Chapter 15.5, Problem 7MRE

To determine

To calculate: The roots of the equation $2x^3-11x^2+4x+5=0$ .

Answer to Problem 7MRE

The roots of the equation $2x^3-11x^2+4x+5=0$ are $1,\frac{-1}{2}\text{ and }5$ .

Explanation of Solution

Given information:

One of the root of the equation is $5$ .

Formula used:

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

Calculation:

Consider theequation $2x^3-11x^2+4x+5=0$ . One of the root of the equation is $5$ .

So, $x-5$ is a factor of the equation $2x^3-11x^2+4x+5=0$ .

Use long division to divide $2x^3-11x^2+4x+5=0$ by its factor $x-5$ .

  $x-5\begin{array}{c}\hfill 2x^2-x-1\\ \hfill \overline{)\begin{array}{l}\begin{array}{ccccccc}2x^3& -& 11x^2& +& 4x& +& 5\end{array}\\ \begin{array}{ccccccc}\begin{array}{cc}2x^3& -10\end{array}{x}^2& & & & & & \end{array}\\ \begin{array}{ccccccc}& & -x^2& +& 4x+5& & \end{array}\\ \begin{array}{ccccccc}& & -x^2& +5x& & & \end{array}\\ \begin{array}{ccccccc}& & & -x+5& & & \end{array}\\ \begin{array}{ccccc}& & & -x+5& \end{array}\\ \begin{array}{ccccc}& & & 0& \end{array}\end{array}}\end{array}$

The depressed polynomial is $2x^2-x-1$ .

Use the quadratic formula to factor the above polynomial,

  $\begin{array}{c}x=\frac{1\pm \sqrt{1+8}}{4}\\ x=\frac{1\pm 3}{4}\\ x=1,\frac{-1}{2}\end{array}$

Therefore, roots are $1,\frac{-1}{2}\text{ and }5$ .

Thus, the roots of the equation $2x^3-11x^2+4x+5=0$ are $1,\frac{-1}{2}\text{ and }5$ .