Chapter 15.5, Problem 6MRE

To determine

To calculate: The roots of the equation $x^3-4x^2+3x-12=0$ .

Answer to Problem 6MRE

The roots of the equation $x^3-4x^2+3x-12=0$ are $4\text{ and }\pm i\sqrt{3}$ .

Explanation of Solution

Given information:

One of the root of the equation is $-i\sqrt{3}$ .

Formula used:

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

Calculation:

Consider theequation $x^3-4x^2+3x-12=0$ . One of the root of the equation is $-i\sqrt{3}$ . Then $x=i\sqrt{3}$ 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, $\left(x+i\sqrt{3}\right)\left(x-i\sqrt{3}\right)$ is a root of the equation $x^3-4x^2+3x-12=0$ .

So, $x^2+3$ is a root of the equation $x^3-4x^2+3x-12=0$ .

Use long division to divide $x^3-4x^2+3x-12=0$ by its factor $x^2+3$ .

  $x^2+3\begin{array}{c}\hfill x-4\\ \hfill \overline{)\begin{array}{l}\begin{array}{ccccccc}{x}^3& -& 4x^2& +& 3x& -& 12\end{array}\\ \begin{array}{ccccccc}\begin{array}{cc}{x}^3& +3\end{array}x& & & & & & \end{array}\\ \begin{array}{ccccccc}& & -4x^2& -& 12& & \end{array}\\ \begin{array}{ccccccc}& & -4x^2& -12& & & \end{array}\\ \begin{array}{ccccccc}& & 0& & & & \end{array}\end{array}}\end{array}$

The depressed polynomial is $x-4$ .

The root of the above linear polynomial $x-4=0$ is $x=4$ .

Therefore, roots are $4\text{ and }\pm i\sqrt{3}$ .

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