Chapter 4.6, Problem 11E

To determine

To find: all zeros of the polynomial function.

Answer to Problem 11E

The zeros are $-2,-2,1$ and3.

Explanation of Solution

Given information:

Given function,

  $g\left(x\right)=x^4-9x^2-4x+12$

Calculation:

Consider the equation $g\left(x\right)=x^4-9x^2-4x+12$ :

Because $g\left(x\right)$ is a polynomial function of degree 4, so it has 4 zeros.

The possible rational zeros are $\pm 1,\pm 2,\pm 3,\pm 4,\pm 6\text{and}\pm 12.$

Choose reasonable values from the list above to test using the graph of the function.

  

From the above graph the values are $x=-2,x=1\text{and}x=3$ .

Test out the polynomial using synthetic division until a zero is found.

  $\begin{array}{cccccc}1& 1& 0& -9& -4& 12\\ & & 1& 1& -8& -12\\ & 1& 1& -8& -12& 0\end{array}$

Factor out the polynomial using the result of above synthetic division.

  $\begin{array}{c}g\left(x\right)=x^4-9x^2-4x+12\\ =\left(x-1\right)\left(x^3+x^2-8x-12\right)\end{array}$

Repeat the steps above for $f\left(x\right)=x^3+x^2-8x-12$ .

The possible rational are $\pm 1,\pm 2,\pm 3,\pm 4,\pm 6\text{and}\pm 12.$ Test the value using synthetic division until a zero is found.

  $\begin{array}{ccccc}3& 1& 1& -8& -12\\ & & 3& 12& 12\\ & 1& 4& 4& 0\end{array}$

Factor out the polynomial using the result of above synthetic division.

  $\begin{array}{c}f\left(x\right)=\left(x-1\right)\left(x^3+x^2-8x-12\right)\\ =\left(x-1\right)\left(x-3\right)\left(x^2+4x+4\right)\\ =\left(x-1\right)\left(x-3\right){\left(x+2\right)}^2\end{array}$

Solve $f\left(x\right)=0$ .

  $\begin{array}{c}\left(x-1\right)\left(x-3\right){\left(x+2\right)}^2=0\\ x=-2,-2,1,3\end{array}$

Therefore, the zeros are $-2,-2,1$ and3.