1. Warm Up Lesson Presentation Lesson Quiz
Solve Polynomial Equations in Factored Form
Warm Up
Lesson Presentation
Lesson Quiz

2. Warm-Up
1. Find the GCF of 12 and 28.
ANSWER 4
2. Find the GCF of 18 and 42.
ANSWER 6
The number (in hundreds) of sunscreen and sun tanning products sold at a pharmacy from 1999–2005 can be modeled by –0.8t t + 107, where t is the number of years since About how many products were sold in 2002? 3.
ANSWER
about 10,070

3. The solutions of the equation are 4 and –2.
Example 1
Solve (x – 4)(x + 2) = 0.
(x – 4)(x + 2) = 0
Write original equation.
x – 4 = 0 or
x + 2 = 0
Zero-product property
x = 4 or
x = – 2
Solve for x.
ANSWER
The solutions of the equation are 4 and –2.

4. CHECK Substitute each solution into the original equation to check.
Example 1
CHECK Substitute each solution into the original equation to check.
(4  4)(4 + 2) = 0
0  6 = 0
0 = 0 ?
(2  4)(2 + 2) = 0
6  0 = 0
0 = 0 ?

5. Guided Practice
1. Solve the equation (x – 5)(x – 1) = 0.
ANSWER
The solutions of the equation are 5 and 1.

6. Example 2
Factor out the greatest common monomial factor.
a x + 42y
b. 4x4 + 24x3
SOLUTION a.
The GCF of 12 and 42 is 6. The variables x and y have no common factor. So, the greatest common monomial factor of the terms is 6.
ANSWER
12x + 42y = 6(2x + 7y)

7. Example 2
SOLUTION b.
The GCF of 4 and 24 is 4. The GCF of x4 and x3 is x3. So, the greatest common monomial factor of the terms is 4x3.
ANSWER
4x4 + 24x3 = 4x3(x + 6)

8. Guided Practice
2. Factor out the greatest common monomial factor
from 14m + 35n.
ANSWER
14m + 35n = 7(2m + 5n)

9. The solutions of the equation are 0 and – 4.
Example 3
Solve 2x2 + 8x = 0.
2x2 + 8x = 0
Write original equation.
2x(x + 4) = 0
Factor left side.
2x = 0 or
x + 4 = 0
Zero-product property
x = 0 or
x = – 4
Solve for x.
ANSWER
The solutions of the equation are 0 and – 4.

10. The solutions of the equation are 0 and 5 2 .
Example 4
Solve 6n2 = 15n.
6n2 – 15n = 0
Subtract 15n from each side.
3n(2n – 5) = 0
Factor left side.
3n = 0 or
2n – 5 = 0
Zero-product property
n = 5 2
n = 0 or
Solve for n.
ANSWER
The solutions of the equation are 0 and 5 2 .

11. Guided Practice
Solve the equation.
3. a2 + 5a = 0
ANSWER
0 and – 5
s2 – 9s = 0
ANSWER
0 and 3
x2 = 2x
ANSWER
0 and 1 2

12. Example 5
ARMADILLO
A startled armadillo jumps straight into the air with an initial vertical velocity of 14 feet per second. After how many seconds does it land on the ground?

13. Write a model for the armadillo’s height above the ground.
Example 5
SOLUTION
STEP 1
Write a model for the armadillo’s height above the ground.
h = –16t2 + vt + s
Vertical motion model
h = –16t2 + 14t + 0
Substitute 14 for v and 0 for s.
h = –16t2 + 14t
Simplify.

14. Example 5
STEP 2
Substitute 0 for h. When the armadillo lands, its height above the ground is 0 feet. Solve for t.
0 = –16t2 + 14t
Substitute 0 for h.
0 = 2t(–8t + 7)
Factor right side.
2t = 0 or
–8t + 7 = 0
Zero-product property
t = 0 or
t = 0.875
Solve for t.
ANSWER
The armadillo lands on the ground second after the armadillo jumps.

15. Guided Practice
6. WHAT IF? In Example 5, suppose the initial vertical velocity is 12 feet per second. After how many seconds does armadillo land on the ground?
ANSWER
The armadillo lands on the ground 0.75 second after the armadillo jumps.

16. Lesson Quiz
Solve the equation.
1. (y + 5)(y – 9) = 0
2. (2n + 3)(n – 4) = 0
ANSWER
– 5, 9
ANSWER 3 2 –
, 4
x2 = 20x
x2 = 18x
ANSWER 3 2 0,
ANSWER 10 3 0,

17. Lesson Quiz 5.
A dog jumps in the air with an initial velocity of 18 feet per second to catch a flying disc. How long does the dog remain in the air?
ANSWER
1.125 sec