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Zero Product Principle Andrew Gloag Eve Rawley Anne Gloag Say Thanks to the Authors Click http://www.ck12.org/saythanks (No sign in required) To access a customizable version of this book, as well as other interactive content, visit www.ck12.org CK-12 Foundation is a non-profit organization with a mission to reduce the cost of textbook materials for the K-12 market both in the U.S. and worldwide. Using an open-content, web-based collaborative model termed the FlexBook®, CK-12 intends to pioneer the generation and distribution of high-quality educational content that will serve both as core text as well as provide an adaptive environment for learning, powered through the FlexBook Platform®. Copyright © 2014 CK-12 Foundation, www.ck12.org The names “CK-12” and “CK12” and associated logos and the terms “FlexBook®” and “FlexBook Platform®” (collectively “CK-12 Marks”) are trademarks and service marks of CK-12 Foundation and are protected by federal, state, and international laws. Any form of reproduction of this book in any format or medium, in whole or in sections must include the referral attribution link http://www.ck12.org/saythanks (placed in a visible location) in addition to the following terms. Except as otherwise noted, all CK-12 Content (including CK-12 Curriculum Material) is made available to Users in accordance with the Creative Commons Attribution-Non-Commercial 3.0 Unported (CC BY-NC 3.0) License (http://creativecommons.org/ licenses/by-nc/3.0/), as amended and updated by Creative Commons from time to time (the “CC License”), which is incorporated herein by this reference. Complete terms can be found at http://www.ck12.org/terms. Printed: December 11, 2014 AUTHORS Andrew Gloag Eve Rawley Anne Gloag www.ck12.org C HAPTER Chapter 1. Zero Product Principle 1 Zero Product Principle Here you’ll learn how to apply the zero-product property and how to factor polynomials to solve for their unknown variables. What if you had a polynomial equation like 3x2 + 4x − 4 = 0? How could you factor the polynomial to solve the equation? After completing this Concept, you’ll be able to solve polynomial equations by factoring and by using the zero-product property. Watch This MEDIA Click image to the left or use the URL below. URL: http://www.ck12.org/flx/render/embeddedobject/133003 CK-12 Foundation: 0907S Factoring to Solve Polynomials Guidance The most useful thing about factoring is that we can use it to help solve polynomial equations. Example A Consider an equation like 2x2 + 5x − 42 = 0. How do you solve for x? Solution: There’s no good way to isolate x in this equation, so we can’t solve it using any of the techniques we’ve already learned. But the left-hand side of the equation can be factored, making the equation (x + 6)(2x − 7) = 0. How is this helpful? The answer lies in a useful property of multiplication: if two numbers multiply to zero, then at least one of those numbers must be zero. This is called the Zero-Product Property. What does this mean for our polynomial equation? Since the product equals 0, then at least one of the factors on the left-hand side must equal zero. So we can find the two solutions by setting each factor equal to zero and solving each equation separately. Setting the factors equal to zero gives us: (x + 6) = 0 OR (2x − 7) = 0 Solving both of those equations gives us: 1 www.ck12.org 2x − 7 = 0 x+6 = 0 x = −6 2x = 7 OR x= 7 2 Notice that the solution is x = −6 OR x = 27 . The OR means that either of these values of x would make the product of the two factors equal to zero. Let’s plug the solutions back into the equation and check that this is correct. 7 2 (x + 6)(2x − 7) = 7 7 +6 2· −7 = 2 2 19 (7 − 7) = 2 19 (0) = 0 2 Check : x = −6; Check : x = (x + 6)(2x − 7) = (−6 + 6)(2(−6) − 7) = (0)(−19) = 0 Both solutions check out. Factoring a polynomial is very useful because the Zero-Product Property allows us to break up the problem into simpler separate steps. When we can’t factor a polynomial, the problem becomes harder and we must use other methods that you will learn later. As a last note in this section, keep in mind that the Zero-Product Property only works when a product equals zero. For example, if you multiplied two numbers and the answer was nine, that wouldn’t mean that one or both of the numbers must be nine. In order to use the property, the factored polynomial must be equal to zero. Example B Solve each equation: a) (x − 9)(3x + 4) = 0 b) x(5x − 4) = 0 c) 4x(x + 6)(4x − 9) = 0 Solution Since all the polynomials are in factored form, we can just set each factor equal to zero and solve the simpler equations separately a) (x − 9)(3x + 4) = 0 can be split up into two linear equations: x−9 = 0 x=9 2 3x + 4 = 0 or 3x = −4 4 x=− 3 www.ck12.org Chapter 1. Zero Product Principle b) x(5x − 4) = 0 can be split up into two linear equations: 5x − 4 = 0 x=0 5x = 4 4 x= 5 or c) 4x(x + 6)(4x − 9) = 0 can be split up into three linear equations: 4x = 0 0 x= 4 4x − 9 = 0 x+6 = 0 or x = −6 or 4x = 9 x= x=0 9 4 Solve Simple Polynomial Equations by Factoring Now that we know the basics of factoring, we can solve some simple polynomial equations. We already saw how we can use the Zero-Product Property to solve polynomials in factored form—now we can use that knowledge to solve polynomials by factoring them first. Here are the steps: a) If necessary, rewrite the equation in standard form so that the right-hand side equals zero. b) Factor the polynomial completely. c) Use the zero-product rule to set each factor equal to zero. d) Solve each equation from step 3. e) Check your answers by substituting your solutions into the original equation Example C Solve the following polynomial equations. a) x2 − 2x = 0 b) 2x2 = 5x c) 9x2 y − 6xy = 0 Solution a) x2 − 2x = 0 Rewrite: this is not necessary since the equation is in the correct form. Factor: The common factor is x, so this factors as x(x − 2) = 0. Set each factor equal to zero: x=0 or x−2 = 0 Solve: 3 www.ck12.org x=0 x=2 or Check: Substitute each solution back into the original equation. x = 0 ⇒ (0)2 − 2(0) = 0 works out x = 2 ⇒ (2)2 − 2(2) = 4 − 4 = 0 works out Answer: x = 0, x = 2 b) 2x2 = 5x Rewrite: 2x2 = 5x ⇒ 2x2 − 5x = 0 Factor: The common factor is x, so this factors as x(2x − 5) = 0. Set each factor equal to zero: x=0 2x − 5 = 0 or Solve: x=0 2x = 5 5 x= 2 or Check: Substitute each solution back into the original equation. x = 0 ⇒ 2(0)2 = 5(0) ⇒ 0 = 0 2 5 5 5 25 25 25 25 x= ⇒2 = 5· ⇒ 2· = ⇒ = 2 2 2 4 2 2 2 Answer: x = 0, x = works out works out 5 2 c) 9x2 y − 6xy = 0 Rewrite: not necessary Factor: The common factor is 3xy, so this factors as 3xy(3x − 2) = 0. Set each factor equal to zero: 3 = 0 is never true, so this part does not give a solution. The factors we have left give us: x=0 Solve: 4 or y=0 or 3x − 2 = 0 www.ck12.org Chapter 1. Zero Product Principle x=0 y=0 2 x= 3 or or 3x = 2 Check: Substitute each solution back into the original equation. x = 0 ⇒ 9(0)y − 6(0)y = 0 − 0 = 0 2 y = 0 ⇒ 9x (0) − 6x(0) = 0 − 0 = 0 2 2 2 4 2 y − 6 · y = 9 · y − 4y = 4y − 4y = 0 x = ⇒ 9· 3 3 3 9 Answer: x = 0, y = 0, x = works out works out works out 2 3 Watch this video for help with the Examples above. MEDIA Click image to the left or use the URL below. URL: http://www.ck12.org/flx/render/embeddedobject/133004 CK-12 Foundation: Factoring to Solve Polynomials Vocabulary • Polynomials can be written in expanded form or in factored form. Expanded form means that you have sums and differences of different terms: • The factored form of a polynomial means it is written as a product of its factors. • Zero Product Property: The only way a product is zero is if one or more of the terms are equal to zero: a · b = 0 ⇒ a = 0 or b = 0. Guided Practice Solve the following polynomial equation. 9x2 − 3x = 0 Solution: 9x2 − 3x = 0 Rewrite: This is not necessary since the equation is in the correct form. 5 www.ck12.org Factor: The common factor is 3x, so this factors as: 3x(3x − 1) = 0. Set each factor equal to zero. 3x = 0 or x−2 = 0 Solve: x=0 x=2 or Check: Substitute each solution back into the original equation. x=0 (0)2 − 2(0) = 0 x=2 (2)2 − 2(2) = 0 Answer x = 0, x = 2 Explore More Solve the following polynomial equations. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 6 x(x + 12) = 0 (2x + 1)(2x − 1) = 0 (x − 5)(2x + 7)(3x − 4) = 0 2x(x + 9)(7x − 20) = 0 x(3 + y) = 0 x(x − 2y) = 0 18y − 3y2 = 0 9x2 = 27x 4a2 + a = 0 b2 − 53 b = 0 4x2 = 36 x3 − 5x2 = 0