YOU TRY 3 EXAMPLE 4 In-Class Example 4 Solve. a) (7t 2)(t 2 15t 54) 0 b) 9y3 900y 0 Answer: a) e 2 7 , 6, 9f b) {0, 10, 10} What do you think the “factored out variable” will always be equal to? YOU TRY 4 Solve. a) 5c2 6c 8 b) 3q2 18q c) (m 5)(m 10) 6 d) z(z 3) 40 3 Solve Higher-Degree Equations by Factoring Sometimes, equations that are not quadratics can be solved by factoring as well. Solve each equation. a) (4x 1)(x2 8x 20) 0 b) 12n3 108n 0 Solution a) This is not a quadratic equation because if we multiplied the factors on the left we would get 4x3 33x2 72x 20 0. This is a cubic equation because the degree of the polynomial on the left is 3. The original equation contains the product of two factors so we can use the zero product rule. (4x 1)(x2 8x 20) 0 (4x 1)(x 10)(x 2) 0 Factor. b T R 4x 1 0 or x 10 0 or x 2 0 Set each factor equal to zero. 4x 1 x 1 4 or x 10 or x 2 Solve. The check is left to the student. The solution set is e2, 1 4 , 10 f . b) The GCF of the terms in the equation 12n3 108n 0 is 12n. Remember, however, that we can divide an equation by a constant but we cannot divide an equation by a variable. Dividing by a variable may eliminate a solution and may mean we are dividing by zero. So let’s begin by dividing each term by 12. 12n3 12 108n 12 0 12 Divide by 12. n3 9n 0 Simplify. n(n2 9) 0 Factor out n. n(n 3)(n 3) 0 Factor n2 9. n 0 or n 3 0 or n 3 0 Set each factor equal to zero. n 3 n 3 Solve. Check. The solution set is {0, 3, 3}. Solve. a) (5y 3)(y2 10y 21) 0 b) 8k3 32k 0 428 CHAPTER 7 Factoring Polynomials www.mhhe.com/messersmith
messersmith_power_introductory_algebra_1e_ch4_7_10
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