Chapter 4.6, Problem 9E

To determine

To find: all zeros of the polynomial function.

Answer to Problem 9E

The zeros are $-1,1,2$ and 4.

Explanation of Solution

Given information:

Given function,

  $f\left(x\right)=x^4-6x^3+7x^2+6x-8$

Calculation:

Consider the equation $f\left(x\right)=x^4-6x^3+7x^2+6x-8$

Step 1:

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

The possible rational zeros are $\pm 1,\pm 2,\pm 4\text{and}\pm 8.$

Step 2:

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

  

From the above graph the values are $x=-1,x=1,x=2\text{and}x=4$ .

Step 3:

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

  $\begin{array}{cccccc}-1& 1& -6& 7& 6& -8\\ & & -1& 7& -14& 8\\ & 1& -7& 14& -8& 0\end{array}$

Step 4:

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

  $\begin{array}{c}f\left(x\right)=x^4-6x^3+7x^2+6x-8\\ =\left(x+1\right)\left(x^3-7x^2+14x-8\right)\end{array}$

Repeat the steps above for $g\left(x\right)=x^3-7x^2+14x-8$ .

The possible rational are $\pm 1,\pm 2,\pm 4\text{and}\pm 8.$ Test the value using synthetic division until a zero is found.

  $\begin{array}{ccccc}1& 1& -7& 14& -8\\ & & 1& -6& 8\\ & 1& -6& 8& 0\end{array}$

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

  $\begin{array}{c}f\left(x\right)=\left(x+1\right)\left(x^3-7x^2+14x-8\right)\\ =\left(x+1\right)\left(x-1\right)\left(x^2-6x+8\right)\\ =\left(x+1\right)\left(x-1\right)\left(x-2\right)\left(x-4\right)\end{array}$

Step 6:

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

  $\begin{array}{c}\left(x+1\right)\left(x-1\right)\left(x-2\right)\left(x-4\right)=0\\ x=-1,1,2,4\end{array}$

Therefore, the zeros are $-1,1,2$ and 4.