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