Q37. - Algebra 1

Question

Find each quotient.

$\left(48y^2+8y+7\right)÷\left(12y-1\right)$

Step-by-Step Solution

Answer

The quotient is $4y+1+\frac{8}{12y-1}$.

Step 1. Define the concept.

Method of long division is preferably used when the polynomial cannot be factored or if there are no common factors by which to divide.

Step 2. Use method of long division.

First divide the first term of the dividend, $48y^2$, by the first term of the divisor, $12y$.

 

$12y-14y\begin{array}{l}48y^2+8y+7\\ \frac{48y^2-4y}{12y+7}\end{array}\left(\begin{array}{l}48y^2÷12y=4y\\ \text{Multiply\hspace{0.17em}\hspace{0.17em}}12y-1\text{\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}}4y\\ \text{Subtract.}\end{array}\right)$

Step 3. Simplify the expression.

To further simplify, divide the first term of the partial dividend, $12y$, by the first term of the divisor, $12y$. Repeat the process as many times required.

$12y-14y+1\begin{array}{l}48y^2+8y+7\\ \frac{48y^2-4y}{\begin{array}{l}12y+7\\ \underline{12y-1}\\ 8\end{array}}\end{array}$

Therefore, $\left(48y^2+8y+7\right)÷\left(12y-1\right)$ is $4y+1$ with a remainder of 8. This answer can be written as $4y+1+\frac{8}{12y-1}$.