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CHAPTER 7: FACTORING POLYNOMIALS

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CHAPTER 7: FACTORING POLYNOMIALS

FACTOR (noun) –Any of two or more quantities which form a product when

multiplied together.

12 can be rewritten as 3*4, where 3 and 4 are FACTORS of 12.

FACTOR (verb) - To factor an expression is to rewrite it as a product of 2 or

more quantities. Factoring is sometimes called FACTORIZATION. 12 can be FACTORED into the product 3*4.

Variable Expressions can also be factored.

35x2 can be factored into 7(5x2) or 5x(7x) or 5(7x2), etc… How about this polynomial expression?

5x3 – 35x2 + 10x

To factor a polynomial, the FIRST STEP is to look for a GREATEST COMMON FACTOR.

The GCF of a polynomial is the GREATEST number (or variable expression) that is a factor of every term in the expression. That is, it is the variable expression that is the GCF of the coefficients, and the GCF of each of the variables.

How many terms are there?___ What are they? _______________

What is the GCF of these terms? ______

Let’s take another look at the polynomial: 5x3 – 35x2 + 10x

Coefficients:

The Coefficients are 5, -35, and 10. The Coefficient GCF is 5. Variables:

The only variable is x, the first term has x3, the 2nd term has x2, the third term has x.

The GCF is the greatest x power that can go into ALL of those terms, but practically this means it will be the variable term with the smallest power: x

The GCF of 5x3 – 35x2 + 10x is 5x

Now to factor the polynomial:

Rewrite the polynomial as a product with 5x as one factor, and the remaining expression (after dividing each term by 5x).

7 2

5 5 10 5 35 5 5 5 2 2 3            x x x x x x x x x

x This polynomial cannot be

factored any more. It is “prime.” We’ll see why later.

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:

GCF

,

,

:

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:

GCF

,

,

:

'

:

GCF

12

and

16,8,

:

ts

Coefficien

12

8

16

:

Example

5 2 4 4 2 5 4 2 4 2

y

y

y

s

y

x

x

x

s

x

y

x

y

x

y

x

Put it all together and the GCF of the polynomial is: ____

NOW YOU TRY:

Factor this: 6x4y2 – 9x3y2 + 12x2y4

2 2 4

2 2 5 4 2 2 4 2 2 2 5 4 2 4 2

3

2

4

4

4

12

4

8

4

16

4

12

8

16

y

x

y

x

y

x

y

x

y

x

y

x

y

x

y

x

y

x

y

x

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x

y

x

y

x





___ ___ __

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FACTORING BY GROUPING

Example:

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

Remember, a FACTOR is something being multiplied in a product. Do you see a common factor in this expression?

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

Example 4:

2x(x - 5) + y(5 - x)

At first, it looks like there is no common factor, but notice that

x-5 and 5-x are very similar.

In fact, - (5-x) = -5+x = x-5

So we can rewrite (5-x) has –(x-5) 2x(x - 5) + y(-(x - 5))

= 2x(x - 5) –y(x - 5) =(x – 5)(2x – y)

Try factoring 3y(5x-2) – 4(2-5x)

COMMON FACTOR

)

3

)(

2

(

2

)

2

(

3

2

)

2

(

)

2

(

y

x

x

x

x

x

y

x

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Example 5:

3y3 – 4y2 – 6y + 8

If there is not a common factor of ALL the terms, you can factor by GROUPING the terms in to groups that DO have a common factor.

(3y3 – 4y2) + (– 6y + 8) =y2(3y - 4) + -2(3y – 4)

= (3y - 4)(y2 – 2)

YOU TRY FACTORING y5- 5y3 + 4y2 - 20

Factor out a -2 instead of 2 so that this group will have a common factor to the other group (3y – 4).

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FACTORING POLYNOMIALS OF THE FORM x2 + bx + c

Example 1:

Factor the polynomial: x2 + 18x + 32

STEP 1: Is there a GCF of all three terms? NO

STEP 2: Is this polynomial a trinomial with degree 2? YES Since this is a trinomial with degree 2, it is possible that this polynomial can be factored into 2 binomials:

( x + a )( x + b )

Remember, the FOIL method for (x + a)(x + b)

= x2 + ax + bx + ab = x2 + (a + b)x + ab

So from this general form, we see that the factors of the Last Term (ab), must add up to make the Middle Term’s coefficient.

STEP 3: Try different factor’s of the Last Term that will add up to the Middle Term’s coefficient.

Polynomial: x2 + 18x + 32

Last Term: __ Middle Term’s Coefficient: __

Factors of Last Term Sum of Those Factors

1, 32 1+32 = 33

4,8 4+8 = 12

2,16 2+16 = 18

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Example 2: Factor

x

2

– 6x – 16

STEP 1: Is there a GCF of all three terms? __

STEP 2: Is it a trinomial with degree 2? __

Last Term: __Middle Term’s Coefficient: __

Factors of Last Term Sum of Those Factors

1, -16 1 + -16 = -15 -1, 16 -1 + 16 = 15 2, -8 2 + -8 = -6 -2, 8 -2 + 8 = 6 4, -4 4 + -4 = 0 -4,4 -4 + 4 = 0 FACTORIZATION: (x – 2)(x +8) UNFACTORABLE TRINOMIALS: x2 – 6x – 8

There are no factors of -8 that can add up to -6, So this is considered a “prime” polynomial and is “nonfactorable over the integers.”

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Example:

Factor 3a2b – 18ab – 81b

STEP 1: Is there a GCF of all three terms? YES GCF is ________

Factor out the GCF:

STEP 2: After factoring out the GCF is one of the factors a trinomial with degree 2? ___

STEP 3:

Find factors of the last term of the trinomial that add up to the middle term’s coefficient and factor into two binomials.

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EXAMPLE 4:

Factor x2 + 9xy + 20y2

STEP 1: Is there a GCF of all terms? NO

STEP 2: Is this a trinomial with degree 2? YES

STEP 3: Find factors of the last term of the trinomial that add up to the middle term’s coefficient and factor into two binomials. In this case, make sure the factors of the the last term are “like terms” that can be combined. (eg. 20 and 1y2 are not like terms but are factors of 20y2)

FACTORIZATION: (x + 4y)(x + 5y)

Factors of Last Term, 20y2

Sum of Those Factors

20y, 1y 20y+1y =21y 2y, 10y 2y + 10y = 12y

References

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