Stitz-Zeager_College_Algebra_e-book

1 and concern ourselves only with the nonzero

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Unformatted text preview: 5 3 ↓2 2 −3 2 2 −3 0 We get a remainder of 0 which verifies that, indeed, p(1) = 0. Our quotient polynomial is a second degree polynomial with coefficients 2, 2, and −3. So q (x) = 2x2 + 2x − 3. Theorem 3.4 tells us p(x) = (x − 1) 2x2 + 2x − 3 . To find the remaining real zeros of p, we need to solve 2x2 + 2x − 3 = 0 for x. Since this doesn’t factor nicely, we use the quadratic formula to √ find that the remaining zeros are x = −1± 7 . 2 In Section 3.1, we discussed the notion of the multiplicity of a zero. Roughly speaking, a zero with multiplicity 2 can be divided twice into a polynomial; multiplicity 3, three times and so on. This is illustrated in the next example. Example 3.2.3. Let p(x) = 4x4 − 4x3 − 11x2 + 12x − 3. Given that x = 1 is a zero of multiplicity 2 2, find all of the real zeros of p. Solution. We set up for synthetic division. Since we are told the multiplicity of 1 is two, we 2 continue our tableau and divide 1 into the quotient polynomial 2 1 2 1 2 4 −4 −11 12 −3 ↓ 2 −1 −6 3 4 −2 −12 60 ↓ 2 0 −6 4 0 −1...
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This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

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