1. Solving Quadratic Equations by Factoring 8-6
Warm Up
Lesson Presentation
Lesson Quiz
Holt McDougal Algebra 1
Holt Algebra 1

2. Objective
Solve quadratic equations by factoring.

3. If a quadratic equation is written in standard form, ax2 + bx + c = 0, then to solve the equation, you may need to factor before using the Zero Product Property.

4. Example 2A: Solving Quadratic Equations by Factoring
Solve the quadratic equation by factoring. Check your answer.
x2 – 6x + 8 = 0
(x – 4)(x – 2) = 0
Factor the trinomial.
x – 4 = 0 or x – 2 = 0
Use the Zero Product Property.
x = 4 or x = 2
The solutions are 4 and 2.
Solve each equation.
x2 – 6x + 8 = 0
(4)2 – 6(4)
16 –
0 0 
Check
x2 – 6x + 8 = 0
(2)2 – 6(2)
4 –
0 0 
Check

5. Example 2B: Solving Quadratic Equations by Factoring
Solve the quadratic equation by factoring. Check your answer.
x2 + 4x = 21
The equation must be written in standard form. So subtract 21 from both sides.
x2 + 4x = 21
–21 –21
x2 + 4x – 21 = 0
(x + 7)(x –3) = 0
Factor the trinomial.
x + 7 = 0 or x – 3 = 0
Use the Zero Product Property.
x = –7 or x = 3
The solutions are –7 and 3.
Solve each equation.

6. Example 2C: Solving Quadratic Equations by Factoring
Solve the quadratic equation by factoring. Check your answer.
x2 – 12x + 36 = 0
(x – 6)(x – 6) = 0
Factor the trinomial.
x – 6 = 0 or x – 6 = 0
Use the Zero Product Property.
x = or x = 6
Solve each equation.
Both factors result in the same solution, so there is one solution, 6.

7. Example 2D: Solving Quadratic Equations by Factoring
Solve the quadratic equation by factoring. Check your answer.
–2x2 = 20x + 50
+2x x2
0 = 2x2 + 20x + 50
–2x2 = 20x + 50
The equation must be written in standard form. So add 2x2 to both sides.
2x2 + 20x + 50 = 0
Factor out the GCF 2.
2(x2 + 10x + 25) = 0
Factor the trinomial.
2(x + 5)(x + 5) = 0
2 ≠ 0 or x + 5 = 0
Use the Zero Product Property.
x = –5
Solve the equation.

8. Example 2D Continued
Solve the quadratic equation by factoring. Check your answer.
–2x2 = 20x + 50
Check
–2x2 = 20x + 50
–2(–5) (–5) + 50
– –
– –50 
Substitute –5 into the original equation.

9. Check It Out! Example 2a
Solve the quadratic equation by factoring. Check your answer.
x2 – 6x + 9 = 0
(x – 3)(x – 3) = 0
Factor the trinomial.
x – 3 = 0 or x – 3 = 0
Use the Zero Product Property.
x = 3 or x = 3
Solve each equation.
Both equations result in the same solution, so there is one solution, 3.
x2 – 6x + 9 = 0
(3)2 – 6(3)
9 –
0 0 
Check
Substitute 3 into the original equation.

10. Solve the quadratic equation by factoring. Check your answer.
Check It Out! Example 2b
Solve the quadratic equation by factoring. Check your answer.
x2 + 4x = 5
Write the equation in standard form. Add – 5 to both sides.
x2 + 4x = 5
–5 –5
x2 + 4x – 5 = 0
(x – 1)(x + 5) = 0
Factor the trinomial.
x – 1 = 0 or x + 5 = 0
Use the Zero Product Property.
x = or x = –5
Solve each equation.
The solutions are 1 and –5.

11. Check It Out! Example 2b Continued
Solve the quadratic equation by factoring. Check your answer.
x2 + 4x = 5
Check Graph the related quadratic function. The zeros of the related function should be the same as the solutions from factoring. ●
The graph of y = x2 + 4x – 5 shows that the two zeros appear to be 1 and –5, the same as the solutions from factoring. 

12. Check It Out! Example 2d
Solve the quadratic equation by factoring. Check your answer.
3x2 – 4x + 1 = 0
(3x – 1)(x – 1) = 0
Factor the trinomial.
3x – 1 = 0 or x – 1 = 0
Use the Zero Product Property.
or x = 1
Solve each equation.
The solutions are and x = 1.

13. Check It Out! Example 2d Continued
Solve the quadratic equation by factoring. Check your answer.
3x2 – 4x + 1 = 0
3x2 – 4x + 1 = 0 –
0 0 
Check
3x2 – 4x + 1 = 0
3(1)2 – 4(1)
3 –
0 0 
Check

14. Example 3: Application
The height in feet of a diver above the water can be modeled by h(t) = –16t2 + 8t + 8, where t is time in seconds after the diver jumps off a platform. Find the time it takes for the diver to reach the water.
h = –16t2 + 8t + 8
The diver reaches the water when h = 0.
0 = –16t2 + 8t + 8
0 = –8(2t2 – t – 1)
Factor out the GFC, –8.
0 = –8(2t + 1)(t – 1)
Factor the trinomial.

15.   Example 3 Continued Use the Zero Product Property.
–8 ≠ 0, 2t + 1 = 0 or t – 1= 0
2t = –1 or t = 1
Solve each equation.
Since time cannot be negative, does not make sense in this situation. 
It takes the diver 1 second to reach the water.
Check 0 = –16t2 + 8t + 8
0 –16(1)2 + 8(1) + 8
0 –
Substitute 1 into the original equation. 

16. Lesson Quiz: Part II
5. 2x2 + 12x – 14 = 0
1, –7
6. x2 + 18x + 81 = 0 –9
7. –4x2 = 16x + 16 –2
8. The height of a rocket launched upward from a 160 foot cliff is modeled by the function h(t) = –16t2 + 48t + 160, where h is height in feet and t is time in seconds. Find the time it takes the rocket to reach the ground at the bottom of the cliff.
5 s