Chapter 6.8, Problem 6PT

To determine

To calculate: The simplified form of the expression $\left(3x^3-5x^2-23x+24\right)÷\left(x-3\right)$ .

Answer to Problem 6PT

The simplified form of the expression $\left(3x^3-5x^2-23x+24\right)÷\left(x-3\right)$ is $\left(x-3\right)\left(3x^2+4x-11\right)-9$ .

Explanation of Solution

Given information:

The expression $\left(3x^3-5x^2-23x+24\right)÷\left(x-3\right)$ .

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 $\left(3x^3-5x^2-23x+24\right)÷\left(x-3\right)$ .

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 $x-3$ by $3x^2$ . Subtract it from the dividend and then change the signs as positive to negative and negative to positive at the end of every step.

Next multiply $x-3$ by $4x$ . Subtract the expression obtained.

Next multiply $x-3$ by $-11$ . Subtract the expression obtained.

  $x-3\begin{array}{c}\hfill 3x^2+4x-11\\ \hfill \overline{)\begin{array}{l}\begin{array}{ccccccc}3x^2& -& 5x^2& -& 23x& +& 24\end{array}\\ \begin{array}{ccc}3x^2& +& 9x^2\end{array}\\ \begin{array}{ccc}4x^2& -& 23x\end{array}\\ \begin{array}{ccc}4x^2& +& 12x\end{array}\\ \begin{array}{cccc}-& 11x& +& 24\end{array}\\ \begin{array}{ccc}11x& -& 23\end{array}\\ \begin{array}{cc}-& 9\end{array}\end{array}}\end{array}$

Now, since $\text{Dividend}=\text{Quotient}\times \text{Divisor}+\text{Remainder}$ . Here, dividend is $3x^3-5x^2-23x+24$ quotient is $3x^2+4x-11$ , divisor is $x-3$ and remainder is $-9$ .

Therefore, $\left(3x^3-5x^2-23x+24\right)=\left(x-3\right)\left(3x^2+4x-11\right)-9$ ,

Thus, the simplified form of the expression $\left(3x^3-5x^2-23x+24\right)÷\left(x-3\right)$ is $\left(x-3\right)\left(3x^2+4x-11\right)-9$ .