Chapter 4.6, Problem 14E

To determine

To find: all zeros of the polynomial function.

Answer to Problem 14E

The zeros are $-1,2$ and $-3i,3i$ .

Explanation of Solution

Given information:

Given function,

  $h\left(x\right)=x^4-x^3+7x^2-9x-18$

Calculation:

Consider the equation $h\left(x\right)=x^4-x^3+7x^2-9x-18$ :

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

The possible rational zeros are $\pm 1,\pm 2,\pm 3,\pm 6,\pm 9\text{and}\pm 18.$

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

  

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

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

  $\begin{array}{cccccc}-1& 1& -1& 7& -9& -18\\ & & -1& 2& -9& 18\\ & 1& -2& 9& -18& 0\end{array}$

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

  $\begin{array}{c}h\left(x\right)=x^4-x^3+7x^2-9x-18\\ =\left(x+1\right)\left(x^3-2x^2+9x-18\right)\end{array}$

Repeat the steps above for $f\left(x\right)=x^3-2x^2+9x-18$ .

The possible rational are $\pm 1,\pm 2,\pm 3,\pm 6,\pm 9\text{and}\pm 18.$ Test the value using synthetic division until a zero is found.

  $\begin{array}{ccccc}2& 1& -2& 9& -18\\ & & 2& 0& 18\\ & 1& 0& 9& 0\end{array}$

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

  $\begin{array}{c}h\left(x\right)=\left(x+1\right)\left(x^3-2x^2+9x-18\right)\\ =\left(x+1\right)\left(x-2\right)\left(x^2+9\right)\end{array}$

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

  $\begin{array}{c}\left(x+1\right)\left(x-2\right)\left(x^2+9\right)=0\\ x+1=0orx-2=0or{x}^2+9=0\\ x=-1orx=3or{x}^2+9=0\end{array}$

Now solve the equation $x^2+9=0$

  $\begin{array}{c}{x}^2=-9\\ x=\pm \sqrt{-9}\\ x=\pm 3i\end{array}$

Therefore, the zeros are $-1,2$ and $-3i,3i$ .