1. Factoring and Solving Equations
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2. 6.1 Factoring Polynomials with Common Factors

3. What You Will Learn
Find the greatest common factor of two or more expressions
Factor out the greatest common monomials factor from polynomials
Factor polynomials by grouping

4. Greatest Common Factor

5. Greatest Common Factor
You have used the Distributive Property to multiply
polynomials.
In this chapter, you will study the reverse process, which is
factoring.
Multiplying Polynomials Factoring Polynomials

6. Greatest Common Factor
To factor an expression efficiently, you need to understand
the concept of the greatest common factor of two (or more)
integers or terms.
You have learned that the greatest common factor of two
or more integers is the greatest integer that is a factor of
each integer.
For example, the greatest common factor of 12 = 2  2  3
and 30 = 2  3  5 is 2  3 = 6.

7. Example 1 – Finding the Greatest Common Factor
To find the greatest common factor of 5x2y2 and 30x3y, first factor each term.
5x2y2 = 5  x  x  y  y
= (5x2y)(y)
30x3y = 2  3  5  x  x  x  y
= (5x2y)(6x)
So, you can conclude that the greatest common factor is 5x2y.

8. Example 2 – Finding the Greatest Common Factor
To find the greatest common factor of 8x5, 20x3, and 16x4 first factor each term.
8x5 = 2  2  2  x  x  x  x  x
= (4x3)(4x2)
20x3 = 2  2  5  x  x  x
= (4x3)(5)
16x4 = 2  2  2  2  x  x  x  x
= (4x3)(4x)
So, you can conclude that the greatest common factor is 4x3.

11. Common Monomial Factor

12. Dividing a Polynomial by a Binomial
Consider the polynomial
8x5 + 16x4 + 20x3.
The greatest common factor, 4x3, of these terms is the greatest common monomial factor of the polynomial.
When you use the Distributive Property to remove this factor from each term of the polynomial, you are factoring out the greatest common monomial factor.
8x5 + 16x4 + 20x3 = 4x3(2x2) + 4x3(4x) + 4x3(5)
= 4x3(2x2 + 4x + 5)
Factor each term.
Factor out common
monomial factor.

13. Example 3 – Greatest Common Monomial Factor
Factor out the greatest common monomial factor from x – 18.
Solution:
The greatest common integer factor of 6x and 18 is 6. There is no common variable factor.
6x – 18 = 6(x) – 6(3)
= 6(x – 3)
Greatest common monomial factor is 6.
Factor 6 out of each term.

14. Example 4 – Greatest Common Monomial Factor
Factor out the greatest common monomial factor from y3 – 25y2.
Solution:
For the terms 10y3 – 25y2, 5 is the greatest common integers factor and y2 is the highest-power common variable factor.
10y3 – 25y2 = 5y2(2y) – 5y2(5)
= 5y2(2y – 5)
Greatest common factor
is 5y2.
Factor 5y2 out of each term.

15. Greatest Common Monomial
The greatest common monomial factor of the terms of a
polynomial is usually considered to have a positive
coefficient.
However, sometimes it is convenient to factor a negative
number out of a polynomial.

16. Example 8 – A Negative Common Monomial Factor
Factor the polynomial –2x2 + 8x – 12 in two ways.
a. Factor out a common monomial factor of 2.
b. Factor out a common monomial factor of –2.
Solution:
a. To factor out the common monomial factor of 2, write the following.
–2x2 + 8x – 12 = 2(–x2) + 2(4x) + 2(–6)
= 2(–x2 + 4x – 6)
Factor each term.
Factored form

17. Example 8 – A Negative Common Monomial Factor
cont’d
b. To factor –2 out of the polynomial, write the following.
–2x2 + 8x – 12 = –2(x2) + (–2)(–4x) + (–2)(6)
= –2(x2 – 4x + 6)
Check this result by multiplying (x2 – 4x + 6) by –2. When
you do, you will obtain the original polynomial.
Factor each term.
Factored form

18. Greatest Common Monomial
With experience, you should be able to omit writing the first
step shown in Example 8.
For instance, to factor –2 out of –2x2 + 8x – 12, you could
simply write
–2x2 + 8x – 12 = –2(x2 – 4x + 6).

19. Worksheet

20. Factoring by Grouping

21. Factoring by Grouping There are occasions when the common factor of an
expression is not simply a monomial. For instance, the
expression
x2(x – 2) + 3(x – 2)
has the common binomial factor (x – 2). Factoring out this
common factor produces
x2(x – 2) + 3(x – 2) = (x – 2)(x2 + 3).
This type of factoring is part of a more general procedure
called factoring by grouping.

22. Example 9 – Common Binomial Factors
Factor each expression.
a. 5x2(7x – 1) – 3(7x – 1)
b. 2x(3x – 4) + (3x – 4)
c. 3y2(y – 3) + 4(3 – y)
Solution:
a. Each of the terms of this expression has a binomial factor of (7x – 1).
5x2(7x – 1) – 3(7x – 1) = (7x – 1)(5x2 – 3)

23. Example 9 – Common Binomial Factors
cont’d
b. Each of the terms of this expression has a binomial factor of (3x – 4).
2x(3x – 4) + (3x – 4) = (3x – 4)(2x + 1)
Be sure you see that when (3x – 4) is factored out of itself, you are left with the factor 1.
This follows from the fact that (3x – 4)(1) = (3x – 4).
c. 3y2(y – 3) + 4(3 – y) = 3y2(y – 3) – 4(y – 3)
= (y – 3)(3y2 – 4)
Write 4(3 – y) as –4(y – 3).
Common factor is (y – 3).

24. Factoring by Grouping
In Example 9, the polynomials were already grouped so that it was easy to determine the common binomial factors.
In practice, you will have to do the grouping as well as the factoring.
To see how this works, consider the expression
x3 + 2x2 + 3x + 6
and try to factor it. Note first that there is no common monomial factor to take out of all four terms.

25. Factoring by Grouping
But suppose you group the first two terms together and the last two terms together.
x3 + 2x2 + 3x + 6 = (x3 + 2x2) + (3x + 6)
= x2(x + 2) + 3(x + 2)
= (x + 2)(x2 + 3)
When factoring by grouping, be sure to group terms that have a common monomial factor.
For example, in the polynomial above, you should not group the first term x3 with the fourth term 6.
Group terms.
Factor out common monomial factor in each group.
Factored form

26. Example 10 – Factoring by Grouping
Factor x3 + 2x2 + x + 2.
Solution:
x3 + 2x2 + x + 2 = (x3 + 2x2) + (x + 2)
= x2(x + 2) + (x + 2)
= (x + 2)(x2 + 1)
Group terms.
Factor out common monomial factor in each group.
Factored form

27. Factoring by Grouping
Note that in Example 10 the polynomial is factored by grouping the first and second terms and the third and fourth terms.
You could just as easily have grouped the first and third terms and the second and fourth terms, as follows.
x3 + 2x2 + x + 2 = (x3 + x) + (2x2 + 2)
= x(x2 + 1) + 2(x2 + 1) = (x2 + 1)(x + 2)
You can always check to see that you have factored an expression correctly by multiplying the factors and comparing the result with the original expression.

28. Example 12 – Geometry: Area of a Rectangle
The area of a rectangle of width (2x – 1) feet is (2x3 + 2x – x2) square feet, as shown below. Factor this expression to determine the length of the rectangle.

29. Example 12 – Geometry: Area of a Rectangle
cont’d
Solution
Verbal Model:
Labels: Area = 2x3 + 2x – x (square feet)
Width = 2x – (feet)
Equation:
2x3 + 4x – x2 – 2 = (2x3 + 4x) + (–x2 – 2)
= 2x(x2 + 2) + (x2 + 2)
= (x2 + 2)(2x – 1)
The length of the rectangle is (x2 + 2) feet.
Group terms.
Factor out common monomial factor in each group.
Factored form

31. Factor the Polynomials