Chapter 2.5, Problem 38E

To determine

Find the zeros of the function $h(x)=x^4+6x^3+10x^2+6x+9$ and write the polynomial as a product of linear factors and verify zeros using graphical method.

Answer to Problem 38E

  $-3,-3,i,-i.$

Explanation of Solution

Given:

Function: $h(x)=x^4+6x^3+10x^2+6x+9$

Calculation:

The leading coefficient of given polynomial is 1 and the constant term is 9.

So, possible rational zeros are: $\frac{\text{Factors of}9}{\text{Factors of 1}}=\frac{\pm 1,\pm 3,\pm 9}{\pm 1}=\pm 1,\pm 3,\pm 9$

Using synthetic division to check whether $x=-3$ , is a zero or not.

  $\begin{array}{cccccc}-3& 1& 6& 10& 6& 9\\ & & -3& -9& -3& -9\\ & 1& 3& 1& 3& \boxed{0}\end{array}\begin{array}{c}\\ \\ \hspace{1em}\leftarrow \text{Remainder}\end{array}$

Here, remainder is 0. So, $x=-3$ is a zero of the given polynomial.

So, $h(x)$ factors as

  $h(x)=\left(x+3\right)(x^3+3x^2+x+3)$

Now, consider $x^3+3x^2+x+3.$

The leading coefficient of given polynomial is 1 and the constant term is 3.

So, possible rational zeros are: $\frac{\text{Factors of}3}{\text{Factors of 1}}=\frac{\pm 1,\pm 3}{\pm 1}=\pm 1,\pm 3$

Using synthetic division to check whether $x=-3$ , is a zero or not.

  $\begin{array}{ccccc}-3& 1& 3& 1& 3\\ & & -3& 0& -3\\ & 1& 0& 1& \boxed{0}\end{array}\begin{array}{c}\\ \\ \hspace{1em}\leftarrow \text{Remainder}\end{array}$

Here, remainder is 0. So, $x=-3$ is a zero of the given polynomial.

So, $h(x)$ factors as

  $h(x)=\left(x+3\right)\left(x+3\right)(x^2+1)$

Now, consider $x^2+1.$

The above polynomial is a second degree polynomial.

So, zeroes of a second degree polynomial can be found using the formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}.$

  $\begin{array}{l}\Rightarrow x=\frac{-0\pm \sqrt{{(0)}^2-4(1)(1)}}{2(1)}\\ \Rightarrow x=\frac{0\pm \sqrt{0-4}}{2}\\ \Rightarrow x=\frac{\pm \sqrt{-4}}{2}\\ \Rightarrow x=\frac{\pm 2\sqrt{-1}}{2}\\ \Rightarrow x=\pm i\\ \Rightarrow x=i,-i\end{array}$

Now, writing the given function as product of linear factors,

  $\Rightarrow h(x)=\left(x+3\right)\left(x+3\right)\left(x-i\right)\left(x+i\right)$

Calculation for graph:

Consider $h(x)=x^4+6x^3+10x^2+6x+9$

Values of x	Values of g(x)
0	9
1	32
-1	8
2	125
-2	5

By taking different values of x, the graph can be plotted.

Graph:

  

Interpretation:

Only real zeros of the function can be verified through graph.

By observing graph, it is clear that the curve of the function meets x-axis at $x=-3.$

Hence, the real zeros of the function are $x=-3.$

Conclusion:

Therefore, the zeros of given function are $x=-3,-3,i\text{and}-i.$