Math 10C

Notes Chapter 3

Section 3.3 Common Factors of a Polynomial Math 10C

In algebra, we have factors in the same manner as we did with arithmetic. The difference is that we have to account for variables.

4m + 12

Notice there are 2 parts to this: 4m and +12 Look at the “numbers”…… 4 and 12….. is there a common factor? Look at the “variables” …… m …… is there a common factor?

Take out the common factor and place it in front….open up a set of brackets…..divide each term by the common factor and write what is left…..

4m + 12 = 4(m + 3)

You know you have done it correctly IF when you multiply your factor back into the brackets (distribute) your answer is the same as the expression you started with.

What happens if you have a more complicated expression due to variables?

6x + 4x2

You do the same thing….. just remember to deal with the variables separately.

Look at the 6 and the 4….. what is the common factor? (2) Look at the x and the x2 ….. what is the common factor? (x)

Bring the common factors out in front, and then leave the remaining factors in the brackets.

6x + 4x2

= 2x(3 + 2x)

What if you have more than 2 terms? When dealing with common factors of a polynomial of multiple terms, you do the exact same thing as you did above….you just have to make sure that you find factors that are common in ALL terms.

5 + 5x – 10x2

= 5(1 + x – 2x2) *there isn’t an “x” in all terms, so it is not common

6x2 + 9x – 12x3

= 3x(2x + 3 – 4x2)

You Try:

3g + 6 8d + 12d2 5 – 15x – 5x2 16 + 12z – 20z3

What if you have multiple variables?

-12x3y – 20xy2 – 16x2y2

Look at the coefficients and see if there is a common factor….also, include the negative. (-4)

Look at each individual variable…the x’s… what is the lowest number of x’s in each term? (x) …the y’s… (y)

Put the common factor out front of the brackets and fill in the brackets with the remaining factors

= -4xy(3x2 + 5y + 4xy)

Notice that the number of terms inside the brackets is the same as the number of terms in the original polynomial.

You Try-20a4b – 30a3b2 + 25ab

Reflect:

In a couple of sentences, what have you learned in this class?

Pg. 155 #7 – 10, 12, 14 - 18

Section 3.5 Part 1 FOIL Math 10C

FOIL is a method of distribution. It is used when wanting to expand and simplify the product of 2 binomials.

F – multiply the terms in the Front of each bracket

O – multiply the terms on the outside of the brackets

I – multiply the terms on the inside of the brackets

L – multiply the terms in the back of each bracket

(x – 4)(x + 3) (8 – x)( 3 – x) (x + 3)(x – 7)

(x + 5)(x – 7) (x – 3)(x – 6) (x + 5)(x + 9)

(x + 9)(x – 4) (x + 3)(x + 3) (x – 6)(x – 6)

Have you noticed that the middle term is always going to have an “x”? Have you also noticed that when you multiply the outside and then multiply the inside, both those terms will always need to be combined? I would like you to try and do that in your head….. Let’s talk through some examples and see if you can finish them with the new “one step” multiplication….

1. (2x + 11) (x + 10) 2. (x + 9) (3x + 5)

3. (2x + 1) (x – 1) 4. (2x + 7) (x + 4)

5. (3x – 5) (x + 12) 6. (x + 11) (x – 6)

7. (x + 3) (x + 2) 8. (x + 7) (5x - 4)

9. (x + 11) (x + 5) 10. (x + 5) (x – 6)

11. (x - 13) (x + 11) 12. (x – 9) (x - 8)

13. (x – 2) (x – 6) 14. (x – 11) (x + 9)

15. (x – 12) (x – 15) 16. (x + 11) (x + 9)

17. (x – 11) (x – 14) 18. (x + 7) (2x – 9)

19. (2x – 13) (2x + 1) 20. (x + 3) ( 2x + 5)

Reflect:

In a couple of sentences, what have you learned in this class?

Section 3.5 Part 2 Factoring Trinomials Math 10C

Last class we spend quite a bit of time “putting 2 binomials together”. One thing you should have recognized is that almost always, the result was a trinomial. Today we are going to start with the trinomial, and break it down (factor it) into its 2 binomials. There are many ways to do this. Today we are going to use INSPECTION.

Factor

x2 + 8x + 15 x2 – 8x + 15

What do you notice is the same about these trinomials? What do you notice about the factors?

x2 + 2x – 15 x2 – 2x – 15

What do you notice is the same about these trinomials? What do you notice about the factors?

What if there is a common factor? Take the common factor out and continue with the steps.

3x2 + 18x + 24

Reflect:

In a couple of sentences, what have you learned in this class?

Pg. 167 #14, 17, 21, 23

Section 3.8 Factoring Special Polynomials Math 10C

There are 2 special cases that you are required to aware of:

1. Perfect Square Trinomialsa. Characteristics

i. First term is a perfect squareii. Last term is a perfect squareiii. Last term is POSITIVEiv. Middle term is double the product of the root of the first term and

the root of the last term.

b. Factoring

4x2 + 12x + 9 4x2 – 12x + 9 4x2 + 12x – 9 = (2x + 3)(2x + 3) = (2x – 3)(2x – 3) = not factorable

You Try

25x2 – 20x + 4 36x2 + 12x + 1 49x2 – 56x + 16

8x2 + 8x + 4 -27x2 – 36x – 12 40x2 – 120x + 90

**always foil to double check that you have factored correctly!!!

2. Difference of Squaresa. Characteristics

i. Only 2 termsii. Both terms are perfect squaresiii. MUST be subtractioniv. Factors are the roots of each term, one is addition, one is

subtraction.b. Factoring

25x2 – 4 36y2 – 81 16x2 + 9= (5x – 2)(5x + 2) = (6y – 9)(6y + 9) = Not factorable

You Try

81m2 – 25 81m2 – 25n2 12x2 – 27

4x2 – 1 4x2 – 4 5x4 – 80

Reflect:

In a couple of sentences, what have you learned in this class?

Pg. 194 #4, 5, 8, 10

Section 3.7 Part 1 Multiplying Polynomials Math 10C

When multiplying polynomials, all you need to do is DISTRIBUTE and then collect like terms. The key to success with this, is that you need to stay organized.

(2x + 5)(x2 + 3x – 4) (3x2 + 2x – 4)(4x2 – x – 6)

(3x + 4)(x2 – 2x – 7) (2m2 + 4m – 3)(5m2 – 2m + 1)

(2x + 5y)2 (2x + 5y)3

(3x – 2y)(4x + 3y – 5) (2v – 5w)(3v + 2w – 7)

Reflect:

In a couple of sentences, what have you learned in this class?

Pg. 186 #4, 5, 6, 8, 9, 10, 12

Section 3.7 Part 2 Sum and Difference of Polynomials Math 10C

Adding and subtracting polynomials is about distribution and then collecting like terms. Again, organization is a key element to being successful.

AddingSimply expand to remove the brackets, then collect like terms.

(2x – 3)(x + 5) + (x – 3)(3x + 1)= 2x2 + 10x – 3x – 15 + 3x2 + x – 9x – 3 = 5x2 – x – 18

(2x – 4)(3x + y – 1) + 2(3x – y)2

SubtractingWorks the almost the same way as adding, with one small area of focus….DO NOT FORGET TO DISTRIBUTE THE NEGATIVE!!!!. The most common error is that the negative does not get distributed to all values in the brackets behind the minus sign.

(2x – 3)(x + 5) - (x – 3)(3x + 1)= 2x2 + 10x – 3x – 15 – (3x2 + x – 9x – 3) *keep the second part in brackets until you = 2x2 + 10x – 3x – 15 – 3x2 – x + 9x + 3 distribute the negative= -x2 + 15x – 12

(2x – 4)(3x + y – 1) – 2(3x – y)2

Reflect:In a couple of sentences, what have you learned in this class?

Pg. 186 #15, 17

Section 3.6 Factoring Trinomials where a > 1 Math 10C

We approach this one a bit differently. We will go through 2 different methods, and you are free to choose which ever method you feel most comfortable with.

Inspection

3x2 – 13x – 10

Step 1 – set up your set of bracketsStep 2 – look at the first term…what are the only ways that you can multiply to get 3?Step 3 – fill these factors into the front of your brackets.

(3x )(x )

Step 4 – look at the last term… what are the possible factors? …. 1 and 10 or 2 and 5Step 5 – choose a pair (I typically start with the middle pair) and see if they work…

(3x 5)(x 2) ** do the “outside, inside” of foil and see if you get the middle term. Outside = 6 Inside = 5 cannot get 13

Step 6 – if step 5 did not work….switch the placement of the last terms

(3x 2)(x 5) ** do the “outside, inside” of foil and see if you get the middle term. Outside = 15 Inside = 5 -15 + 2 = -13

Step 7 – if step 5 works, put in your signs and FOIL in your head to verify.

3x2 – 13x – 10 = (3x + 2)(x – 5)

2x2 – 7x + 3 2x2 + 7x + 3 2x2 – x – 3 2x2 + 5x + 3

What if you have a common factor first?

4x2 – 12x – 14 = 2(2x2 – 6x – 7) *now factor the trinomial as you did before… = 2 ( )( )

20m2 + 70m + 60 15d2 – 65d + 20

8x2 + 18x – 5 8x2 – 14x + 3

The second method is the Grouping Method, we will go through the same examples as we did with inspection using the grouping method. Please make notes on the side if you wish to help you remember the steps.

3x2 – 13x – 10 2x2 – 7x + 3 6x2 – x – 15

You Try:

3x2 – 2x – 5 5x2 + 19x + 12 2x2 + 5x + 2

Now using either method, which ever one you’re more comfortable with, do the following:

5x2 – 18x + 9 4x2 – 15x – 25

Reflect:In a couple of sentences, what have you learned in this class?

Pg. 177 #12, 13, 15, 18, 21, 22

What you are required to know before your test:

**we did not work with algebra tiles, you will not be required to deal with them.

REVIEWPg. 199 # 11 – 22, 27 – 35

Pg. 201 #2, 6, 7abdf