Question

find the zeros of the given function.
$f(x)=x^{4}-2x^{3}+6x^{2}+22x + 13$
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: Rational Root Theorem check

Possible rational roots: $\pm1, \pm13$. Test $x=-1$:
$(-1)^4 - 2(-1)^3 + 6(-1)^2 + 22(-1) +13 = 1+2+6-22+13=0$

Step2: Perform polynomial division

Divide $x^4 -2x^3 +6x^2 +22x +13$ by $(x+1)$:
$\frac{x^4 -2x^3 +6x^2 +22x +13}{x+1}=x^3 -3x^2 +9x +13$

Step3: Test root for cubic

Test $x=-1$ on cubic: $(-1)^3 -3(-1)^2 +9(-1)+13=-1-3-9+13=0$

Step4: Divide cubic by $(x+1)$

$\frac{x^3 -3x^2 +9x +13}{x+1}=x^2 -4x +13$

Step5: Solve quadratic equation

Use quadratic formula $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ for $x^2-4x+13=0$, where $a=1,b=-4,c=13$:
$x=\frac{4\pm\sqrt{(-4)^2-4(1)(13)}}{2(1)}=\frac{4\pm\sqrt{16-52}}{2}=\frac{4\pm\sqrt{-36}}{2}=\frac{4\pm6i}{2}=2\pm3i$

Answer:

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