Question

Find the complex zeros of each polynomial function f(x). Write f in factored form.

$f\left(x\right)=2x^4+2x^3-11x^2+x-6$

Short answer

The factored form is$f\left(x\right)=(x+3)(x-2)(x+\frac{\sqrt{2}}{2}i)(x-\frac{\sqrt{2}}{2}i)$

Step-by-step solution

Step 1. Given Information

The given function is$f\left(x\right)=2x^4+2x^3-11x^2+x-6$

Step 2. Explanation 

The given function has a degree 4. So, there will be a maximum of 4 complex zeroes.

We will list the integers which are factors of the constant term $-6$.

$p=\pm 1,\pm 2,\pm 3,\pm 6$

Now, we will list the integers which are factors of the leading coefficient 2.

$q=\pm 1,\pm 2$

Next, list the possible potential rational zeroes which is the ratio $\frac{p}{q}$

$$\begin{gathered} \frac{p}{q}=\pm \frac{6}{2},\pm \frac{3}{2},\pm \frac{2}{2},\pm \frac{1}{2},\pm \frac{6}{1},\pm \frac{3}{1},\pm \frac{2}{1},\pm \frac{1}{1} \\ =\pm 6,\pm 2,\pm 3,\pm 1,\pm \frac{3}{2},\pm \frac{1}{2} \end{gathered}$$

Step 3. Graph

Graph the function.

From the above graph, the function intersects x-axis at two distinct points which are $x=-3andx=2$

Thus, the two factors are$(x+3),(x-2)$

Step 4. Calculation   

Use the synthetic division and divide the polynomial by $(x+3)$

Thus, the depressed equation is $2x^3-4x^2+x-2=0$

Now, factorize the above equation by grouping.

$$\begin{gathered} 2x^3-4x^2+x-2=0 \\ 2x^2(x-2)+1(x-2)=0 \\ (2x^2+1)(x-2)=0 \\ x-2=0\Rightarrow x=2 \\ 2x^2+1\Rightarrow x=\sqrt{-\frac{1}{2}} \\ \Rightarrow x=\pm \frac{\sqrt{2}}{2}i \end{gathered}$$