Question

In Problems 31– 40, find the complex zeros of each polynomial function. Write f in factored form.

$f\left(x\right)=2x^4+x^3-35x^2-113x+65$

Short answer

$f\left(x\right)=(x-5)(2x-1)(x+3-2i)(x+3+2i)$

Complex zeros are$-3\pm 2i.$

Step-by-step solution

Step 1. Given Information 

The given polynomial is

$f\left(x\right)=2x^4+x^3-35x^2-113x+65$

Step 2. Finding potential roots

The factors of constant terms are $\pm 1,\pm 5,\pm 13,\pm 65$and factor of leading term are $\pm 1,\pm 2$.

The potential roots of$f\left(x\right)$are$\pm 1,\pm \frac{1}{2},\pm 5,\pm \frac{5}{2},\pm 13,\pm \frac{13}{2},\pm 15,\pm \frac{15}{2}$

Step 3. Finding factors by checking zeros

Let us check at $\frac{1}{2},$$f\left(\frac{1}{2}\right)=0$

So, $f\left(x\right)=\left(x-\frac{1}{2}\right)\left(2x^3+2x^2-34x-130\right)$

Now, factor

$$\begin{gathered} 2x^3+2x^2-34x-130=0 \\ 2\left(x-5\right)\left(x^2+6x+13\right)=0 \end{gathered}$$

Where$x^2+6x+13$can not factorable.

Step 4. Factor x2+6x+13 by quadratic formula

$$\begin{gathered} x=\frac{-6\pm \sqrt{36-52}}{2}=\frac{-6\pm \sqrt{-16}}{2} \\ x=\frac{-6\pm 4i}{2} \\ x=-3+2i,-3-2i \end{gathered}$$