Question

find the zeros of the given function
$f(x)=x^{4}-4x^{3}-x^{2}+14x + 10$

the zeros are
(simplify your answer use a comma to separate answers as needed. use integers or fractions for any numbers in the expressi

Explanation:

Step1: List possible rational zeros

By Rational Root Theorem, possible zeros are $\pm1, \pm2, \pm5, \pm10$.

Step2: Test $x=-1$

$f(-1)=(-1)^4 -4(-1)^3 -(-1)^2 +14(-1)+10=1+4-1-14+10=0$. So $x=-1$ is a zero, and $(x+1)$ is a factor.

Step3: Polynomial division

Divide $x^4 -4x^3 -x^2 +14x+10$ by $(x+1)$:
$$\frac{x^4 -4x^3 -x^2 +14x+10}{x+1}=x^3-5x^2+4x+10$$

Step4: Test $x=-1$ on cubic

$f(-1)$ for cubic: $(-1)^3-5(-1)^2+4(-1)+10=-1-5-4+10=0$. So $x=-1$ is a repeated zero, $(x+1)$ is another factor.

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

$$\frac{x^3-5x^2+4x+10}{x+1}=x^2-6x+10$$

Step6: Solve quadratic equation

Use quadratic formula for $x^2-6x+10=0$:
$x=\frac{6\pm\sqrt{(-6)^2-4(1)(10)}}{2(1)}=\frac{6\pm\sqrt{36-40}}{2}=\frac{6\pm\sqrt{-4}}{2}=3\pm i$

Answer:

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