Question
Answer
$$\dfrac{4q^3+16q^2}{2q^3+6q^2-8q}=\dfrac{2q}{q-1}$$
Explanation
| $$ \begin{aligned}\frac{4q^3+16q^2}{2q^3+6q^2-8q}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2q}{q-1}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{4q^3+16q^2}{2q^3+6q^2-8q} $ to $ \dfrac{2q}{q-1} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{2q^2+8q}$. $$ \begin{aligned} \frac{4q^3+16q^2}{2q^3+6q^2-8q} & =\frac{ 2q \cdot \color{blue}{ \left( 2q^2+8q \right) }}{ \left( q-1 \right) \cdot \color{blue}{ \left( 2q^2+8q \right) }} = \\[1ex] &= \frac{2q}{q-1} \end{aligned} $$ |
This page was created using
Simplify Rational Expressions