Question
Answer
$$\dfrac{64k^3-1}{4k^2-13k+3}=\dfrac{16k^2+4k+1}{k-3}$$
Explanation
| $$ \begin{aligned}\frac{64k^3-1}{4k^2-13k+3}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{16k^2+4k+1}{k-3}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{64k^3-1}{4k^2-13k+3} $ to $ \dfrac{16k^2+4k+1}{k-3} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{4k-1}$. $$ \begin{aligned} \frac{64k^3-1}{4k^2-13k+3} & =\frac{ \left( 16k^2+4k+1 \right) \cdot \color{blue}{ \left( 4k-1 \right) }}{ \left( k-3 \right) \cdot \color{blue}{ \left( 4k-1 \right) }} = \\[1ex] &= \frac{16k^2+4k+1}{k-3} \end{aligned} $$ |
This page was created using
Simplify Rational Expressions