Chapter 15.5, Problem 5MRE

To determine

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

Answer to Problem 5MRE

The roots of the equation $x^3+x^2-2x-2=0$ are $1\text{ and }\pm \sqrt{2}$ .

Explanation of Solution

Given information:

One of the root of the equation is $\sqrt{2}$ .

Formula used:

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

Calculation:

Consider theequation $x^3+x^2-2x-2=0$ . One of the root of the equation is $\sqrt{2}$ . Then $x=-\sqrt{2}$ will also be the root of the equation, as irrational numbers exist in pairs.

If $x=\sqrt{c}$ is a root of an equation then $x=-\sqrt{c}$ is also a root of the equation.

Therefore, $\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)$ is a root of the equation $x^3+x^2-2x-2=0$ .

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

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

  $x^2-2\begin{array}{c}\hfill x+1\\ \hfill \overline{)\begin{array}{l}\begin{array}{ccccccc}{x}^3& +& x^2& -& 2x& -& 2\end{array}\\ \begin{array}{ccccccc}\begin{array}{cc}{x}^3& -2\end{array}x& -& \sqrt{2}{x}^2& & & & \end{array}\\ \begin{array}{ccccccc}& & x^2& -2& & & \end{array}\\ \begin{array}{ccccccc}-& & x^2& -2& & & \end{array}\\ \begin{array}{ccccccc}& & 0& & & & \end{array}\end{array}}\end{array}$

The depressed polynomial is $x+1$ .

The root of the above linear polynomial $x+1=0$ is $x=-1$ .

Therefore, roots are $1\text{ and }\pm \sqrt{2}$ .

Thus, the roots of the equation $x^3+x^2-2x-2=0$ are $1\text{ and }\pm \sqrt{2}$ .