 # Holt Algebra 1 9-6 Solving Quadratic Equations by Factoring Warm Up Find each product. 1. (x + 2)(x + 7)2. (x – 11)(x + 5) 3. (x – 10) 2 Factor each polynomial.

## Presentation on theme: "Holt Algebra 1 9-6 Solving Quadratic Equations by Factoring Warm Up Find each product. 1. (x + 2)(x + 7)2. (x – 11)(x + 5) 3. (x – 10) 2 Factor each polynomial."— Presentation transcript:

Holt Algebra 1 9-6 Solving Quadratic Equations by Factoring Warm Up Find each product. 1. (x + 2)(x + 7)2. (x – 11)(x + 5) 3. (x – 10) 2 Factor each polynomial. 4. x 2 + 12x + 355. x 2 + 2x – 63 6. x 2 – 10x + 167. 2x 2 – 16x + 32 x 2 + 9x + 14 x 2 – 6x – 55 x 2 – 20x + 100 (x + 5)(x + 7)(x – 7)(x + 9) (x – 2)(x – 8)2(x – 4 )2

Holt Algebra 1 9-6 Solving Quadratic Equations by Factoring 9-6 Solving Quadratic Equations by Factoring Holt Algebra 1

9-6 Solving Quadratic Equations by Factoring

Holt Algebra 1 9-6 Solving Quadratic Equations by Factoring Example 1A: Use the Zero Product Property Use the Zero Product Property to solve the equation. Check your answer. (x – 7)(x + 2) = 0 x – 7 = 0 or x + 2 = 0 x = 7 or x = –2 The solutions are 7 and –2. Use the Zero Product Property. Solve each equation.

Holt Algebra 1 9-6 Solving Quadratic Equations by Factoring Example 1A Continued Use the Zero Product Property to solve the equation. Check your answer. Substitute each solution for x into the original equation. Check (x – 7)(x + 2) = 0 (7 – 7)(7 + 2) 0 (0)(9) 0 0 Check (x – 7)(x + 2) = 0 (–2 – 7)(–2 + 2) 0 (–9)(0) 0 0

Holt Algebra 1 9-6 Solving Quadratic Equations by Factoring Example 1B: Use the Zero Product Property Use the Zero Product Property to solve each equation. Check your answer. (x – 2)(x) = 0 x = 0 or x – 2 = 0 x = 2 The solutions are 0 and 2. Use the Zero Product Property. Solve the second equation. Substitute each solution for x into the original equation. Check (x – 2)(x) = 0 (0 – 2)(0) 0 (–2)(0) 0 0 (x – 2)(x) = 0 (2 – 2)(2) 0 (0)(2) 0 0 0 (x)(x – 2) = 0

Holt Algebra 1 9-6 Solving Quadratic Equations by Factoring Example 2A: Solving Quadratic Equations by Factoring Solve the quadratic equation by factoring. Check your answer. x 2 – 6x + 8 = 0 (x – 4)(x – 2) = 0 x – 4 = 0 or x – 2 = 0 x = 4 or x = 2 The solutions are 4 and 2. Factor the trinomial. Use the Zero Product Property. Solve each equation. x 2 – 6x + 8 = 0 (4) 2 – 6(4) + 8 0 16 – 24 + 8 0 0 0 Check x 2 – 6x + 8 = 0 (2) 2 – 6(2) + 8 0 4 – 12 + 8 0 0 0 Check

Holt Algebra 1 9-6 Solving Quadratic Equations by Factoring Example 2B: Solving Quadratic Equations by Factoring Solve the quadratic equation by factoring. Check your answer. x 2 + 4x = 21 –21 x 2 + 4x – 21 = 0 (x + 7)(x –3) = 0 x + 7 = 0 or x – 3 = 0 x = –7 or x = 3 The solutions are –7 and 3. The equation must be written in standard form. So subtract 21 from both sides. Factor the trinomial. Use the Zero Product Property. Solve each equation.

Holt Algebra 1 9-6 Solving Quadratic Equations by Factoring Example 2D: Solving Quadratic Equations by Factoring Solve the quadratic equation by factoring. Check your answer. –2x 2 = 20x + 50 The equation must be written in standard form. So add 2x 2 to both sides. Factor out the GCF 2. +2x 2 0 = 2x 2 + 20x + 50 –2x 2 = 20x + 50 2x 2 + 20x + 50 = 0 2(x 2 + 10x + 25) = 0 Factor the trinomial. 2(x + 5)(x + 5) = 0 2 ≠ 0 or x + 5 = 0 x = –5 Use the Zero Product Property. Solve the equation.

Holt Algebra 1 9-6 Solving Quadratic Equations by Factoring Check It Out! Example 2c Solve the quadratic equation by factoring. Check your answer. 30x = –9x 2 – 25 –9x 2 – 30x – 25 = 0 –1(3x + 5)(3x + 5) = 0 –1(9x 2 + 30x + 25) = 0 –1 ≠ 0 or 3x + 5 = 0 Write the equation in standard form. Factor the trinomial. Use the Zero Product Property. – 1 cannot equal 0. Solve the remaining equation. Factor out the GCF, –1.

Holt Algebra 1 9-6 Solving Quadratic Equations by Factoring Check It Out! Example 3 What if…? The equation for the height above the water for another diver can be modeled by h = –16t 2 + 8t + 24. Find the time it takes this diver to reach the water. h = –16t 2 + 8t + 24 0 = –16t 2 + 8t + 24 0 = –8(2t 2 – t – 3) 0 = –8(2t – 3)(t + 1) The diver reaches the water when h = 0. Factor out the GFC, –8. Factor the trinomial.

Holt Algebra 1 9-6 Solving Quadratic Equations by Factoring –8 ≠ 0, 2t – 3 = 0 or t + 1= 0 Use the Zero Product Property. 2t = 3 or t = –1 Solve each equation. Since time cannot be negative, –1 does not make sense in this situation. It takes the diver 1.5 seconds to reach the water. Check 0 = –16t 2 + 8t + 24 Substitute 1 into the original equation. 0 –16(1.5) 2 + 8(1.5) + 24 0 –36 + 12 + 24 0 Check It Out! Example 3 Continued t = 1.5 

Holt Algebra 1 9-6 Solving Quadratic Equations by Factoring

Holt Algebra 1 9-6 Solving Quadratic Equations by Factoring 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. x 2 + 16x + 48 = 0 4. x 2 – 11x = –24 10, –5 –5, 0 –4, –12 3, 8

Holt Algebra 1 9-6 Solving Quadratic Equations by Factoring Lesson Quiz: Part II 1, –7 –9 –2 5 s 5. 2x 2 + 12x – 14 = 0 6. x 2 + 18x + 81 = 0 7. –4x 2 = 16x + 16 8. The height of a rocket launched upward from a 160 foot cliff is modeled by the function h(t) = –16t 2 + 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.

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