Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 3 - 3.4 - Zeroes of Polynomial Functions - 3.4 Exercises - Page 284: 44



Work Step by Step

If $3-2i$ is a zero of $f$ then the complex conjugate $3+2i$ is also a zero. $f(x)=a[x-(-1)](x-5)[x-(3-2i)][x-(3+2i)]$ $f(x)=a(x+1)(x-5)[(x-3)+2i][(x-3)-2i]$ $f(x)=a(x^2-5x+x-5)[(x-3)^2-(2i)^2]$ $f(x)=a(x^2-4x-5)(x^2-6x+9+4)$ $f(x)=a(x^2-4x-5)(x^2-6x+13)$ $f(x)=a(x^4-6x^3+13x^2-4x^3+24x^2-52x-5x^2+30x-65)$ We can choose any value for $a$. So, let's make $a=1$ $f(x)=x^4-10x^3+32x^2-22x-65$
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