Chapter 11.3, Problem 12PPE

To determine

To divide $\left(8q^2-32q\right)÷2q^2$ .

Answer to Problem 12PPE

The quotient of given division is 4 and remainder is $-16$ .

Explanation of Solution

Given:

  $\left(8q^2-32q\right)÷2q^2$

Concept Used:

To conduct a division, $\frac{R\left(x\right)}{Q\left(x\right)}$ , cancel out the common factor of numerator and denominator.

Calculation:

In order to divide $\left(8q^2-32q\right)÷2q^2$ first factorize the numerator by taking common factor $2q$ out, as

  $\begin{array}{c}8q^2-32q=2q\cdot 4q-2q\cdot 16\\ =2q\left(4q-16\right)\end{array}$

Now, the given divisor can be proceed as:

  $\begin{array}{c}\left(8q^2-32q\right)÷2q^2=\frac{2q\left(4q-16\right)}{2q^2}\\ =\frac{4q-16}{q}\left(\begin{array}{l}\text{dividing common}\\ \text{factor 2}q\text{.}\end{array}\right)\end{array}$

Now, to divide $\frac{4q-16}{q}$ use long division algorithm, as

  $\begin{array}{l}q)4q-16(4\\ \underset{\_}{4q}\\ \underset{\_}{-16}\end{array}$

Thus, the quotient of given division is 4 and remainder is $-16$ .