Copyright © 2015, 2011, 2007 Pearson Education, Inc. 1 1 Chapter 6 Factoring

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 2 Factoring 6.1Greatest Common Factor and Factoring by Grouping 6.2Factoring Trinomials 6.3Factoring Special Products and Factoring Strategies 6.4Solving Equations by Factoring CHAPTER 6

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 3 Solving Equations by Factoring 1.Use the zero-factor theorem to solve equations by factoring. 2.Solve problems involving quadratic equations. 6.4

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 4 Zero-Factor Theorem If a and b are real numbers and ab = 0, then a = 0 or b = 0. Polynomial equation: An equation that equates two polynomials.

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 5 Example Solve. (x + 4)(x + 5) = 0 Solution Since we have a product of two factors, x + 4 and x + 5, equal to zero, one or both factors must equal 0. x + 4 = 0 or x + 5 = 0 Solve each equation. x =  4 x =  5 Check Verify that  4 and  5 satisfy the original equation, (x + 4)(x + 5) = 0. For x =  4: For x =  5: (  4 + 4)(  4 + 5) = 0 (  5 + 4)(  5 + 5) = 0 0(1) = 0 (  1)(0) = 0

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 6 Polynomial equation in standard form: P = 0, where P is a polynomial in terms of one variable written in descending order of degree.

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 7 Example Solve. 2x 2 – 5x = 3 Solution First, we need the equation in standard form. 2x 2 – 5x – 3 = 0 (2x + 1)(x – 3) = 0 Factor. 2x + 1 = 0 or x – 3 = 0 Use the zero-factor theorem to solve. The solutions are and 3. To check, we verify that the solutions satisfy the original equations.

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 8 Quadratic equation in one variable: An equation that can be written in the form ax 2 + bx + c = 0, where a, b, and c are all real numbers and a  0.

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 9 Example Solve. x 3 – 4x 2 = 21x Solution x 3 – 4x 2 – 21x = 0 Write the equation in standard form. x(x 2 – 4x – 21) = 0 x(x – 7 )(x + 3) = 0 x = 0 or x – 7 = 0 or x + 3 = 0 The solutions are 0, 7, and  3.

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 10 Cubic equation in one variable: An equation that can be written in the form ax 3 + bx 2 + cx + d = 0, where a, b, c, and d, are all real numbers and a  0. Solving Polynomial Equations Using Factoring To solve a polynomial equation using factoring, 1. Write the equation in standard form (Set one side equal to 0 with the other side in descending order of degree when possible). 2. Write the polynomial in factored form. 3. Use the zero-factor theorem to solve.

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 11 Example Solve. y(y – 7y) = –12 Solution Write the equation in standard form.

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 12 Example Solve 14x 2 + 9x + 2 = 10x + 6. Solution 14x 2 + 9x + 2 = 10x x 2 + 9x  10x + 2  6 = 0 14x 2  x  4 = 0 (7x  4)(2x + 1) = 0 7x  4 = 0 or 2x + 1 = 0 7x = 4 or 2x =  1 x = 4/7 or x =  1/2 The solutions are 4/7 and  1/2.

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 13 Example The equation h(t) =  16t 2 + v o t + h 0 describes the height, h, in feet of an object t seconds after being thrown upwards with an initial velocity of v 0 feet per second from an initial height of h 0 feet. Suppose a cannonball is fired from a platform that is 384 feet above the ground. The initial velocity of the object is 160 feet per second. How long will it take for the cannonball to land on the ground? Understand We are given a formula, the initial velocity of 160 feet per second, and the initial height of 384 feet. We are to find the time it takes the cannonball to reach the ground, which is at 0 feet.

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 14 continued 0 = –16t t = –16(t 2 – 10t – 24) = –16(t – 12)(t + 2) t – 12 = 0 or t + 2 = 0 t = 12 or t =  2 Plan Replace the variables in the formula with given values for h 0 and v 0, then solve for t. Execute Answer Our answer must describe the amount of time, so only the positive value, 12 makes sense. The cannonball takes 12 seconds to reach the ground. Check h = 16(12) 2  160(12)  384 = 2304 – 1920 – 384 = 0

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 15 The Pythagorean Theorem Given a right triangle, where a and b represent the lengths of the legs and c represents the length of the hypotenuse, then a 2 + b 2 = c 2. c (hypotenuse) b (leg) a (leg)

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 16 Example The figure shows a garden to be created. Find the length, in feet, of each side. (x) 2 + (x + 2) 2 = (x + 4) 2 (x + 2)(x – 6) = 0 x + 2 = 0 or x – 6 = 0 x = –2 or x = 6 x x + 2 x + 4 Understand We are given expressions for the lengths of the three sides of a right triangle and we are to find those lengths. Pan and Execute Use the Pythagorean theorem. x 2 + x 2 + 4x + 4 = x 2 + 8x + 16 x 2 – 4x – 12 = 0

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 17 continued (6) 2 + (8) 2 = (10) 2 Answer Because x describes a length in feet, only the positive solution is sensible so x must be 6 feet. This means x + 2 is 8 feet and x + 4 is 10 feet. Check = = 100