## College Algebra (6th Edition)

$f(x)=x^{4}+10x^{2}+9$
The Linear Factorization Theorem... in words: An nth-degree polynomial can be expressed as the product of a nonzero constant and $n$ linear factors, where each linear factor has a leading coefficient of 1. Imaginary roots, if they exist, occur in conjugate pairs. ------------------------- If $(i)$ is a zero, so is $(-i).$ If $(3i)$ is a zero, so is $(-3i).$ $f(x)=a_{n}(x-i)(x+i)(x-3i)(x+3i)$ $f(x)=a_{n}(x^{2}-i^{2})(x^{2}-9i^{2})$ ...$i^{2}=-1$ ... $f(x)=a_{n}(x^{2}+1)(x^{2}+9)$ Using the given $f(-1),$we find $a_{n}:$ $f(-1)=a_{n}(1+1)(1+9)$ $20=a_{n}(20)$ $a_{n}=1$ $f(x)=1\cdot(x^{2}+1)(x^{2}+9)$ $f(x)=x^{4}+10x^{2}+9$