Chapter 6, Problem 9MCQ

To determine

To calculate: The simplified form of the expression $\frac{3x^3-16x^2+9x-24}{x-5}$ .

Answer to Problem 9MCQ

The simplified form of the expression $\frac{3x^3-16x^2+9x-24}{x-5}$ is $\left(3x^2-x+19\right)+\frac{71}{\left(x-5\right)}$ .

Explanation of Solution

Given information:

The expression $\frac{3x^3-16x^2+9x-24}{x-5}$ .

Formula used:

When a polynomial is divided by its factor then dividend is the product of divisor and quotient increased by remainder.

Calculation:

Consider the expression $\frac{3x^3-16x^2+9x-24}{x-5}$ .

When a polynomial is divided by its factor then dividend is the product of divisor and quotient increased by remainder.

Apply the method of long division. First multiply the expression $x-5$ by $3x^2$ . Subtract it from the dividend. Next multiply $x-5$ by $-x$ . Subtract the expression obtained. Again multiply the expression $x-5$ by $19.$ Subtract the expression obtained.

  $\begin{array}{l}x-5\begin{array}{c}\hfill 3x^2-x+19\\ \hfill \overline{)3x^3-16x^2+9x-24}\end{array}\\ \begin{array}{ccc}& & 3x^3-15x^2\end{array}\\ \begin{array}{ccc}& & \begin{array}{cc}& -x^2+9x\end{array}\end{array}\\ \begin{array}{cc}& \begin{array}{ccc}& & -x^2-10x\end{array}\end{array}\\ \begin{array}{cccccc}& & & & & 19x-24\end{array}\\ \begin{array}{cccccc}& & & & & 19x-95\end{array}\\ \overline{\begin{array}{ccc}\begin{array}{ccc}\begin{array}{cc}\begin{array}{cc}\begin{array}{cc}& \end{array}& \end{array}& \end{array}& & \end{array}& & 71\end{array}}\end{array}$

Now, since $\text{Dividend}=\text{Quotient}\times \text{Divisor}+\text{Remainder}$ . Here, dividend is $3x^3-16x^2+9x-24$ quotient is $3x^2-x+19$ , divisor is $x-5$ and remainder is $71$ .

Therefore, $3x^3-16x^2+9x-24=\left(x-5\right)\left(3x^2-x+19\right)+71$ that is,

  $\frac{3x^3-16x^2+9x-24}{x-5}=\left(3x^2-x+19\right)+\frac{71}{\left(x-5\right)}$

Thus, the simplified form of the expression $\frac{3x^3-16x^2+9x-24}{x-5}$ is $\left(3x^2-x+19\right)+\frac{71}{\left(x-5\right)}$ .