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**Solving Quadratic Equations by Factoring 8-6**

Warm Up Lesson Presentation Lesson Quiz Holt McDougal Algebra 1 Holt Algebra 1

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Objective Solve quadratic equations by factoring.

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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.

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**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

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**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.

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**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.

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**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.

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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.

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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.

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**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.

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**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.

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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.

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**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

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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.

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** 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.

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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

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