Chapter 3 - 3.4 - Zeroes of Polynomial Functions - 3.4 Exercises - Page 284: 43

$f(x)=x^4-6x^3+14x^2-16x+8$

Work Step by Step

If $1+i$ is a zero of $f$ then the complex conjugate $1-i$ is also a zero. $f(x)=a(x-2)(x-2)[x-(1+i)][x-(1-i)]$ $f(x)=a(x-2)^2[(x-1)-i][(x-1)+i]$ $f(x)=a(x^2-4x+4)[(x-1)^2-i^2]$ $f(x)=a(x^2-4x+4)(x^2-2x+1+1)$ $f(x)=a(x^2-4x+4)(x^2-2x+2)$ $f(x)=a(x^4-2x^3+2x^2-4x^3+8x^2-8x+4x^2-8x+8)$ We can choose any value for $a$. So, let's make $a=1$ $f(x)=x^4-6x^3+14x^2-16x+8$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.