Chapter 6.7, Problem 42PPS

To determine

To find: The zeros of the function $f\left(x\right)=9x^4+9x^3+4x^2+4x$ .

Answer to Problem 42PPS

The zeros are 0, $-1$ , $\frac{2}{3}i$ and $-\frac{2}{3}i$ .

Explanation of Solution

Given:

The function is $f\left(x\right)=9x^4+9x^3+4x^2+4x$ .

Calculation:

Take out the Greatest common factor $x$ as follows.

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

Set $x$ equal to 0 to get $x=0$ .

Group the remaining terms as follows.

  $\left(9x^3+9x^2\right)+\left(4x+4\right)$

Find the GCF, which is $\left(x+1\right)$ in this case.

  $9x^2\left(x+1\right)+4\left(x+1\right)$

Take out the GCF and group the remaining term.

  $\left(x+1\right)\left(9x^2+4\right)$

Set the expression equal to zero.

  $\left(x+1\right)\left(9x^2+4\right)=0$

Solve for $x$ .

  $\begin{array}{c}x+1=0\\ x=-1\\ 9x^2+4=0\\ x=\pm \frac{2}{3}i\end{array}$

Therefore, the zeros are 0, $-1$ , $\frac{2}{3}i$ and $-\frac{2}{3}i$ .