e) It is tempting to solve (z 8)(z 4) 5 like this: (z 8)(z 4) 5 b R z 8 5 or z 4 5 This is incorrect! One side of the equation must equal zero in order to set each factor equal to zero. Begin by multiplying on the left. (z 8)(z 4) 5 z2 12z 32 5 Multiply using FOIL. z2 12z 27 0 Standard form (z 9)(z 3) 0 Factor. b R z 9 0 or z 3 0 Set each factor equal to zero. z 9 or z 3 Solve. The check is left to the student. The solution set is {3, 9}. YOU TRY 3 Solve. a) w2 4w 5 0 b) 29b 5(b2 4) c) (a 6)(a 4) 3 d) t2 8t e) (2y 1)2 5 y2 2( y 7) 3 Solve Higher Degree Equations by Factoring Sometimes, equations that are not quadratics can be solved by factoring as well. EXAMPLE 4 Solve each equation. a) (2x 1)(x2 9x 22) 0 b) 4w3 100w 0 Solution a) This is not a quadratic equation because if we multiplied the factors on the left we would get 2x3 19x2 35x 22 0. This is a cubic equation because the degree of the polynomial on the left is 3. The original equation is the product of two factors so we can use the zero product rule. (2x 1)(x2 9x 22) 0 (2x 1)(x 11)(x 2) 0 Factor. b T R 2x 1 0 or x 11 0 or x 2 0 Set each factor equal to zero. 2x 1 x 1 2 or x 11 or x 2 Solve. The check is left to the student. The solution set is e2, 1 2 , 11 f . The degree of the polynomial is the same as the greatest number of solutions the equation might have. www.mhhe.com/messersmith SECTION 7.4 Solving Quadratic Equations by Factoring 397
messersmith_power_intermediate_algebra_1e_ch4_7_10
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