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