Lesson 8-2
Example 1 Use the Distributive Property
Use the Distributive Property to factor each polynomial.
a. 21xy – 18x2
Find the GCF of each term.
21xy = 3 7 x y Factor each term.
18x = 2 3 3 x x
2
Circle the common factors.
GCF = 3 x or 3x
Write each term as the product of the GCF and the remaining factors. Use the Distributive
Property to factor out the GCF.
21xy – 18x2 = 3x(7y) – 3x(6x) Rewrite each term using the GCF.
= 3x(7y – 6x) Distributive Property
b. 64r3s – 32r2s3 + 8r2s2
64r3s = 2 2 2 2 2 2 r r r s Factor each term.
32r2s3 = 2 2 2 2 2 r r s s s Circle the common factors.
8r2s2 = 2 2 2 r r s s
GCF = 2 2 2 r r s or 8r2s.
64r3s – 32r2s3 + 8r2s2 = 8r2s (8r) – 8r2s(4s2) + 8r2s(s) Rewrite each term using the GCF.
= 8r2s(8r – 4s2 + s) Distributive Property
Example 2 Factor by Grouping
Factor 2x2+ x – 8x – 4.
2x 2 + x – 8x – 4 Original expression
= (2x 2 + x) + (–8x – 4) Group terms with common factors.
= x(2x + 1) + (–4)(2x + 1) Factor the GCF from each group.
Notice that (2x + 1) is common in both groups, so it becomes the GCF.
= [x + (–4)](2x + 1) Distributive Property
= (x – 4)(2x + 1) Simplify.
Example 3 Factor by Grouping with Additive Inverses
Factor 3a2 – 3a + 4 – 4a.
3a2 – 3a + 4 – 4a
= (3a2 – 3a) + (4 – 4a) Group terms with common factors.
= 3a(a – 1) + 4(1 – a) Factor the GCF from each group.
= 3a(a – 1) + 4[(-1)(a – 1)] 1 – a = -1(a – 1)
= 3a(a – 1) – 4(a – 1) Associative Property
= (3a – 4)(a – 1) Distributive Property
Example 4 Solve Equations
Solve each equation. Check your solutions.
a. (q + 2)(2q – 1) = 0
(q + 2)(2q – 1) = 0 Original equation
q + 2 = 0 or 2q – 1 = 0 Zero Product Property
q = -2 2q = 1 Solve each equation.
1
q=2 Divide.
1
The roots are –2 and 2.
1
Check Substitute –2 and 2 for q in the original equation.
(q + 2)(2q – 1) = 0 (q + 2)(2q – 1) = 0
? 1 1 ?
(-2 + 2)[2(-2) – 1] 0 (2 + 2)[2(2) – 1] 0
? 1 ?
(-2 + 2)(-4 – 1) 0 (2 + 2)(1 – 1) 0
? 5 ?
(0)(-5) 0 ( )(0) 0
2
0=0 0=0
b. 3x2 + 9x = 0
3x2 + 9x = 0 Original equation
3x(x + 3) = 0 Factor by using the GCF.
3x = 0 or x + 3 = 0 Zero Product Property
x=0 x = -3 Solve each equation.
The roots are –3 and 0. Check by substituting 0 and -3 for x.
Real-World Example 5 Use Factoring
SOCCER The path of a soccer ball kicked upward can be modeled by the equation
h = 64t – 16t2, where t represents the time in seconds and h is the height in feet. Find
the values of t when h = 0.
h = 64t – 16t2 Original equation
0 = 64t – 16t2 Substitution, h = 0
0 = 16t(4 – t) Factor by using the GCF.
16t = 0 or 4 – t = 0 Zero Product Property
t=0 –t = –4 Solve each equation.
t=4 Divide each side by -1.
The ball’s height is 0 feet at 0 seconds and 4 seconds.