring

These advanced factoring techniques are generally not used until College Algebra or beyond.

6.1 Basic Synthetic Division

There is one special technique with polynomials that can help in advanced factoring thatrequires our attention by the time you take College Algebra. In lower levels of Algebra, wecovered division with polynomials, but most students do not get any exposure to a niftytime-saving technique called “synthetic division.” If you have seen it before, it’s still good toreview it and make sure that you are practiced up on it.

Another reason to look at it now is that most instructors and textbooks that deal with syn-thetic division say that there are many problems that it cannot be used on, because they aretoo involved. These notes describe how to use synthetic division on any polynomial divisionwith a single variable. If there are two or more different variables, then it may be true thatyou can’t use synthetic division. But that should be the only thing that must stop you if youwant to use the method.

You might want to review the section on synthetic evaluation (see page ??), because syntheticdivision uses the same steps. As strange as it sounds, dividing a polynomial by x− 3 can beaccomplished by evaluating the polynomial at x = 3. All we need to do is take the non-xpart of the downstairs divisor, change the sign, and evaluate.

Procedure Synthetic Division

For ease in explanation, we will be working an example as we go through the procedure. We willdivide 2x2 + 3x− 14 by x− 2.

1. Arrange the terms in order of descending powers (from highest power of x down tolowest). Our example is already good to go.

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2. Put in a placeholder for any “missing terms.” For example, x2 + x + 1 has no missingterms, because from highest to lowest, the exponent on the x doesn’t skip anything.But x3 + x + 1 has a missing term: there is no x2 term in it. Placeholders can be insertedwith a zero times the missing term. So x3 + x + 1 should be written as x3 + 0x2 + x + 1.Our example has no missing terms.

3. Write the coefficents in a row with a little space between each one. Leave space formore numbers above them, and draw a line below them. Determine the degree of thedivisor. Draw two vertical lines as shown. You need to make the number of columnsto the left of the first vertical line match the number of columns to the right of thesecond line, and both these should be equal to the degree of the divisor. Our examplehas a divisor of degree 1, so we need one column on the left of the first vertical line and also onecolumn to the right of the second vertical line. The setup looks like this:

+ 2 3 −14

÷

4. Look at the polynomial you are dividing by (the downstairs if written as a fraction).The term with the highest power of x (or other variable) determines whether we needthe division row or not. If its leading coefficient is 1, omit the division row. But if it isanything else, put ÷ and the coefficient in the row just below the horizontal line. Nowwrite the opposite of the coefficient of all the lower power terms on the bottom row.If there are any missing terms in the divisor, put a 0 in their place. Our example has aleading coefficient of 1, so we will omit the division row. Your setup should look like this:

+ 2 3 −14

2

Note: when dividing by x− 2, you are actually evaluating at x = 2, and for x + 2, youare evaluating at x = −2.

5. Now we begin a pattern. It goes like this: starting at the left after the first verticalline, add all the numbers in the leftmost column of the addition area together, andput the answer under the line (in the division row if there is one; otherwise in themultiplication row), divide by the number in the division row (if there is one) and writethe answer below that in the multiplication row, multiply by each of the numbers in themultiplication row, and put the answer in the next column in the addition area abovethe line. If there are additional numbers to multiply by, put their products steppingover to the right. This pattern repeats until all the non-remainder columns are full.Here is what your work should look like at each step of the way:

Important: Do not multiply or divide in the remainder columns! Do ONLY the additionfor remainders!

44

+ 2 3 −14

2 2

first column only has a 2, so we drop it down

4

+ 2 3 −14

2 2

Multiplied (2)(2), answer goes into a higher row in the next column.

4

+ 2 3 −14

2 2 7

Added 4+3

4 14

+ 2 3 −14

2 2 7

Multiplied 2 times 7

4 14

+ 2 3 −14

2 2 7 0

Added -14+14.

1. The bottom row represents the answer to the division. You can read it from right toleft, kind of backwards. Keep going with higher powers of x as needed.

4 14

+ 2 3 −14

2 2 7 0

N/A x’s plain number remainder

Answer: 2x + 7 (Since the remainder is zero, we can drop that part)

Example Divide3x3 + 2x2 − 7x− 4

x + 3using synthetic division.

Here is the setup for the division:

+ 3 2 -7 −4

−3

Now we fill in all the blanks, using the usual pattern of adding down, then multiplyingto begin the next column:

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−9 21 −42

+ 3 2 −7 −4

−3 3 −7 14 −46

This time we did not get a zero remainder, so we will express it as a fraction (over thex + 3 that we are dividing by).

Answer: 3x2 − 7x + 14 +−46x + 3

6.2 Advanced Synthetic Division

Any student who is already familiar with synthetic division will notice that the setup shownherein has a different appearance from what many other books use. The reason for that is thatmost other books wimp out on the tougher problems and say that they can’t be done withsynthetic division. I changed the appearance (but not the method behind the procedure) tomake it more fluid for all types of problems. You are free to use the more conventional setupif you want, but it is less graceful and you will need to adapt it somewhat for the tougherproblems.

Here is the generic setup for a problem:ax3 + bx2 + cx + d

mx2 + nx + p.

addition area, number

+ a b c d of rows varies

÷ m division row (sometimes not needed)

−n −p multiplication row

The top rows, the “addition area” above the top horizontal line, are just like we had forsynthetic evaluation. Those numbers are added together in columns. Depending on theproblem, there may be more rows going up higher and higher, but they are all for addition,no matter how high we go. The division row (marked with the division symbol ÷) is notalways needed — we only include it when the leading coefficient of the divisor is not 1(m 6= 1). The multiplication row on the bottom, is just like for synthetic evaluation, butsometimes we need more columns on the left side of the line. The number of columns onthe left side of the first vertical line will match the degree of the divisor (the polynomial weare dividing by). The second vertical line marks the cutoff point for the remainder, and wealways need the same number of columns to the right of it as we have to the left of the firstvertical line. So there is always a symmetry between left and right.

Note: For the simplest division problems (which are very common), the setup we usematches exactly what we used for synthetic evaluation (see page 9).

Example Divide2x3 − 5x2 − 22x− 15

2x + 3using synthetic division.

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Note: This example requires a division row, since the leading coefficient downstairs isnot equal to 1. Here is the setup:

+ 2 −5 −22 −15

÷2

−3

Now we fill in the blanks, by taking in order: add, divide, and multiply.

−3 12 15

+ 2 −5 −22 −15

÷2 2 −8 −10 0

−3 1 −4 −5 *

Watch out! * Remember to never, ever divide or multiply in the remainder column(s).

Answer: x2 − 4x− 5

Example Divide6x4 − 17x3 − 61x2 − 23x + 16

2x2 − 7x− 15

Note: This example requires not only a division row, since the leading coefficient down-stairs is not equal to 1, but also two columns on the left and on the right. Here is thesetup:

+ 6 −17 −61 −23 16

÷2

7 15

Now we fill in the blanks, by taking in order: add, divide, and multiply.

21 45

+ 6 −17 −61 −23 16

÷2 6

7 15 3

Notice the 3 · 15 goes into the next column after the 3 · 7. The vertical position is notimportant, so just stack it on top of the heap.

47

14

21 45 30

+ 6 −17 −61 −23 16

÷2 6 4

7 15 3 2

This time the 2 · 7 and the 2 · 15 do not end up in the same row. That’s OK. Only thecolumn position is important.

14 -7

21 45 30 -15

+ 6 −17 −61 −23 16

÷2 6 4 −2 0 1

7 15 3 2 −1 * *

Now that we have completed the non-remainder part, all that is left is to do ONLYthe addition for the remainder. Also, note that the remainder part has the same kindof power business going on as the non-remainder part. In this example, we have theplain number part as the right-most column, and the x-term coefficient to the left of it.Since it’s zero in this example, we don’t have any x-term in the top of the remainderfraction.

Watch out! * Remember to never, ever divide or multiply in the remainder column(s).

Answer: 3x2 + 2x− 1 +1

2x2 − 7x− 15

Practice Problems

Try synthetic division on these puppies:

1.3x3 − 2x2 + 5x− 4

x + 2

2.y5 − 1y− 1

3.2k3 + 5k− 2

2k + 1

6.3 Using Verified Roots to Factor Polynomials

This section is missing a lot of detail, and it is doubtful that any student would be able tomaster all the techniques mentioned just from this material. However, it is hoped that it will

48

be sufficient for a review; it may help suggest enough that you might remember how to getthrough some of the techniques. Remember to get as much help as you need from othersources, including coming to class!

6.3.1 Listing all possible rational roots

There are a couple of things you should routinely do when confronted with a higher orderpolynomial to factor. Always check to see if there are common factors to factor out. Thenarrange the terms in descending order of powers if they are not already. Then you are readyto begin making the list. Here is how to make your list of all possible rational roots:

Procedure Listing all possible rational roots

1. Make a list of all the factors of the plain number term, just like what you do withthe key number method. These are the possible numerators of the roots.

2. Make the same kind of list of all the factors of the coefficient of the highest powerterm (the first term, right?). These are the possible denominators of the roots.

3. Combine the numerators from step 1 with each possible denominator from step 2,and list them twice: once positive and once negative. This makes the entire list ofall possible rational roots for the polynomial. Nothing else is possible.

Example Find all possible rational roots for x2 + 5x + 6.

1. All possible factors for 6 are: 1, 6, 2, 3. These will be the upstairs parts.

2. All possible factors for 1 are: 1. The only downstairs part is 1, so in effect, we don’thave fractions for this example.

3. The final list will be the list from step 1, with plus and minus: 1, −1, 2, −2, 3,−3,6, −6

Remember that this list is not a list of all the roots. It is only a list that must contain all therational roots. There may be irrational roots or complex roots, and they will not be in the list.And, almost always, there will be some duds in the list that are not roots at all. The list caneven be all duds! All we know is that there are not going to be any rational roots that are notin the list.

Example Find all possible rational roots for 2x3 + 15x2 − 32x + 7

1. All possible factors of 7: 1, 7

2. All possible factors of 2: 1, 2

3. The final list is: 1, 7, 12 , 7

2 , −1, −7, −12 , −7

2 .

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6.3.2 Using a Graphing Calculator to Narrow the Choices

If you graph a polynomial function on a graphing calculator (aka “grakulator”), the realroots will be obvious on the graph as points where the graph touches the x-axis. Use thegrakulator to calculate the “zeros” (which is what it calls the roots), but remember that itonly gives decimal approximations.

Figure out which of the possible rational roots would be approximately the same value aswhat the grakulator says, and then use synthetic division to verify that the ± p

qvalue is

indeed an exact root.

Example Find the real roots of 3x4 + 4x3 − x2 + 4x− 4

In the shortest cheater’s way of expressing it, the possible rational roots are: ±1, 2, 41, 3

. When

we look at the graph, we see two places that the graph touches the x-axis:

We ignore the y values, which should be zero or very close to it. So the grakulator says thatwe have two roots: one near –2 and the other near 0.67.

Which of the ± pq

values are near those? –2 is obvious, but 0.67 is not. However, the only

way to get a positive fraction less than 1 is to choose13

or23

. The grakulator can be used to

find decimal equivalents of these fractions if you need help, and we find that13≈ 0.33 and

23≈ 0.67. Therefore,

23

is the guy to try.

Use synthetic evaluation to check that you get exactly zero for –2 and again for23

. That nastyold 0.67 will NOT get you a zero remainder. There is more on this in the next section.

6.4 Analyzing graphs to find roots

Since the exponent of the leading term tells us exactly how many roots to account for, thereare many cases where we can find all the roots for a polynomial. We use a combination oftechniques: analyzing the graph, synthetic division, finding zero roots — and sometimespeople use Descartes’ Rule of Signs to figure out how many positive roots and how manynegative roots there are.

As you account for a root, you will be able to factor out the factor associated with that root.And once you have accounted for all the roots except the last two, you can use the quadratic

50

formula or any other method for solving quadratic equations to find the last two roots prettyeasily.

We have several tools at our disposal to accomplish this task. Remember that every root wefind will make it easier to find the others. Just take it one step at a time, and start with theeasiest steps first.

Watch out! When you are finding the roots of a polynomial function, remember that someroots are multiple roots. So if you know that there must be two real roots, but you can onlyfind one, remember the two roots can have the same value!

6.4.1 Factors associated with roots

Every root has a factor associated with it. It is of the form (x− root). So if you find that 2 isa root, then (x− 2) is the factor that goes with it, and you can factor it out of the polynomial.Watch your signs, especially with negative roots. If you find that−3 is a root, then (x− (−3))is its factor, but we will simplify the double negative and say (x + 3) is the factor.

6.4.2 Number of “zero roots” (roots equal to zero)

The easiest roots to find are the ones that occur where x = 0. These are called “zero roots.” Ifall the terms in the polynomial have a common factor of x to some power, factor the commonpart out. Whatever power of x you can factor out of all the terms gives you a zero root, andthe exponent tells you the multiplicity of it.

Example Find the number of zero roots in P(x) = 2x3 − 3x2.

Since all terms have x to some power in them, we can factor out the smallest power of xfrom all the terms. This is x2. Then P(x) = x2(2x− 3). From this we see there is a zeroroot with multiplicity 2. The remaining root can be found easily by other methods.

Example Find the number of zero roots in P(x) = 2x3 − 3x.

Since all terms have x to some power in them, we can factor out the smallest power ofx from all the terms. This is x. Then P(x) = x(2x2− 3). From this we see there is a zeroroot with no multiplicity. The remaining roots can be found easily by other methods.

Example Find the number of zero roots in P(x) = 2x3 − 3.

Since there is a term that has no x, we can’t factor out any power of x from all the terms.Therefore there is no zero root. The 3 roots must be found by other methods, but thechore may not be easy.

Remember to factor out all the factors associated with zero roots. So, with the second exam-ple above, 2x3 − 3x would be factored out to x(2x2 − 3), and then you would set to workfinding the roots of the leftover part in parentheses. Each root you find makes it easier tofind the other roots.

51

6.4.3 Using a graph to find non-zero real roots

Your graphing calculator will show you approximately where the graph touches or crossesthe x-axis. These places are real roots, but remember that the graphing calculator providesonly approximate values. You will need to use synthetic division to verify them as actual exactroots, and some of them will not be rational.

Therefore, use the ± pq

list of possible rational roots, and see which of them are very close to

the decimal approximations that show up as real roots. Then use synthetic division to verifythe exact value and, as a nice bonus, factor out the associated factor.

Example Find a real root of f (x) = 3x3 + x2 + 12x + 4.

The first thing to look for is zero roots, but there are none. Next we graph the polyno-mial function and examine the graph for where it crosses the x-axis:

We only pay attention to the x value where the graph touches the x-axis. In this case,that is −0.333. That is not an exact value, so we need to look for a ± p

qpossible root

that is approximately equal to the −0.333 that the grakulator came up with. Of all the

possibles: ±1, 2, 41, 3

, only −13≈ −0.333, so that is the most likely suspect of being the

exact root. We need to verify this using synthetic division.

−1 0 −4

+ 3 1 12 4

−13 3 0 12 0

Since the remainder is 0, then −1/3 is the exact value of a real root. Note that −0.333would not have come out with zero remainder, because it is only an approximate root,not exact.

Note: Remember that the middle part of the bottom line will be extremely useful for findingfurther roots. It can be interpreted as 3x2 + 0x + 12 or as 3x2 + 12 (ignoring the 0x term inthe middle). There is no need to go back to the original function to find the other roots. Asa matter of fact, do not go back to the original unless you like to work way too hard all thetime!

52

Once you have found a root and verified it using synthetic division (with 0 remain-der), use the quotient part in the middle of the bottom line to find further roots,and each successive one will become easier to find. If you ignore this, it will likelybecome impossible to finish finding all the roots!

6.4.4 Adjusting for integer coefficients

In the last example, we found a root with a fractional value. If we wrote what we found in

factored form, it became(

x− 13

) (3x2 + 12

). This form has a couple of things we don’t like

to have if we can avoid it: the fraction in the first parentheses, and the common factor of 3in the second parentheses. It is pretty common, as in this case, for these two hiccups to helpeach other out.

If we factor out the common 3 from the second parentheses, and multiply it instead to thecontents of the first parentheses, we get (3x− 1)

(x2 + 4

)and now have integer coefficients

and no common factors within one set of parentheses. That is better and easier to work with.

6.4.5 Looking for multiplicity

The second thing that the graph can tell you about is multiplicity of roots. Look at the areawhere the graph approaches the x-axis. If it flattens and goes horizontal at the axis, then it isa multiple root.

If you see evidence of multiplicity, either even or odd, use synthetic division to factor out thesame root again.

6.4.6 Irrational and Irreal roots always come in pairs

The last stage of finding all the roots and factors of a polynomial function comes when youhave found all of them except the last two roots. If you followed my advice and worked yourway from the easiest roots to find (the zero roots), then the medium ones that grakulators andsynthetic division help you find, then the last two are going to be the hardest kind to find.

The good news is that you have excellent ways to find the last two roots. Most people usethe quadratic formula, but sometimes the square root property or factoring can work. Manytimes, however, you will end up with two irrational roots or two complex roots. Rememberthey always come in conjugate pairs.

53

A Appendix of Reference Materials

This appendix contains materials for easy reference.

54

Prime Factorizations of Natural Numbers 2 through 4092 = 23 = 34 = 22

5 = 56 = 2 • 37 = 78 = 23

9 = 32

10 = 2 • 511 = 1112 = 22 • 313 = 1314 = 2 • 715 = 3 • 516 = 24

17 = 1718 = 2 • 32

19 = 1920 = 22 • 521 = 3 • 722 = 2 • 1123 = 2324 = 23 • 325 = 52

26 = 2 • 1327 = 33

28 = 22 • 729 = 2930 = 2 • 3 • 531 = 3132 = 25

33 = 3 • 1134 = 2 • 1735 = 5 • 736 = 22 • 32

37 = 3738 = 2 • 1939 = 3 • 1340 = 23 • 541 = 4142 = 2 • 3 • 743 = 4344 = 22 • 1145 = 32 • 546 = 2 • 2347 = 4748 = 24 • 349 = 72

50 = 2 • 52

51 = 3 • 1752 = 22 • 13

53 = 5354 = 2 • 33

55 = 5 • 1156 = 23 • 757 = 3 • 1958 = 2 • 2959 = 5960 = 22 • 3 • 561 = 6162 = 2 • 3163 = 32 • 764 = 26

65 = 5 • 1366 = 2 • 3 • 1167 = 6768 = 22 • 1769 = 3 • 2370 = 2 • 5 • 771 = 7172 = 23 • 32

73 = 7374 = 2 • 3775 = 3 • 52

76 = 22 • 1977 = 7 • 1178 = 2 • 3 • 1379 = 7980 = 24 • 581 = 34

82 = 2 • 4183 = 8384 = 22 • 3 • 785 = 5 • 1786 = 2 • 4387 = 3 • 2988 = 23 • 1189 = 8990 = 2 • 32 • 591 = 7 • 1392 = 22 • 2393 = 3 • 3194 = 2 • 4795 = 5 • 1996 = 25 • 397 = 9798 = 2 • 72

99 = 32 • 11100 = 22 • 52

101 = 101102 = 2 • 3 • 17103 = 103

104 = 23 • 13105 = 3 • 5 • 7106 = 2 • 53107 = 107108 = 22 • 33

109 = 109110 = 2 • 5 • 11111 = 3 • 37112 = 24 • 7113 = 113114 = 2 • 3 • 19115 = 5 • 23116 = 22 • 29117 = 32 • 13118 = 2 • 59119 = 7 • 17120 = 23 • 3 • 5121 = 112

122 = 2 • 61123 = 3 • 41124 = 22 • 31125 = 53

126 = 2 • 32 • 7127 = 127128 = 27

129 = 3 • 43130 = 2 • 5 • 13131 = 131132 = 22 • 3 • 11133 = 7 • 19134 = 2 • 67135 = 33 • 5136 = 23 • 17137 = 137138 = 2 • 3 • 23139 = 139140 = 22 • 5 • 7141 = 3 • 47142 = 2 • 71143 = 11 • 13144 = 24 • 32

145 = 5 • 29146 = 2 • 73147 = 3 • 72

148 = 22 • 37149 = 149150 = 2 • 3 • 52

151 = 151152 = 23 • 19153 = 32 • 17154 = 2 • 7 • 11

155 = 5 • 31156 = 22 • 3 • 13157 = 157158 = 2 • 79159 = 3 • 53160 = 25 • 5161 = 7 • 23162 = 2 • 34

163 = 163164 = 22 • 41165 = 3 • 5 • 11166 = 2 • 83167 = 167168 = 23 • 3 • 7169 = 132

170 = 2 • 5 • 17171 = 32 • 19172 = 22 • 43173 = 173174 = 2 • 3 • 29175 = 52 • 7176 = 24 • 11177 = 3 • 59178 = 2 • 89179 = 179180 = 22 • 32 • 5181 = 181182 = 2 • 7 • 13183 = 3 • 61184 = 23 • 23185 = 5 • 37186 = 2 • 3 • 31187 = 11 • 17188 = 22 • 47189 = 33 • 7190 = 2 • 5 • 19191 = 191192 = 26 • 3193 = 193194 = 2 • 97195 = 3 • 5 • 13196 = 22 • 72

197 = 197198 = 2 • 32 • 11199 = 199200 = 23 • 52

201 = 3 • 67202 = 2 • 101203 = 7 • 29204 = 22 • 3 • 17205 = 5 • 41

206 = 2 • 103207 = 32 • 23208 = 24 • 13209 = 11 • 19210 = 2 • 3 • 5 • 7211 = 211212 = 22 • 53213 = 3 • 71214 = 2 • 107215 = 5 • 43216 = 23 • 33

217 = 7 • 31218 = 2 • 109219 = 3 • 73220 = 22 • 5 • 11221 = 13 • 17222 = 2 • 3 • 37223 = 223224 = 25 • 7225 = 32 • 52

226 = 2 • 113227 = 227228 = 22 • 3 • 19229 = 229230 = 2 • 5 • 23231 = 3 • 7 • 11232 = 23 • 29233 = 233234 = 2 • 32 • 13235 = 5 • 47236 = 22 • 59237 = 3 • 79238 = 2 • 7 • 17239 = 239240 = 24 • 3 • 5241 = 241242 = 2 • 112

243 = 35

244 = 22 • 61245 = 5 • 72

246 = 2 • 3 • 41247 = 13 • 19248 = 23 • 31249 = 3 • 83250 = 2 • 53

251 = 251252 = 22 • 32 • 7253 = 11 • 23254 = 2 • 127255 = 3 • 5 • 17256 = 28

257 = 257258 = 2 • 3 • 43259 = 7 • 37260 = 22 • 5 • 13261 = 32 • 29262 = 2 • 131263 = 263264 = 23 • 3 • 11265 = 5 • 53266 = 2 • 7 • 19267 = 3 • 89268 = 22 • 67269 = 269270 = 2 • 33 • 5271 = 271272 = 24 • 17273 = 3 • 7 • 13274 = 2 • 137275 = 52 • 11276 = 22 • 3 • 23277 = 277278 = 2 • 139279 = 32 • 31280 = 23 • 5 • 7281 = 281282 = 2 • 3 • 47283 = 283284 = 22 • 71285 = 3 • 5 • 19286 = 2 • 11 • 13287 = 7 • 41288 = 25 • 32

289 = 172

290 = 2 • 5 • 29291 = 3 • 97292 = 22 • 73293 = 293294 = 2 • 3 • 72

295 = 5 • 59296 = 23 • 37297 = 33 • 11298 = 2 • 149299 = 13 • 23300 = 22 • 3 • 52

301 = 7 • 43302 = 2 • 151303 = 3 • 101304 = 24 • 19305 = 5 • 61306 = 2 • 32 • 17307 = 307

308 = 22 • 7 • 11309 = 3 • 103310 = 2 • 5 • 31311 = 311312 = 23 • 3 • 13313 = 313314 = 2 • 157315 = 32 • 5 • 7316 = 22 • 79317 = 317318 = 2 • 3 • 53319 = 11 • 29320 = 26 • 5321 = 3 • 107322 = 2 • 7 • 23323 = 17 • 19324 = 22 • 34

325 = 52 • 13326 = 2 • 163327 = 3 • 109328 = 23 • 41329 = 7 • 47330 = 2 • 3 • 5 • 11331 = 331332 = 22 • 83333 = 32 • 37334 = 2 • 167335 = 5 • 67336 = 24 • 3 • 7337 = 337338 = 2 • 132

339 = 3 • 113340 = 22 • 5 • 17341 = 11 • 31342 = 2 • 32 • 19343 = 73

344 = 23 • 43345 = 3 • 5 • 23346 = 2 • 173347 = 347348 = 22 • 3 • 29349 = 349350 = 2 • 52 • 7351 = 33 • 13352 = 25 • 11353 = 353354 = 2 • 3 • 59355 = 5 • 71356 = 22 • 89357 = 3 • 7 • 17358 = 2 • 179

359 = 359360 = 23 • 32 • 5361 = 192

362 = 2 • 181363 = 3 • 112

364 = 22 • 7 • 13365 = 5 • 73366 = 2 • 3 • 61367 = 367368 = 24 • 23369 = 32 • 41370 = 2 • 5 • 37371 = 7 • 53372 = 22 • 3 • 31373 = 373374 = 2 • 11 • 17375 = 3 • 53

376 = 23 • 47377 = 13 • 29378 = 2 • 33 • 7379 = 379380 = 22 • 5 • 19381 = 3 • 127382 = 2 • 191383 = 383384 = 27 • 3385 = 5 • 7 • 11386 = 2 • 193387 = 32 • 43388 = 22 • 97389 = 389390 = 2 • 3 • 5 • 13391 = 17 • 23392 = 23 • 72

393 = 3 • 131394 = 2 • 197395 = 5 • 79396 = 22 • 32 • 11397 = 397398 = 2 • 199399 = 3 • 7 • 19400 = 24 • 52

401 = 401402 = 2 • 3 • 67403 = 13 • 31404 = 22 • 101405 = 34 • 5406 = 2 • 7 • 29407 = 11 • 37408 = 23 • 3 • 17409 = 409