1. Using the Distributive Property, Factoring by Grouping (8-2)
Objective: Use the Distributive Property to factor polynomials. Solve quadratic equations of the form ax2 + bx = 0.

2. Use the Distributive Property to Factor
The Distributive Property has been used to multiply a monomial by a polynomial.
You can work backward to express a polynomial as a product of a monomial factor and a polynomial factor.
5z(4z + 7) =
20z2 + 35z =
5z(4z + 7) is the factored form of z2 + 35z.
Factoring a polynomial involves finding the completely factored form.
20z2 + 35z
5z(4z + 7)

3. Example 1 Use the Distributive Property to factor each polynomial.
15x + 25x2
GCF = 5x
5x(
12xy + 24xy2 – 30x2y4
GCF = 6xy
6xy( 3
+ 5x) 2
+ 4y
– 5xy3)

4. Check Your Progress Choose the best answer for the following.
Use the Distributive Property to factor the polynomial x2y + 12xy2.
3xy(x + 4y)
3(x2y + 4xy2)
3x(xy + 4y2)
xy(3x + 2y)
GCF = 3xy

5. Check Your Progress Choose the best answer for the following.
Use the Distributive Property to factor the polynomial ab2 + 15a2b2 + 27ab3.
3(ab2 + 5a2b2 + 9ab3)
3ab(b + 5ab + 9b2)
ab(b + 5ab + 9b2)
3ab2(1 + 5a + 9b)
GCF = 3ab2

6. Factoring by Grouping
Using the Distributive Property to factor polynomials with four or more terms is called factoring by grouping because terms are put into groups and then factored.
The Distributive Property is then applied to a common binomial factor.
A polynomial can be factored by grouping only if all of the following conditions exist.
There are four or more terms.
Terms have common factors that can be grouped together.
There are two common factors that are identical or additive inverses of each other.
ax + bx + ay + by = (ax + bx) + (ay + by)
= x(a + b) + y(a + b)
= (x + y)(a + b)

7. Example 2 Factor 2xy + 7x – 2y – 7. (2xy + 7x) + (-2y – 7)
x(2y + 7) – 1(2y + 7)
(x – 1)(2y + 7)

8. Check Your Progress Choose the best answer for the following.
Factor 4xy + 3y – 20x – 15.
(4x – 5)(y + 3)
(7x + 5)(2y – 3)
(4x + 3)(y – 5)
(4x – 3)(y + 5)
(4xy + 3y) + (-20x – 15)
y(4x + 3) – 5(4x + 3)

9. Factoring by Grouping
It can be helpful to recognize when binomials are additive inverses of each other.
For example 6 – a = -1(a – 6).

10. Example 3 Factor 15a – 3ab + 4b – 20. (15a – 3ab) + (4b – 20)

11. Check Your Progress Choose the best answer for the following.
Factor -2xy – 10x + 3y + 15.
(2x – 3)(y – 5)
(-2x + 3)(y + 5)
(3 + 2x)(5 + y)
(-2x + 5)(y + 3)
(-2xy – 10x) + (3y + 15)
-2x(y + 5) + 3(y + 5)

12. Solve Equations by Factoring
Some equations can be solved by factoring.
Consider the following:
3(0) =
Notice that in each case, at least one of the factors is 0.
These examples are demonstrations of the Zero Product Property.
If the product of two factors is 0, then at least one of the factors must be 0.
For any real numbers a and b, if ab = 0, then a = 0, b = 0, or both a and b equal 0.
0(2 – 2) =
-312(0) =
0(0.25) =

13. Example 4 Solve each equation. Check your solutions.
(x – 2)(4x – 1) = 0
x – 2 = 0
4x – 1 = 0
x = 2
4x = 1
x = ¼
{2, ¼}

14. Example 4 Solve each equation. Check your solutions. 4y = 12y2
{0, 1/3}

15. Check Your Progress Choose the best answer for the following.
Solve (s – 3)(3s + 6) = 0. Then check your solution.
{3, -2}
{-3, 2}
{0, 2}
{3, 0}
s – 3 = 0
3s + 6 = 0
x = 3
3s = -6
s = -2

16. Check Your Progress Choose the best answer for the following.
Solve 5x – 40x2 = 0. Then check your solution.
{0, 8}
{1/8}
{0}
{0, 1/8}
5x(1 – 8x) = 0
5x = 0
1 – 8x = 0
x = 0
-8x = -1
x = 1/8

17. Example 5
A football is kicked into the air. The height of the football can be modeled by the equation h = -16x2 + 16x, where h is the height reached by the ball after x seconds. Find the values of x when h = 0.
0 = -16x2 + 16x
0 = 16x(-x + 1)
16x = 0
-x + 1 = 0
x = 0
-x = -1
x = 1
0 seconds and 1 second

18. Check Your Progress Choose the best answer for the following.
Juanita is jumping on a trampoline in her back yard. Juanita’s jump can be modeled by the equation h = -14t2 + 21t, where h is the height of the jump in feet at t seconds. Find the values of t when h = 0.
0 or 1.5 seconds
0 or 7 seconds
0 or 2.66 seconds
0 or 1.25 seconds
0 = -14t2 + 21t
0 = 7t(-2t + 3)
7t = 0
-2t + 3 = 0
t = 0
-2t = -3
t = 1.5