348 CHAPTER 3 Polynomial and Rational Functions (b) 1)3 -2 1 -8 5 1 ƒ1x2 = 3x5 - 2x4 + x3 - 8x2 + 5x + 1 3 1 2 -6 -1 3 1 2 -6 -1 0 ƒ112 = 0 Because the remainder is 0, x - 1 is a factor. Additionally, we can determine from the coefficients in the bottom row that the other factor is 3x4 + 1x3 + 2x2 - 6x - 1. Thus, we can express the polynomial in factored form. ƒ1x2 = 1x - 1213x4 + x3 + 2x2 - 6x - 12 S Now Try Exercises 9 and 11. We can use the factor theorem to factor a polynomial of greater degree into linear factors of the form ax - b. EXAMPLE 2 Factoring a Polynomial Given a Zero Factor ƒ1x2 = 6x3 + 19x2 + 2x - 3 into linear factors given that -3 is a zero. SOLUTION Because -3 is a zero of ƒ, x - 1-32 = x + 3 is a factor. -3)6 19 2 -3 Use synthetic division to divide ƒ1x2 by x + 3. -18 -3 3 6 1 -1 0 The quotient is 6x2 + x - 1, which is the factor that accompanies x + 3. ƒ1x2 = 1x + 3216x2 + x - 12 ƒ1x2 = 1x + 3212x + 1213x - 12 Factor 6x2 + x - 1. (++++++)++++++* These factors are all linear. S Now Try Exercise 21. Rational Zeros Theorem The rational zeros theorem gives a method to determine all possible candidates for rational zeros of a polynomial function with integer coefficients. Rational ZerosTheorem If p q is a rational number written in lowest terms, and if p q is a zero of ƒ, a polynomial function with integer coefficients, then p is a factor of the constant term and q is a factor of the leading coefficient. Proof ƒ A p qB = 0 because p q is a zero of ƒ1x2. ana p qb n + an-1a p qb n-1 + g+ a1a p qb + a0 = 0 Definition of zero of ƒ ana pn qnb + an-1a pn-1 qn-1b + g+ a1a p qb + a0 = 0 Power rule for exponents an pn + a n-1 pn-1q + g+ a 1 pqn-1 = -a 0 qn Multiply by qn. Subtract a 0 qn. p1an pn-1 + a n-1 pn-2q + g+ a 1qn-12 = -a 0 qn Factor out p. LOOKING AHEAD TO CALCULUS Finding the derivative of a polynomial function is one of the basic skills required in a first calculus course. For the functions ƒ1x2 = x4 - x2 + 5x - 4, g1x2 = -x6 + x2 - 3x + 4, and h1x2 = 3x3 - x2 + 2x - 4, the derivatives are ƒ′1x2 = 4x3 - 2x + 5, g′1x2 = -6x5 + 2x - 3, and h′1x2 = 9x2 - 2x + 2. Notice the use of the “prime” notation. For example, the derivative of ƒ1x2 is denoted ƒ′1x2. Look for the pattern among the exponents and the coefficients. Using this pattern, what is the derivative of F1x2 = 4x4 - 3x3 + 6x - 4? The answer is at the top of the next page.
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