Chapter 2, Problem 115RE

a.

To determine

To find the zeros of the function $f(x)=x^4+34x^2+225$ .

Answer to Problem 115RE

  $x=5i,-5i,3i,-3i.$

Explanation of Solution

Given:

Function: $f(x)=x^4+34x^2+225$

Calculation:

Let $X=x^2$

Then the given polynomial is rewritten as $X^2+34X+225$

Now solve the polynomial as a normal quadratic equation using the formula, $\begin{array}{l}x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ x=\frac{-(34)\pm \sqrt{{(34)}^2-4(1)(225)}}{2(1)}\\ x=\frac{-34\pm \sqrt{1156-900}}{2}\\ x=\frac{-34\pm \sqrt{256}}{2}\\ x=\frac{-34\pm \sqrt{256}}{2}\\ x=\frac{-34\pm 16}{2}\\ x=-25,-9\end{array}$

Therefore,

  $X^2+34X+225=(X+9)(X+25)$

Now substitute for $X=x^2$

Then, $x^4+34x^2+225=(x^2+9)(x^2+25)$

Now, $\begin{array}{l}(x^2+9)=(x+3i)(x-3i)\\ (x^2+25)=(x+5i)(x-5i)\end{array}$

Conclusion:

Therefore, the zeros of given polynomial function are $x=5i,-5i,3i,-3i.$

b.

To determine

To write the factors of the function $f(x)=x^4+34x^2+225$ as a product of linear factors

Answer to Problem 115RE

  $f(x)=(x-5i)(x+5i)(x-3i)(x+3i)$

Explanation of Solution

Given:

Function: $f(x)=x^4+34x^2+225$

Calculation:

From above answers, the polynomial as a product of linear factors is

  $f(x)=(x-5i)(x+5i)(x-3i)(x+3i)$ .

Conclusion:

Therefore, $f(x)=(x-5i)(x+5i)(x-3i)(x+3i)$ is complete factorization to given polynomial function.

c.

To determine

To find the x - intercepts of the function $f(x)=x^4+34x^2+225$ .

Answer to Problem 115RE

No intercepts since the roots are imaginary.

Explanation of Solution

Given:

Function: $f(x)=x^4+34x^2+225$

Calculation:

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

  $\begin{array}{c}\Rightarrow (x-5i)(x+5i)(x-3i)(x+3i)=0\\ \Rightarrow x=5i,-5i,3i,-3i.\end{array}$

Since the roots are imaginary, there are no real roots that can be plotted in the graph.

Therefore, no graph can be drawn.

Conclusion:

Therefore, no real x intercepts can be found.