Chapter 4.6, Problem 15E

To determine

To find: all zeros of the polynomial function.

Answer to Problem 15E

The zeros are $-1,2,4$ and $-\sqrt{2}i,\sqrt{2}i$ .

Explanation of Solution

Given information:

Given function,

  $g\left(x\right)=x^5+3x^4-4x^3-2x^2-12x-16$

Calculation:

Consider the equation $g\left(x\right)=x^5+3x^4-4x^3-2x^2-12x-16$ :

Because $g\left(x\right)$ is a polynomial function of degree 5, so it has 5 zeros.

The possible rational zeros are $\pm 1,\pm 2,\pm 4,\pm 8\text{and}\pm 16.$

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.

  $\begin{array}{ccccccc}-1& 1& 3& -4& -2& -12& -16\\ & & -1& -2& 6& -4& 16\\ & 1& 2& -6& 4& -16& 0\end{array}$

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

  $\begin{array}{c}g\left(x\right)=x^5+3x^4-4x^3-2x^2-12x-16\\ =\left(x+1\right)\left(x^4+2x^3-6x^2+4x-16\right)\end{array}$

Repeat the steps above for $f\left(x\right)=x^4+2x^3-6x^2+4x-16$ .

The possible rational are $\pm 1,\pm 2,\pm 4,\pm 8\text{and}\pm 16.$ Test the value using synthetic division until a zero is found.

  $\begin{array}{cccccc}2& 1& 2& -6& 4& -16\\ & & 2& 8& 4& 16\\ & 1& 4& 2& 8& 0\end{array}$

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

  $\begin{array}{c}g\left(x\right)=\left(x+1\right)\left(x^4+2x^3-6x^2+4x-16\right)\\ =\left(x+1\right)\left(x-2\right)\left(x^3+4x^2+2x+8\right)\end{array}$

Repeat the steps above for $h\left(x\right)=x^3+4x^2+2x+8$ .

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.

  $\begin{array}{ccccc}-4& 1& 4& 2& 8\\ & & -4& 0& -8\\ & 1& 0& 2& 0\end{array}$

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

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

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

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

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

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

Therefore, the zeros are $-1,2,4$ and $-\sqrt{2}i,\sqrt{2}i$ .