i_f. factoring polynomials

8/13/2019 I_F. Factoring Polynomials

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Chapter 13

Factoring

Polynomials


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13.1

The Greatest CommonFactor


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Martin-Gay,Developmental Mathematics 4

Factors

Factors(either numbers or polynomials)

When an integer is written as a product of

integers, each of the integers in the product is a

factorof the original number.When a polynomial is written as a product of

polynomials, each of the polynomials in the

product is a factorof the original polynomial.

Factoringwriting a polynomial as a product of

polynomials.


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Greatest common factorlargest quantity that is afactor of all the integers or polynomials involved.

F inding the GCF of a List of I ntegers or Terms

1) Prime factor the numbers.

2) Identify common prime factors.

3) Take the product of all common prime factors. If there are no common prime factors, GCF is 1.

Greatest Common Factor


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Find the GCF of each list of numbers.

1) 12 and 8

12 = 2 238 = 222

So the GCF is 22= 4.

2) 7 and 20

7 = 1 7

20 = 2 2 5

There are no common prime factors so the

GCF is 1.


Example


7/67Martin-Gay,Developmental Mathematics 7

Find the GCF of each list of numbers.

1) 6, 8 and 46

6 = 2 3

8 = 2 2 2

46 = 2 23

So the GCF is 2.

2) 144, 256 and 300

144 = 2 2 2 3 3

256 = 2 2 2 2 2 2 2 2

300 = 2 2 3 5 5

So the GCF is 2 2= 4.


Example



1) x3andx7

x3

= x x xx7 = x x xxxxx

So the GCF is x x x=x3

2) 6x5

and 4x3

6x5= 2 3 x x x

4x3= 2 2 x x x

So the GCF is 2 x x x= 2x3

Find the GCF of each list of terms.


Example



Find the GCF of the following list of terms.

a3b2, a2b5and a4b7

a3b2= a a a b b

a2b5= a a b b b b b

a4b7= a a a a b b b b b b b

So the GCF is a a b b = a2b2

Notice that the GCF of terms containing variables will use the

smallest exponent found amongst the individual terms for each

variable.


Example



The first step in factoring a polynomial is tofind the GCF of all its terms.

Then we write the polynomial as a product by

factor ing outthe GCF from all the terms.

The remaining factors in each term will form a

polynomial.

Factoring Polynomials



Factor out the GCF in each of the following

polynomials.

1) 6x39x2+ 12x=

3 x 2 x23 x 3 x+ 3 x 4 =

3x(2x23x+ 4)

2) 14x3y+ 7x2y7xy =

7 x y 2 x2+ 7 x y x7 x y 1 =

7xy(2x2+x1)

Factoring out the GCF

Example



Factor out the GCF in each of the following

polynomials.

1) 6(x+ 2)y(x+ 2) =

6 (x+ 2)y(x+ 2) =

(x+ 2)(6y)

2)xy(y+ 1)(y+ 1) =

xy(y+ 1)1 (y+ 1)=

(y+ 1)(xy1)

Factoring out the GCF

Example



Remember that factoring out the GCF from the terms of

a polynomial should always be the first step in factoring

a polynomial.

This will usually be followed by additional steps in the

process.

Factor 90 + 15y218x3xy2.

90 + 15y2

18x3xy2

= 3(30 + 5y2

6xxy2

) =3(5 6 + 5y26 xxy2) =

3(5(6 + y2)x(6 + y2)) =

3(6 + y2

)(5x)

Factoring

Example


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13.2

Factoring Trinomials of theForm x2+ bx+ c



Factoring Trinomials

Recall by using the FOIL method thatF O I L

(x+ 2)(x+ 4) =x2+ 4x+ 2x+ 8

=x2+ 6x+ 8

To factorx2+ bx+ cinto (x+ one #)(x+ another #),note that bis the sum of the two numbers and cis the

product of the two numbers.

So well be looking for 2 numbers whose product iscand whose sum is b.

Note: there are fewer choices for the product, sothats why we start there first.



Factor the polynomialx2+ 13x+ 30.

Since our two numbers must have a product of 30 and a

sum of 13, the two numbers must both be positive.

Positive factors of 30 Sum of Factors

1, 30 31

2, 15 17

3, 10 13

Note, there are other factors, but once we find a pairthat works, we do not have to continue searching.

Sox

2

+ 13x+ 30 = (x+ 3)(x+ 10).


Example



Factor the polynomialx211x+ 24.


sum of -11, the two numbers must both be negative.Negative factors of 24 Sum of Factors

1,24 25

2,

12

14

3,8 11

Sox211x+ 24 = (x3)(x8).


Example



Factor the polynomialx22x35.

Since our two numbers must have a product of35 and a

sum of2, the two numbers will have to have different signs.Factors of35 Sum of Factors

1, 35 34

1,35 34

5, 7 2

5,7 2

Sox22x35 = (x+ 5)(x7).


Example



Factor the polynomialx26x+ 10.


sum of6, the two numbers will have to both be negative.

Negative factors of 10 Sum of Factors

1,10 11

2,5 7Since there is not a factor pair whose sum is6,

x26x+10 is not factorable and we call it a primepolynomial.

Prime Polynomials

Example



You should always check your factoring

results by multiplying the factored polynomial

to verify that it is equal to the original

polynomial.

Many times you can detect computational

errors or errors in the signs of your numbers

by checking your results.

Check Your Result!


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13.3

Factoring Trinomials ofthe Form ax2+ bx+ c



Factoring Trinomials

Returning to the FOIL method,F O I L

(3x + 2)(x + 4) = 3x2+ 12x + 2x + 8

= 3x2

+ 14x + 8To factor ax2+ bx+ cinto (#1x+ #2)(#3x+ #4),note that ais the product of the two first coefficients,cis the product of the two last coefficients and bis

the sum of the products of the outside coefficientsand inside coefficients.

Note that bis the sum of 2 products, not just 2numbers, as in the last section.



Factor the polynomial 25x2+ 20x+ 4.

Possible factors of 25x2are {x, 25x} or {5x, 5x}.

Possible factors of 4 are {1, 4} or {2, 2}.We need to methodically try each pair of factors until we find

a combination that works, or exhaust all of our possible pairs

of factors.

Keep in mind that, because some of our pairs are not identicalfactors, we may have to exchange some pairs of factors and

make 2 attempts before we can definitely decide a particular

pair of factors will not work.


Example

Continued.


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We will be looking for a combination that gives the sum of the

products of the outside terms and the inside terms equal to 20x.

{x, 25x} {1, 4} (x+ 1)(25x+ 4) 4x 25x 29x

(x+ 4)(25x+ 1) x 100x 101x

{x, 25x} {2, 2} (x+ 2)(25x+ 2) 2x 50x 52x

Factorsof 25x2 ResultingBinomials Product ofOutside Terms Product ofInside Terms Sum ofProductsFactorsof 4

{5x, 5x} {2, 2} (5x+ 2)(5x+ 2) 10x 10x 20x


Example Continued

Continued.


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Factor the polynomial 21x241x+ 10.

Possible factors of 21x2are {x, 21x} or {3x, 7x}.

Since the middle term is negative, possible factors of 10

must both be negative: {-1, -10} or {-2, -5}.

We need to methodically try each pair of factors untilwe find a combination that works, or exhaust all of our

possible pairs of factors.


Example

Continued.


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We will be looking for a combination that gives the sum of

the products of the outside terms and the inside terms equal

to 41x.

Factors

of 21x2Resulting

Binomials

Product of

Outside Terms

Product of

Inside Terms

Sum of

Products

Factors

of 10

{x, 21x}{1, 10}(x1)(21x10) 10x 21x 31x

(x

10)(21x1) x 210x 211x

{x, 21x} {2, 5} (x2)(21x5) 5x 42x 47x

(x5)(21x2) 2x 105x 107x


Example Continued

Continued.


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Factors

of 21x2Resulting

Binomials

Product of

Outside Terms

Product of

Inside Terms

Sum of

Products

Factors

of 10

(3x

5)(7x2) 6x 35x 41x

{3x, 7x}{1, 10}(3x

1)(7x10) 30x 7x 37x

(3x10)(7x1) 3x 70x 73x

{3x, 7x} {2, 5} (3x2)(7x5) 15x 14x 29x


Example Continued

Continued.


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Check the resulting factorization using the FOIL method.

(3x

5)(7x2)=

= 21x26x35x+ 10

3x(7x)F

+ 3x(-2)

O

- 5(7x)

I

- 5(-2)

L

= 21x241x+ 10

So our final answer when asked to factor 21x241x+ 10

will be (3x5)(7x2).


Example Continued


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Factor the polynomial 3x27x+ 6.

The only possible factors for 3 are 1 and 3, so we know that, if

factorable, the polynomial will have to look like (3x )(x )in factored form, so that the product of the first two terms in the

binomials will be 3x2.

Since the middle term is negative, possible factors of 6 must both

be negative: {1, 6} or {2, 3}.

We need to methodically try each pair of factors until we find a

combination that works, or exhaust all of our possible pairs of

factors.


Example

Continued.


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We will be looking for a combination that gives the sum of the

products of the outside terms and the inside terms equal to 7x.

{1, 6} (3x1)(x6) 18x x 19x

(3x

6)(x1) Common factor so no need to test.{2, 3} (3x2)(x3) 9x 2x 11x

(3x3)(x2) Common factor so no need to test.

Factorsof 6

ResultingBinomials

Product ofOutside Terms

Product ofInside Terms

Sum ofProducts


Example Continued

Continued.


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Since the product of the last two terms of the binomials

will have to be30, we know that they must be

different signs.

Possible factors of30 are {1, 30}, {1,30}, {2, 15},

{2,15}, {3, 10}, {3,10}, {5, 6} or {5,6}.

We will be looking for a combination that gives the sum

of the products of the outside terms and the inside terms

equal tox.


Example Continued

Continued.


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Factors

of -30

Resulting

Binomials

Product of

Outside Terms

Product of

Inside Terms

Sum of

Products

{-1, 30} (3x1)(x+ 30) 90x -x 89x

(3x + 30)(x1) Common factor so no need to test.

{1, -30} (3x+ 1)(x30) -90x x -89x

(3x30)(x+ 1) Common factor so no need to test.

{-2, 15} (3x2)(x+ 15) 45x -2x 43x

(3x + 15)(x2) Common factor so no need to test.

{2, -15} (3x+ 2)(x15) -45x 2x -43x

(3x15)(x+ 2) Common factor so no need to test.


Example Continued

Continued.


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Check the resulting factorization using the FOIL method.

(3x10)(x+ 3) =

= 3x2+ 9x10x30

3x(x)

F

+ 3x(3)

O

10(x)

I

10(3)

L

= 3x2x30

So our final answer when asked to factor the polynomial

6x2y22xy260y2will be 2y2(3x10)(x+ 3).


Example Continued


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13.4

Factoring Trinomials of

the Form x2+ bx+ cby Grouping


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Factoring polynomials often involves additionaltechniques after initially factoring out the GCF.

One technique is factor ing by grouping.

Factorxy+y+ 2x+ 2 by grouping.

Notice that, although 1 is the GCF for all four

terms of the polynomial, the first 2 terms have a

GCF ofyand the last 2 terms have a GCF of 2.

xy+y+ 2x+ 2 =xy+ 1 y+ 2x+ 21 =

y(x+ 1)+ 2(x+ 1)= (x+ 1)(y+ 2)

Factoring by Grouping

Example


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1) x3+ 4x+x2+ 4 = xx2+ x 4 + 1x2+ 1 4 =

x(x2+ 4)+ 1(x2+ 4)=

(x2+ 4)(x+ 1)

2) 2x3x210x+ 5 = x2 2xx2 15 2x5 (1) =

x2

(2x

1)5(2x

1)=(2x1)(x25)

Factor each of the following polynomials by grouping.


Example


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Factor 2x9y+ 18xyby grouping.

Neither pair has a common factor (other than 1).

So, rearrange the order of the factors.

2x+ 189yxy= 2x+ 2 99 yx y=

2(x+ 9)y(9 +x) =

2(x+ 9)y(x+ 9)= (make sure the factors are identical)(x + 9)(2y)


Example


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13.5

Factoring Perfect Square

Trinomials and theDifference of Two Squares

P f S T i i l


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Recall that in our very first example in Section4.3 we attempted to factor the polynomial

25x2+ 20x+ 4.

The result was (5x+ 2)2, an example of a

binomial squared.

Any trinomial that factors into a singlebinomial squared is called a perfect square

trinomial.

Perfect Square Trinomials

P f S T i i l


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In the last chapter we learned a shortcut for squaring a

binomial

(a+ b)2= a2+ 2ab+ b2

(ab)2= a22ab+ b2

So if the first and last terms of our polynomial to be

factored are can be written as expressions squared, and

the middle term of our polynomial is twice the product

of those two expressions, then we can use these twoprevious equations to easily factor the polynomial.

a2+ 2ab+ b2 =(a+ b)2

a22ab+ b2 = (ab)2


P f t S T i i l


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Factor the polynomial 16x28xy+y2.

Since the first term, 16x2, can be written as (4x)2, and

the last term,y2is obviously a square, we check the

middle term.

8xy= 2(4x)(y) (twice the product of the expressions

that are squared to get the first and last terms of the

polynomial)Therefore 16x28xy+y2= (4xy)2.

Note: You can use FOIL method to verify that the

factorization for the polynomial is accurate.


Example


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Difference of Two Squares

Another shortcut for factoring a trinomial is when we

want to factor the difference of two squares.

a2b2= (a + b)(ab)

A binomial is the difference of two square if

1.both terms are squares and

2.the signs of the terms are different.

9x225y2

c4+ d4


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Difference of Two Squares

Example

Factor the polynomialx29.

The first term is a square and the last term, 9, can be

written as 32. The signs of each term are different, so

we have the difference of two squares

Thereforex29 = (x3)(x+ 3).

Note: You can use FOIL method to verify that thefactorization for the polynomial is accurate.


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13.6

Solving Quadratic

Equations by Factoring

Z F t Th


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Zero Factor Theorem

Quadratic Equations Can be written in the form ax2+ bx+ c= 0.

a, band care real numbers and a0.

This is referred to as standard form.Zero Factor Theorem

If aand bare real numbers and ab= 0, then a= 0

or b= 0. This theorem is very useful in solving quadratic

equations.

S l i Q d ti E ti


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Steps for Solving a Quadratic Equation byFactoring

1) Write the equation in standard form.

2) Factor the quadratic completely.3) Set each factor containing a variable equal to 0.

4) Solve the resulting equations.

5) Check each solution in the original equation.

Solving Quadratic Equations

S l i Q d ti E ti


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Solvex25x= 24.

First write the quadratic equation in standard form.

x2

5x24 = 0 Now we factor the quadratic using techniques from

the previous sections.

x25x24 = (x8)(x+ 3) = 0

We set each factor equal to 0.

x8 = 0 orx+ 3 = 0, which will simplify to

x= 8 orx=3


Example

Continued.

S l i Q d ti E ti


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Check both possible answers in the original

equation.

825(8) = 6440 = 24 true

(3)25(3) = 9(15) = 24 true

So our solutions forxare 8 or3.

Example Continued


S l i Q d ti E ti


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Solve 4x(8x+ 9) = 5

First write the quadratic equation in standard form.

32x2+ 36x= 5

32x2+ 36x5 = 0

Now we factor the quadratic using techniques from the

previous sections.

32x2+ 36x5 = (8x1)(4x+ 5) = 0

We set each factor equal to 0.

8x1 = 0 or 4x+ 5 = 0


Example

Continued.

8x= 1 or 4x =5, which simplifies tox = or5.

4

1

8

S l i Q d ti E ti


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Check both possible answers in the original equation.

1 1 1

4 8 9 4 1 9 4 (10) (10) 58

1

8

1

8 8 2

true

5 54 8 9 4 10 9 4 ( 1) ( 5)( 1) 545 54 44 true

So our solutions for x are or .8

1

4

5

Example Continued



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Finding x intercepts


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Find thex-intercepts of the graph ofy= 4x2+ 11x+ 6.

The equation is already written in standard form, so

we lety= 0, then factor the quadratic inx.

0 = 4x2+ 11x+ 6 = (4x+ 3)(x+ 2)

We set each factor equal to 0 and solve for x.

4x+ 3 = 0 orx+ 2 = 0

4x=3 orx=2

x= orx=2

So thex-intercepts are the points (, 0) and (2, 0).

Finding x-intercepts

Example


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13.7

Quadratic Equations

and Problem Solving

Strategy for Problem Solving


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Strategy for Problem Solving

General Strategy for Problem Solving1) Understand the problem

Read and reread the problem

Choose a variable to represent the unknown

Construct a drawing, whenever possible Propose a solution and check

2) Translate the problem into an equation

3) Solve the equation

4) Interpret the result Check proposed solution in problem

State your conclusion

Finding an Unknown Number


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The product of two consecutive positive integers is 132. Find the

two integers.

1.) Understand

Read and reread the problem. If we let

x= one of the unknown positive integers, then

x+ 1 = the next consecutive positive integer.


Example

Continued



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Example continued

2.) Translate

Continued

two consecutive positive integers

x (x+ 1)

is

=

132

132

The product of



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Example continued3.) Solve

Continued

x(x+ 1) = 132

x2+x= 132 (Distributive property)

x2+x132 = 0 (Write quadratic in standard form)

(x+ 12)(x11) = 0 (Factor quadratic polynomial)

x+ 12 = 0 orx11 = 0 (Set factors equal to 0)

x=12 orx= 11 (Solve each factor forx)



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Example continued

4.) Interpret

Check: Remember thatxis suppose to represent a positive

integer. So, althoughx= -12 satisfies our equation, it cannot be asolution for the problem we were presented.

If we letx= 11, thenx+ 1 = 12. The product of the two numbers

is 11 12 = 132, our desired result.

State: The two positive integers are 11 and 12.

The Pythagorean Theorem


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Pythagorean Theorem

In a right triangle, the sum of the squares of the

lengths of the two legs is equal to the square of the

length of the hypotenuse.

(leg a)2+ (leg b)2= (hypotenuse)2

leg a

hypotenuse

leg b




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Find the length of the shorter leg of a right triangle if the longer leg

is 10 miles more than the shorter leg and the hypotenuse is 10 miles

less than twice the shorter leg.


Example

Continued

1.) Understand

Read and reread the problem. If we let

x= the length of the shorter leg, then

x+ 10 = the length of the longer leg and

2x10 = the length of the hypotenuse.

+ 10

2 - 10x

x



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Example continued2.) Translate

Continued

By the Pythagorean Theorem,

(leg a)2+ (leg b)2= (hypotenuse)2

x2+ (x+ 10)2= (2x10)23.) Solve

x2+ (x+ 10)2= (2x10)2

x2+x2+ 20x+ 100 = 4x240x+ 100 (multiply the binomials)

2x2

+ 20x+ 100 = 4x2

40x+ 100 (simplify left side)

x= 0 orx= 30 (set each factor = 0 and solve)

0 = 2x(x30) (factor right side)

0 = 2x260x (subtract 2x2+ 20x+ 100 from both sides)



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Example continued4.) Interpret

Check: Remember thatxis suppose to represent the length of

the shorter side. So, althoughx= 0 satisfies our equation, it

cannot be a solution for the problem we were presented.

If we letx= 30, thenx+ 10 = 40 and 2x10 = 50. Since 302 +

402 = 900 + 1600 = 2500 = 502, the Pythagorean Theorem

checks out.

State: The length of the shorter leg is 30 miles. (Remember that

is all we were asked for in this problem.)

i_f. factoring polynomials

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