405 3.6 Polynomial and Rational Inequalities EXAMPLE 3 Solving a Polynomial Inequality Using a Graph Solve 4x4 - 6x3 - 3x2 + x Ú 2x3 - 6x + 2. SOLUTION Step 1 Subtract 2x3 - 6x + 2 from each side of the inequality to obtain the equivalent inequality 4x4 - 8x3 - 3x2 + 7x - 2 Ú 0. Step 2 Let ƒ1x2 = 4x4 - 8x3 - 3x2 + 7x - 2 and graph. By the rational zeros theorem, the possible rational zeros of the polynomial are {1, {2, { 1 2 , and { 1 4 . Use synthetic division to show that -1 is a zero. -1) 4 -8 -3 7 -2 -4 12 -9 2 4 -12 9 -2 0 Use the results of the synthetic division to factor as follows. ƒ1x2 = 1x + 1214x3 - 12x2 + 9x - 22 The cubic polynomial has the same possible rational zeros as ƒ1x2. Use synthetic division to show that 2 is a zero of the cubic polynomial. 2) 4 -12 9 -2 8 -8 2 4 -4 1 0 Therefore, ƒ1x2 = 1x + 121x - 2214x2 - 4x + 12 ƒ1x2 = 1x + 121x - 2212x - 122. Factor again. The zeros of the polynomial are -1, 2, and 1 2 (multiplicity 2). The dominating term of ƒ1x2 is 4x4, indicating end behavior . The y-intercept, found by evaluating ƒ102, is 10, -22. See the graph in Figure 64. x y f(x) = 4x4 – 8x3 – 3x2 + 7x – 2 f(x) = (x + 1)(x – 2)(2x – 1)2 5 2 –5 –2 –1 0 (21, 0) (2, 0) 1 2Q , 0R Figure 64 Step 3 Referring to Figure 64, ƒ1x2 Ú 0 —that is, the graph lies on or above the x-axis—for solution set 1-∞, -14 ´ e 1 2f ´ 32, ∞2. Because the inequality is nonstrict, the zeros of ƒ1x2 are included. S Now Try Exercise 39.
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