Section 2: Zeros of Polynomials

Elementary and Intermediate Algebra: Graphs & Models (3rd Edition)

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Section 6.2 Zeros of Polynomials 565 Version: Fall 2007 6.2 Zeros of Polynomials In the previous section we studied the end-behavior of polynomials. We know that a polynomial’s end-behavior is identical to the end-behavior of its leading term. Our focus was concentrated on the far right- and left-ends of the graph and not upon what happens in-between. In this section, our focus shifts to the interior. There are two important areas of concentration: the local maxima and minima of the polynomial, and the location of the x -intercepts or zeros of the polynomial. In this section we concentrate on finding the zeros of the polynomial. Zeros Let’s begin with a formal definition of the zeros of a polynomial. Definition 1. Let p ( x ) = a 0 + a 1 x + a 2 x 2 + ··· + a n x n be a polynomial with real coefficients. We say that a is a zero of the polynomial if and only if p ( a ) = 0 . The definition also holds if the coefficients are complex, but that’s a topic for a more advanced course. For example, 5 is a zero of the polynomial p ( x ) = x 2 + 3 x 10 because p ( 5) = ( 5) 2 + 3( 5) 10 = 25 15 10 = 0 . Similarly, 1 is a zero of the polynomial p ( x ) = x 3 + 3 x 2 x 3 because p ( 1) = ( 1) 3 + 3( 1) 2 ( 1) 3 = 1 + 3 + 1 3 = 0 . Let’s look at a more extensive example. Example 2. Find the zeros of the polynomial defined by p ( x ) = ( x + 3)( x 2)( x 5) . (3) At first glance, the function does not appear to have the form of a polynomial. However, two applications of the distributive property provide the product of the last two factors. p ( x ) = ( x + 3)( x ( x 5) 2( x 5)) = ( x + 3)( x 2 5 x 2 x + 10) = ( x + 3)( x 2 7 x + 10) Copyrighted material. See: http://msenux.redwoods.edu/IntAlgText/ 1
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566 Chapter 6 Polynomial Functions Version: Fall 2007 A third and fourth application of the distributive property reveals the nature of our function. p ( x ) = x ( x 2 7 x + 10) + 3( x 2 7 x + 10) = x 3 7 x 2 + 10 x + 3 x 2 21 x + 30 = x 3 4 x 2 11 x + 30 (4) Hence, p is clearly a polynomial. However, the original factored form provides quicker access to the zeros of this polynomial. Using Definition 1 , we need to find values of x that make p ( x ) = 0. That is, we need to solve the equation p ( x ) = 0 . Of course, p ( x ) = ( x + 3)( x 2)( x 5), so, equivalently, we need to solve the equation ( x + 3)( x 2)( x 5) = 0 . By the zero product property , either x + 3 = 0 or x 2 = 0 or x 5 = 0 . These are linear (first degree) equations, each of which can be solved independently. Thus, either x = 3 or x = 2 or x = 5 . Hence, the zeros of the polynomial
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Section 2: Zeros of Polynomials - Section 6.2 Zeros of...

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