Precalculus: Mathematics for Calculus, 7th Edition

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

Chapter 3 - Section 3.4 - Real Zeros of Polynomials - 3.4 Exercises - Page 286: 107

Answer

a. See proof b. $X^3+17X-\frac{34}{3}$

Work Step by Step

a. Replace x with $X-a/3$ in the equation, we have $x^3+ax^2+bx+c=(X-\frac{a}{3})^3+a(X-\frac{a}{3})^2+b(X-\frac{a}{3})+c =X^3-3\frac{a}{3}X^2+3\frac{a^2}{9}X-\frac{a^3}{27}+aX^2-2\frac{a}{3}X+a\frac{a^2}{9}+bX-\frac{ab}{3}+c =X^3+(\frac{a^2}{3}-\frac{2a}{3}+b)X+(\frac{2a^3}{27}-\frac{ab}{3}+c)$ which shows that the $x^2$ term disappeared. b. Following the results from (a) we have $a=6,b=9,c=4$ and $x^3+6x^2+9x+4=X^3+(6^2/3-2\times6/3+9)X+(2\times6^2/27-6\times9/3+4) =X^3+17X-\frac{34}{3}$
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