1. Factoring, Solving Quadratic Equtions with a  1 (8-4)
Objective: Factor trinomials of the form ax2 + bx + c. Solve equations of the form ax2 + bx + c = 0.

2. Factor ax2 + bx + c
In the last lesson, you factored quadratic expressions of the form ax2 + bx + c, where a = 1.
In this lesson, you will apply the factoring methods to quadratic expressions in which a is not 1.
To factor trinomials for the form ax2 + bx + c, find two integers, m and p, with a sum of b and a product of ac.
Then write ax2 + bx + c as ax2 + mx + px + c, and factor by grouping.

3. Example 1 Factor each trinomial. 5x2 + 27x + 10 5x2 + 2x + 25x + 10
b = 27
ac = 50
5x2 + 2x + 25x + 10
x(5x + 2) + 5(5x + 2)
(x + 5)(5x + 2)
Factors of ac
Sum
1, 50 51
2, 25 27
5, 10 15

4. Example 1 Factor each trinomial. 4x2 + 24x + 32 4(x2 + 2x + 4x + 8)
GCF = 4
4(x2 + 6x + 8)
b = 6
ac = 8
4(x2 + 2x + 4x + 8)
4(x(x + 2) + 4(x + 2))
4(x + 4)(x + 2)
Factors of ac
Sum
1, 8 9
2, 4 6

5. Check Your Progress Choose the best answer for the following.
Factor 3x2 + 26x + 35.
(3x + 7)(x + 5)
(3x + 1)(x + 35)
(3x + 5)(x + 7)
(x + 1)(3x + 7)
b = 26, ac = 105
Factors of ac
Sum
1, 105
106
3, 35 38
5, 21 26
7, 15 22
3x2 + 5x + 21x + 35
x(3x + 5) + 7(3x + 5)

6. Check Your Progress Choose the best answer for the following.
Factor 2x2 + 14x + 20.
(2x + 4)(x + 5)
(x + 2)(2x + 10)
2(x2 + 7x + 10)
2(x + 2)(x + 5)
GCF = 2
2(x2 + 7x + 10)
b = 7, ac = 10
Factors of ac
Sum
1, 10 11
2, 5 7
2(x2 + 2x + 5x + 10)
2(x(x + 2) + 5(x + 2))

7. Example 2 Factor 24x2 – 22x + 3. b = -22 ac = 72 24x2 – 4x – 18x + 3
Factors of ac
Sum
-1, -72
-73
-2, -36
-38
-3, -24
-27
-4, -18
-22
-6, -12
-18
-8, -9
-17

8. Check Your Progress Choose the best answer for the following.
Factor 10x2 – 23x + 12.
(2x + 3)(5x + 4)
(2x – 3)(5x – 4)
(2x + 6)(5x – 2)
(2x – 6)(5x – 2)
b = -23, ac = 120
Factors of ac
Sum
-1, -120
-121
-2, -60
-62
-3, -40
-43
-4, -30
-34
-5, -24
-29
-6, -20
-26
-8, -15
-23
-10, -12
-22
10x2 – 8x – 15x + 12
2x(5x – 4) – 3(5x – 4)

9. Factoring
A polynomial that cannot be written as a product of two polynomials with integral coefficients is called a prime polynomial.

10. Example 3
Factor 3x2 + 7x – 5. If the polynomial cannot be factored using integers, write prime.
b = 7
ac = -15
Prime
Factors of ac
Sum
1, -15
-14
-1, 15 14
3, -5 -2
-3, 5 2

11. Check Your Progress Choose the best answer for the following.
Factor 3x2 – 5x + 3, if possible.
(3x + 1)(x – 3)
(3x – 3)(x – 1)
(3x – 1)(x – 3)
Prime
b = -5, ac = 9
Factors of ac
Sum
-1, -9
-10
-3, -3 -6

12. Solve Equations by Factoring
A model for the height of a projectile is given by h = -16t2 + vt + h0, where h is the height in feet, t is the time in seconds, v is the initial velocity in feet per second, and h0 is the initial height in feet.
Equations of the form ax2 + bx + c = 0 can be solved by factoring and by using the Zero Product Property.

13. Example 4 The rocket was in flight for 3.5 seconds.
Mr. Nguyen’s science class built a model rocket. They launched the rocket outside. It cleared the top of a 60-foot high pole and then landed in a nearby tree. If the launch pad was 2 feet above the ground, the initial velocity of the rocket was 64 feet per second, and the rocket landed 30 feet above the ground, how long was the rocket in flight? Use the equation h = -16t2 + vt + h0.
30 = -16t2 + 64t + 2
16t2 – 64t + 28 = 0
GCF = 4
4(4t2 – 16t + 7) = 0
b = -16
ac = 28
4(4t2 – 2t – 14t + 7) = 0
4(2t(2t – 1) – 7(2t – 1)) = 0
4(2t – 7)(2t – 1) = 0
2t – 7 = 0
Factors of ac
Sum
-1, -28
-29
-2, -14
-16
-4, -7
-11
On the way down.
2t = 7
t = 7/2
2t – 1 = 0
On the way up.
2t = 1
t = 1/2
The rocket was in flight for 3.5 seconds.

14. Check Your Progress Choose the best answer for the following.
When Mario jumps over a hurdle, his feet leave the ground traveling at an initial upward velocity of 12 feet per second. Find the time t in seconds it takes for Mario’s feet to reach the ground again. Use the equation h = -16t2 + vt + h0.
1 second
0 seconds
¾ seconds
½ seconds
0 = -16t2 + 12t + 0
16t2 – 12t = 0
GCF = 4t
4t(4t – 3) = 0
4t = 0
4t – 3 = 0
t = 0
4t = 3
t = ¾
Beginning
End