Chapter 2.5, Problem 43E

a.

To determine

Find the zeros of the polynomial function $f(x)=x^4+25x^2+144$ .

Answer to Problem 43E

  $-3i,3i,-4i,4i$

Explanation of Solution

Given:

Function: $f(x)=x^4+25x^2+144$

Calculation:

  $\begin{array}{l}\Rightarrow f(x)=x^4+25x^2+144\\ \Rightarrow f(x)=x^4+9x^2+16x^2+144\\ \Rightarrow f(x)=x^2(x^2+9)+16(x^2+9)\\ \Rightarrow f(x)=(x^2+9)(x^2+16)\end{array}$

To find the roots of given function, put $f(x)=0,$

  $\begin{array}{l}\Rightarrow (x^2+9)(x^2+16)=0\\ \\ \begin{array}{cc}\begin{array}{l}\Rightarrow x^2+9=0\\ \Rightarrow x^2=-9\\ \Rightarrow x=\pm \sqrt{-9}\\ \Rightarrow x=\pm 3i\\ \Rightarrow x=3i,-3i\end{array}& \begin{array}{l}\Rightarrow x^2+16=0\\ \Rightarrow x^2=-16\\ \Rightarrow x=\pm \sqrt{-16}\\ \Rightarrow x=\pm 4i\\ \Rightarrow x=4i,-4i\end{array}\end{array}\\ \\ \Rightarrow x=-3i,3i,-4i,4i\end{array}$

Conclusion:

Therefore, the zeros of given polynomial function are $x=-3i,3i,-4i\text{and}4i.$

b.

To determine

Write the zeros of polynomial $f(x)=x^4+25x^2+144$ as a product of linear factors.

Answer to Problem 43E

  $f(x)=\left(x+3i\right)\left(x-3i\right)\left(x+4i\right)\left(x-4i\right)$

Explanation of Solution

Given:

Function: $f(x)=x^4+25x^2+144$

Calculation:

From above answer, the zeros of given function are $x=-3i,3i,-4i\text{and}4i.$

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

  $\Rightarrow f(x)=\left(x+3i\right)\left(x-3i\right)\left(x+4i\right)\left(x-4i\right)$

c.

To determine

Determine the x-intercepts of polynomial $f(x)=x^4+25x^2+144$ using factorization and verify whether the real zeros are the only x-intercepts.

Answer to Problem 43E

0

Explanation of Solution

Given:

Function: $f(x)=x^4+25x^2+144$

Calculation:

Factorization of given function is $f(x)=\left(x+3i\right)\left(x-3i\right)\left(x+4i\right)\left(x-4i\right).$

To find x-intercepts, put $f(x)=0.$

  $\begin{array}{l}\Rightarrow \left(x+3i\right)\left(x-3i\right)\left(x+4i\right)\left(x-4i\right)=0\\ \\ \begin{array}{cccc}\begin{array}{l}\Rightarrow x+3i=0\\ \Rightarrow x=-3i\end{array}& \begin{array}{l}\Rightarrow x-3i=0\\ \Rightarrow x=3i\end{array}& \begin{array}{l}\Rightarrow x+4i=0\\ \Rightarrow x=-4i\end{array}& \begin{array}{l}\Rightarrow x-4i=0\\ \Rightarrow x=4i\end{array}\end{array}\\ \\ \Rightarrow x=-3i,3i,-4i,4i\end{array}$

Hence, the x-intercepts are $x=-3i,3i,-4i\text{and}4i.$

Calculation for graph:

Consider $f(x)=x^4+25x^2+144$

Values of x	Values of f (x)
0	144
1	170
-1	170
2	260
-2	260

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

Graph:

Interpretation:

From graph, it is clear that, the number of x-intercepts is equal to 0.

The number of real zeros of the function is equal to 0.

So, the number of real zeros and number of x-intercepts are equal to each other.