349 3.3 Zeros of Polynomial Functions This result shows that -a0qn equals the product of the two factors p and 1an pn-1 + g+ a 1qn-12. For this reason, p must be a factor of -a 0qn. Because it was assumed that p q is written in lowest terms, p and q have no common factor other than 1, so p is not a factor of qn. Thus, p must be a factor of a 0. In a similar way, it can be shown that q is a factor of an. EXAMPLE 3 Using the Rational ZerosTheorem Consider the polynomial function. ƒ1x2 = 6x4 + 7x3 - 12x2 - 3x + 2 (a) List all possible rational zeros. (b) Find all rational zeros and factor ƒ1x2 into linear factors. SOLUTION (a) For a rational number p q to be a zero, p must be a factor of a0 = 2, and q must be a factor of a4 = 6. Thus, p can be {1 or {2, and q can be {1, {2, {3, or {6. The possible rational zeros p q are {1, {2, { 1 2 , { 1 3 , { 1 6 , and { 2 3 . (b) Use the remainder theorem to show that 1 is a zero. 1)6 7 -12 -3 2 6 13 1 -2 6 13 1 -2 0 ƒ112 = 0 The 0 remainder shows that 1 is a zero. The quotient is 6x3 + 13x2 + x - 2. ƒ1x2 = 1x - 1216x3 + 13x2 + x - 22 Begin factoring ƒ1x2. Now, use the quotient polynomial and synthetic division to find that -2 is a zero. -2)6 13 1 -2 -12 -2 2 6 1 -1 0 ƒ1-22 = 0 The new quotient polynomial is 6x2 + x - 1. Therefore, ƒ1x2 can now be completely factored as follows. ƒ1x2 = 1x - 121x + 2216x2 + x - 12 ƒ1x2 = 1x - 121x + 2213x - 1212x + 12 Setting 3x - 1 = 0 and 2x + 1 = 0 yields the zeros 1 3 and - 1 2 . In summary, the rational zeros are 1, -2, 1 3 , and - 1 2 . These zeros correspond to the x-intercepts of the graph of ƒ1x2 in Figure 18. The linear factorization of ƒ1x2 is as follows. ƒ1x2 = 6x4 + 7x3 - 12x2 - 3x + 2 ƒ1x2 = 1x - 121x + 2213x - 1212x + 12 S Now Try Exercise 39. ( ) , 0 1 3 , 0 ( ) – 1 2 f(x) = 6x4 + 7x3 – 12x2 – 3x + 2 (–2, 0) (1, 0) x 1 2 –1 –8 y 0 2 Figure 18 Use “trial and error” to find zeros. Check by multiplying these factors. Answer to Looking Ahead to Calculus: F′1x2 = 16x3 - 9x2 + 6
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