Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 3 - Section 3.3 - Dividng Polynomials - 3.3 Exercises - Page 274: 69

Answer

$P(x)=3x^4-9x^2+6$

Work Step by Step

RECALL: If c is a zero of a polynomial, then $x-c$ is a factor of the polynomial. Since -1, 1, and $\sqrt2$ are zeros of the polynomial, and radical zeros always come in conjugate pairs, then (x+1),(x-1), $(x-\sqrt2)$, and $(x+\sqrt2)$ are factors of the polynomial. Thus, the polynomial of degree 4 with the given zeros is: $\\P(x)=a(x+1)(x-1)(x-\sqrt2)(x+\sqrt2) \\P(x)=a(x^2-1)(x^2-2) \\P(x)=a(x^4-3x^2+2)$ where $a$ is a real number. Since the constant term is 6, then $a$ must be: $a(2)=6 \\a=\frac{6}{2} \\a=3$ Thus, the polynomial with the given zeros whose constant term is 6 is: $\\P(x)=3(x^4-3x^2+2) \\P(x)=3x^4-9x^2+6$
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