Question

find the zeros of the given function.
$f(x) = x^4 - 4x^3 + 2x^2 + 20x + 13$
the zeros are
(simplify your answer. use a comma to separate answers as needed. use integers or fractions for any numbers in the expression. express complex numbers in terms of $i$)

Explanation:

Step1: Assume quadratic form

Let $y = x^2$, rewrite the quartic as a quadratic in $y$:
$y^2 - 4y^2 + 2y + 20x + 13$ is incorrect; instead, use rational root theorem first. Test possible rational roots $\pm1,\pm13$:
$f(1)=1-4+2+20+13=32
eq0$, $f(-1)=1+4+2-20+13=0$.

Step2: Factor out $(x+1)$

Use polynomial division or synthetic division:
Divide $x^4-4x^3+2x^2+20x+13$ by $(x+1)$:

$$\begin{align*} &x^4-4x^3+2x^2+20x+13\\ =&(x+1)(x^3-5x^2+7x+13) \end{align*}$$

Step3: Factor the cubic

Test $x=-1$ again for the cubic: $(-1)^3-5(-1)^2+7(-1)+13=-1-5-7+13=0$. Factor out $(x+1)$:

$$\begin{align*} &x^3-5x^2+7x+13\\ =&(x+1)(x^2-6x+13) \end{align*}$$

Step4: Factor the quadratic

Solve $x^2-6x+13=0$ using quadratic formula $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$, where $a=1,b=-6,c=13$:

$$ x=\frac{6\pm\sqrt{36-52}}{2}=\frac{6\pm\sqrt{-16}}{2}=\frac{6\pm4i}{2}=3\pm2i $$

Answer:

$-1, -1, 3+2i, 3-2i$