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Stitz-Zeager_College_Algebra_e-book

# a1 xa0 is a polynomial a0 a1 of degree n with n 1 let

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Unformatted text preview: exists? The answer comes from our old friend, polynomial division. Dividing x3 + 4x2 − 5x − 14 by x − 2 gives x2 + 6x + 7 x− 2 x3 + 4x2 − 5x − 14 − x3 − 2x2 6x2 − 5x − 6x2 − 12x) 7x − 14 − (7x − 14) 0 As you may recall, this means x3 + 4x2 − 5x − 14 = (x − 2) x2 + 6x + 7 , and so to ﬁnd the zeros of f , we now can solve (x − 2) x2 + 6x + 7 = 0. We get x − 2 = 0 (which gives us our known zero, x = 2) as well as x2 + 6x + 7 = 0. The latter doesn’t factor nicely, so we apply the Quadratic √ Formula to get x = −3 ± 2. The point of this section is to generalize the technique applied here. First up is a friendly reminder of what we can expect when we divide polynomials. Theorem 3.4. Polynomial Division: Suppose d(x) and p(x) are nonzero polynomials where the degree of p is greater than or equal to the degree of d. There exist two unique polynomials, q (x) and r(x), such that p(x) = d(x) q (x) + r(x), where either r(x) = 0 or the degree of r is strictly less than the degree of d. 1 2 and probably forgot pun intended 198 Polynomial Functions As you may recall, a...
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