# Solving Quadratic Equations by Factoring 8-6

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

Warm Up Find each product. 1. (x + 2)(x + 7) 2. (x – 11)(x + 5)
Factor each polynomial. 4. x2 + 12x x2 + 2x – 63 6. x2 – 10x x2 – 16x + 32 x2 + 9x + 14 x2 – 6x – 55 x2 – 20x + 100 (x + 5)(x + 7) (x – 7)(x + 9) (x – 2)(x – 8) 2(x – 4)2

Objective Solve quadratic equations by factoring.

You have solved quadratic equations by graphing
You have solved quadratic equations by graphing. Another method used to solve quadratic equations is to factor and use the Zero Product Property.

Example 1A: Use the Zero Product Property
Use the Zero Product Property to solve the equation. Check your answer. (x – 7)(x + 2) = 0 Use the Zero Product Property. x – 7 = 0 or x + 2 = 0 Solve each equation. x = 7 or x = –2 The solutions are 7 and –2.

Example 1A Continued Use the Zero Product Property to solve the equation. Check your answer. Check (x – 7)(x + 2) = 0 (7 – 7)(7 + 2) 0 (0)(9) 0 0 0 Substitute each solution for x into the original equation. Check (x – 7)(x + 2) = 0 (–2 – 7)(–2 + 2) 0 (–9)(0) 0 0 0

  Example 1B: Use the Zero Product Property
Use the Zero Product Property to solve each equation. Check your answer. (x – 2)(x) = 0 (x)(x – 2) = 0 Use the Zero Product Property. x = 0 or x – 2 = 0 Solve the second equation. x = 2 The solutions are 0 and 2. (x – 2)(x) = 0 (2 – 2)(2) 0 (0)(2) 0 Check (x – 2)(x) = 0 (0 – 2)(0) 0 (–2)(0) 0 Substitute each solution for x into the original equation.

Substitute each solution for x into the original equation.
Check It Out! Example 1a Use the Zero Product Property to solve each equation. Check your answer. (x)(x + 4) = 0 Use the Zero Product Property. x = 0 or x + 4 = 0 Solve the second equation. x = –4 The solutions are 0 and –4. Check (x)(x + 4) = 0 (0)(0 + 4) 0 (0)(4) 0 0 0 (x)(x +4) = 0 (–4)(–4 + 4) 0 (–4)(0) 0 Substitute each solution for x into the original equation.

Check It Out! Example 1b Use the Zero Product Property to solve the equation. Check your answer. (x + 4)(x – 3) = 0 Use the Zero Product Property. x + 4 = 0 or x – 3 = 0 x = –4 or x = 3 Solve each equation. The solutions are –4 and 3.

Check It Out! Example 1b Continued
Use the Zero Product Property to solve the equation. Check your answer. (x + 4)(x – 3) = 0 Check (x + 4)(x – 3 ) = 0 (–4 + 4)(–4 –3) 0 (0)(–7) 0 Substitute each solution for x into the original equation. Check (x + 4)(x – 3 ) = 0 (3 + 4)(3 –3) 0 (7)(0) 0

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.

To review factoring techniques, see lessons 8-3 through 8-5.

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

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.

Example 2B Continued Solve the quadratic equation by factoring. Check your answer. x2 + 4x = 21 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 – 21 shows that two zeros appear to be –7 and 3, the same as the solutions from factoring. 

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.

Example 2C Continued Solve the quadratic equation by factoring. Check your answer. x2 – 12x + 36 = 0 Check Graph the related quadratic function. The graph of y = x2 – 12x + 36 shows that one zero appears to be 6, the same as the solution from factoring. 

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.

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.

(x – 3)(x – 3) is a perfect square
(x – 3)(x – 3) is a perfect square. Since both factors are the same, you solve only one of them. Helpful Hint

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.

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.

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.

Check It Out! Example 2c Solve the quadratic equation by factoring. Check your answer. 30x = –9x2 – 25 –9x2 – 30x – 25 = 0 Write the equation in standard form. –1(9x2 + 30x + 25) = 0 Factor out the GCF, –1. –1(3x + 5)(3x + 5) = 0 Factor the trinomial. –1 ≠ 0 or 3x + 5 = 0 Use the Zero Product Property. – 1 cannot equal 0. Solve the remaining equation.

Check It Out! Example 2c Continued
Solve the quadratic equation by factoring. Check your answer. 30x = –9x2 – 25 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 = –9x2 – 30x – 25 shows one zero and it appears to be at , the same as the solutions from factoring.

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.

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

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.

  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.

Check It Out! Example 3 What if…? The equation for the height above the water for another diver can be modeled by h = –16t2 + 8t Find the time it takes this diver to reach the water. h = –16t2 + 8t + 24 The diver reaches the water when h = 0. 0 = –16t2 + 8t + 24 0 = –8(2t2 – t – 3) Factor out the GFC, –8. 0 = –8(2t – 3)(t + 1) Factor the trinomial.

Check It Out! Example 3 Continued
Use the Zero Product Property. –8 ≠ 0, 2t – 3 = 0 or t + 1= 0 2t = 3 or t = –1 Solve each equation. Since time cannot be negative, –1 does not make sense in this situation. t = 1.5 It takes the diver 1.5 seconds to reach the water. Check 0 = –16t2 + 8t + 24 0 –16(1.5)2 + 8(1.5) + 24 0 – Substitute 1 into the original equation.

Lesson Quiz: Part I Use the Zero Product Property to solve each equation. Check your answers. 1. (x – 10)(x + 5) = 0 2. (x + 5)(x) = 0 Solve each quadratic equation by factoring. Check your answer. 3. x2 + 16x + 48 = 0 4. x2 – 11x = –24 10, –5 –5, 0 –4, –12 3, 8

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