Question
Answer
$$\dfrac{4x^2-7x-2}{64x^3+1}=\dfrac{x-2}{16x^2-4x+1}$$
Explanation
| $$ \begin{aligned}\frac{4x^2-7x-2}{64x^3+1}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{x-2}{16x^2-4x+1}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{4x^2-7x-2}{64x^3+1} $ to $ \dfrac{x-2}{16x^2-4x+1} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{4x+1}$. $$ \begin{aligned} \frac{4x^2-7x-2}{64x^3+1} & =\frac{ \left( x-2 \right) \cdot \color{blue}{ \left( 4x+1 \right) }}{ \left( 16x^2-4x+1 \right) \cdot \color{blue}{ \left( 4x+1 \right) }} = \\[1ex] &= \frac{x-2}{16x^2-4x+1} \end{aligned} $$ |
This page was created using
Simplify Rational Expressions