Chapter 15.5, Problem 8MRE

To determine

To calculate: The roots of the equation $x^3+6x^2+16x+96=0$ .

Answer to Problem 8MRE

The roots of the equation $x^3+6x^2+16x+96=0$ are $-6\text{ and }\pm 4i$ .

Explanation of Solution

Given information:

One of the root of the equation is $4i$ .

Formula used:

In long division of polynomials $\text{Dividend}=\text{Divisor}\times \text{Quotient}+\text{Remainder}$ .

Calculation:

Consider theequation $x^3+6x^2+16x+96=0$ . One of the root of the equation is $4i$ . Then $x=-4i$ will also be the root of the equation, as complex number exist in pairs.

If $x=a+ib$ is a root of an equation then $x=a-ib$ is also a root of the equation.

Therefore, $\left(x+4i\right)\left(x-4i\right)$ is a root of the equation $x^3+6x^2+16x+96=0$ .

So, $x^2+16$ is a root of the equation $x^3+6x^2+16x+96=0$ .

Use long division to divide $x^3+6x^2+16x+96=0$ by its factor $x^2+16$ .

  $x^2+16\begin{array}{c}\hfill x+6\\ \hfill \overline{)\begin{array}{l}\begin{array}{ccccccc}{x}^3& +& 6x^2& +& 16x& +& 96\end{array}\\ \begin{array}{ccccccc}\begin{array}{cc}{x}^3& +16\end{array}x& & & & & & \end{array}\\ \begin{array}{ccccccc}& & 6x^2& +& 96& & \end{array}\\ \begin{array}{ccccccc}& & 6x^2& +96& & & \end{array}\\ \begin{array}{ccccccc}& & 0& & & & \end{array}\end{array}}\end{array}$

The depressed polynomial is $x+6$ .

The root of the above linear polynomial $x+6=0$ is $x=-6$ .

Therefore, roots are $-6\text{ and }\pm 4i$ .

Thus, the roots of the equation $x^3+6x^2+16x+96=0$ are $-6\text{ and }\pm 4i$ .