Question

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

$f\left(x\right)=3x^4-x^3-9x^2+159x-52$

Short answer

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

Complex zeros are$2\pm 3i.$

Step-by-step solution

Step 1. Given Information

The given polynomial is$f\left(x\right)=3x^4-x^3-9x^2+159x-52$

Since $a_0=-52$and $a_n=3$.

The factors of $-52are\pm 1,\pm 2,\pm 3,\pm 4,\pm 13,\pm 26,\pm 52$and factors of $3$are $\pm 1,\pm 3$

Therefore, potential roots of $f\left(x\right)$are$\pm 1,\pm \frac{1}{3},\pm 2,\pm \frac{2}{3},\pm 4,\pm \frac{4}{3},\pm 13,\pm \frac{13}{3},\pm 26,\pm \frac{26}{3},\pm 52,\pm \frac{52}{3}$

Step 2. Check for zeros

First, we will use synthetic division to determine the zero of $f\left(x\right),$let us check at $\frac{1}{3}$. We see that

. $f\left(\frac{1}{3}\right)=0$That is $\frac{1}{3}$is a zero of polynomial.

$f\left(x\right)=\left(x-\frac{1}{3}\right)\left(3x^3-9x+156\right)$

Step 3. Factor 3x3-9x+156

$$\begin{gathered} 3x^3-9x+156=0 \\ 3\left(x+4\right)\left(x^2-4x+13\right)=0 \end{gathered}$$

where $x^2-4x+13$is not factorable.

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

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