Example: Given that $$f_0(x)=x^4+2x^3-9x^2-10x+50$$ has a root (aka zero) $$x=-3-i$$ Find all zeros (aka roots) of \(f_0(x)\) and factor \(f_0(x)\) completely.
Solution:
- By the Factor Theorem, \( (x-[-3-i])\) is a factor of \(f_0(x)\).
- All coefficients of \(f_0(x)\) are real, so other factor of \(f_0(x)\) is \((x-[-3+i])\), that is the roots are conjugate pairs.
- Now we have two factors of \(f_0(x)\), and they are
\begin{align*}
(x-[-3-i])(x-[-3+i]) &=([x+3]+i)([x+3]-i) \\
&=[x+3]^2-i^2\\
&=x^2+6x+9+1 \\
&=x^2+6x+10 \,\,\,\,\,\, \text{ is a factor of } f_0(x)
\end{align*}
- Use long division to find the other factor(s)
$$
\require{enclose}
\begin{array}{r}
x^2-4x+5\hspace{31mm} \\[-3pt]
x^2+6x+10 \enclose{longdiv}{x^4+2x^3-\,\,9x^2-10x+50}\kern-.2ex \\[-3pt]
\underline{\,x^4+6x^3+10x^2\phantom{-10x+50}} \\[-3pt]
\phantom{x^4}-4x^3-19x^2-10x+50 \\[-3pt]
\underline{\phantom{x^4}-4x^3-24x^2-40x\phantom{.+50}} \\[-3pt]
\phantom{x^4-4x^3-2}5x^2+30x+50 \\[-3pt]
\underline{\phantom{x^4-4x^3-2}5x^2+30x+50} \\[-3pt]
\phantom{00}0
\end{array}
$$
Hence $$f_0(x)=x^4+2x^3-\,\,9x^2-10x+50 = (x^2+6x+10)(x^2-4x+5)$$
- Find the zeros of the \((x^2-4x+5)\) factor using quadratic formula
\[
x_{1,2}= \frac{4\pm\sqrt{16-4(5)}}{2}=\frac{4\pm\sqrt{-4}}{2}=\frac{4\pm2i}{2}=2\pm i
\]
Now we know all four zeros of \(f_0(x)\), and they are \(\{-3-i, -3+i, 2-i, 2+i\}\)
- Finally, we declare victory by writing complete factorization of \(f_0(x)\) as
\[
f_0(x)=\underbrace{\left(x-[-3-i]\right)\left(x-[-3+i]\right)}_{x^2+6x+10}\underbrace{\left(x-[2-i]\right)\left(x-[2+i]\right)}_{x^2-4x+5}
\]