Question

find the zeros of the given function.\\( f(x) = x^4 - 2x^3 + 13x^2 + 36x + 20 \\)\\( \\)\\( \\)the zeros are \\( \square \\)\\( \\) (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: List possible rational zeros

Using Rational Root Theorem, possible rational zeros are $\pm1, \pm2, \pm4, \pm5, \pm10, \pm20$.

Step2: Test $x=-1$

Substitute $x=-1$ into $f(x)$:
$$f(-1)=(-1)^4 - 2(-1)^3 +13(-1)^2 +36(-1)+20=1+2+13-36+20=0$$
So $(x+1)$ is a factor.

Step3: Polynomial division

Divide $x^4 -2x^3 +13x^2 +36x+20$ by $(x+1)$:
$$\frac{x^4 -2x^3 +13x^2 +36x+20}{x+1}=x^3-3x^2+16x+20$$

Step4: Test $x=-1$ again on cubic

Substitute $x=-1$ into $x^3-3x^2+16x+20$:
$$(-1)^3-3(-1)^2+16(-1)+20=-1-3-16+20=0$$
So $(x+1)$ is a factor again.

Step5: Divide cubic by $(x+1)$

$$\frac{x^3-3x^2+16x+20}{x+1}=x^2-4x+20$$

Step6: Solve quadratic equation

Set $x^2-4x+20=0$, use quadratic formula $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ where $a=1,b=-4,c=20$:
$$x=\frac{4\pm\sqrt{(-4)^2-4(1)(20)}}{2(1)}=\frac{4\pm\sqrt{16-80}}{2}=\frac{4\pm\sqrt{-64}}{2}=\frac{4\pm8i}{2}=2\pm4i$$

Answer:

$-1, -1, 2+4i, 2-4i$