Chapter 4.6, Problem 13E

To determine

To find: all zeros of the polynomial function.

Answer to Problem 13E

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

Explanation of Solution

Given information:

Given function,

  $g\left(x\right)=x^4+4x^3+7x^2+16x+12$

Calculation:

Consider the equation $g\left(x\right)=x^4+4x^3+7x^2+16x+12$ :

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 $x=-3\text{and}x=-1$ .

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

  $\begin{array}{cccccc}-1& 1& 4& 7& 16& 12\\ & & -1& -3& -4& -12\\ & 1& 3& 4& 12& 0\end{array}$

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

  $\begin{array}{c}g\left(x\right)=x^4+4x^3+7x^2+16x+12\\ =\left(x+1\right)\left(x^3+3x^2+4x+12\right)\end{array}$

Repeat the steps above for $f\left(x\right)=x^3+3x^2+4x+12$ .

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.

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

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

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

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

  $\begin{array}{c}\left(x+1\right)\left(x+3\right)\left(x^2+4\right)=0\\ x+1=0orx+3=0or{x}^2+4=0\\ x=-1orx=-3or{x}^2+4=0\end{array}$

Now solve the equation $x^2+4=0$

  $\begin{array}{c}{x}^2=-4\\ x=\pm \sqrt{-4}\\ x=\pm 2i\end{array}$

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