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Quadratic Equations: Factoring and Zero Product Property

Learn how to solve quadratic equations by factoring and using the Zero Product Property. Practice solving equations and check your answers.

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Quadratic Equations: Factoring and Zero Product Property

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  1. Preview Warm Up California Standards Lesson Presentation

  2. Warm Up • Find each product. • 1. (x + 2)(x + 7) 2. (x – 11)(x + 5) • 3. (x – 10)2 • Factor each polynomial. • 4. x2 + 12x + 35 5. x2 + 2x – 63 • 6. x2 – 10x + 16 7. 2x2 – 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

  3. California Standards 14.0 Students solve a quadratic equation by factoring or completing the square. 23.0 Students apply quadratic equations to physical problems, such as the motion of an object under the force of gravity.

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

  5. Additional 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

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

  7. Check (x – 2)(x) = 0 (x – 2)(x) = 0 (0 – 2)(0) 0 (2 – 2)(2) 0 (0)(2) 0 (–2)(0) 0   0 0 0 0 Additional 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 Substitute each solution for x in the original equation.

  8. Check(x)(x + 4) = 0 (x)(x +4) = 0 (0)(0 + 4) 0 (–4)(–4 + 4) 0 (–4)(0) 0 (0)(4) 0  0 0 0 0 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 Substitute each solution for x in the original equation. 

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

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

  11. You may need to factor before using the Zero Product Property. You can check your answers by substituting into the original equation or by graphing. If the factored form of the equation has two different factors, the graph of the related function will cross the x-axis in two places. If the factored form has two identical factors, the graph will cross the x-axis in one place.

  12. Helpful Hint To review factoring techniques, see Lessons 8-3 through 8-5.

  13. x =4or x =2 Check Check x2 – 6x + 8 = 0 x2 – 6x + 8 = 0 (4)2 – 6(4) + 8 0 (2)2 – 6(2) + 8 0 16 – 24 + 8 0 4 – 12 + 8 0   0 0 0 0 Additional 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. Solve each equation.

  14. x2 + 4x = 21 –21 –21 x2 + 4x – 21 = 0 Additional 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. (x + 7)(x –3) = 0 Factor the trinomial. x + 7 = 0 or x – 3 = 0 Use the Zero Product Property. x = –7or x = 3 Solve each equation.

  15. ● ● Additional Example 2B Continued CheckGraph the related quadratic function. Because there are two solutions found by factoring, the graph should cross the x-axis in two places. The graph of y = x2 + 4x – 21 intersects the x-axis at x = –7 and x = 3, the same as the solutions from factoring. 

  16. Additional 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 = 6 or x = 6 Solve each equation. Both factors result in the same solution, so there is one solution, 6.

  17. The graph of y = x2 – 12x + 36 shows one zero at 6, the same as the solution from factoring.  ● Additional Example 2C Continued CheckGraph the related quadratic function.

  18. –2x2 = 20x + 50 +2x2 +2x2 0 = 2x2 + 20x + 50 Additional Example 2D: Solving Quadratic Equations by Factoring Solve the quadratic equation by factoring. Check your answer. –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.

  19. –2x2 = 20x + 50 –2(–5)2 20(–5) + 50 –50 –100 + 50 –50 –50  Additional Example 2D Continued Solve the quadratic equation by factoring. Check your answer. –2x2 = 20x + 50 Check Substitute –5 into the original equation.

  20. Check x2 – 6x + 9 = 0 (3)2 – 6(3) + 9 0 9 – 18 + 9 0  0 0 Check It Out! Example 2a Solve the quadratic equation by factoring. Check your answer. x2 – 6x + 9 = 0 Factor the trinomial. (x – 3)(x – 3) = 0 Use the Zero Product Property. x – 3 = 0 or x – 3 = 0 x =3or x =3 Solve each equation. The only solution is 3. Substitute 3 into the original equation.

  21. x2 + 4x = 5 –5 –5 x2 + 4x – 5 = 0 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. (x – 1)(x + 5) = 0 Factor the trinomial. x – 1 = 0 or x + 5 = 0 Use the Zero Product Property. x = 1 or x = –5 Solve each equation.

  22. The graph of y = x2 + 4x – 5 intersects the x-axis at x = 1 and x = –5, the same as the solutions from factoring. ● ●  Check It Out! Example 2b Continued CheckGraph the related quadratic function. Because there are two solutions found by factoring, the graph should cross the x-axis in two places.

  23. Check It Out! Example 2c Solve the quadratic equation by factoring. Check your answer. 30x = –9x2 – 25 Write the equation in standard form. –9x2 – 30x – 25 = 0 –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.

  24. The graph of y = –9x2 – 30x – 25 intersects the x-axis at x = , the same as the solution from factoring.  ● Check It Out! Example 2c Continued Check Graph the related quadratic function. Because there is one solution found by factoring, the graph should cross the x-axis in one place.

  25. or x = 1 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. Solve each equation.

  26. Check Check 3x2 – 4x + 1 = 0 3x2 – 4x + 1 = 0 3(1)2 – 4(1) + 1 0 3 – 4 + 1 0 3 – 4 + 1 0  0 0  0 0 Check It Out! Example 2d Continued

  27. Additional Example 3: Application The height in feet of a diverabove 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 GCF, –8. 0 = –8(2t + 1)(t – 1) Factor the trinomial.

  28. Since time cannot be negative, does not make sense in this situation.  0 –16(1)2 + 8(1) + 8 0 –16 + 8 + 8 0 0 Additional 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. It takes the diver 1 second to reach the water. Check 0 = –16t2 + 8t + 8 Substitute 1 into the original equation. 

  29. 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 + 24. 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 GCF, –8. 0 = –8(2t – 3)(t + 1) Factor the trinomial.

  30. 0 –16(1.5)2 + 8(1.5) + 24 0 –36 + 12 + 24 0 0 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 Substitute 1.5 into the original equation. 

  31. 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

  32. Lesson Quiz: Part II 1, –7 5. 2x2 + 12x – 14 = 0 6. x2 + 18x + 81 = 0 –9 –2 7. –4x2 = 16x + 16 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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