Question
Answer
$$\dfrac{40x^3-40x^2-80x}{56x^3+88x^2+32x}=\dfrac{5x-10}{7x+4}$$
Explanation
| $$ \begin{aligned}\frac{40x^3-40x^2-80x}{56x^3+88x^2+32x}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{5x-10}{7x+4}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{40x^3-40x^2-80x}{56x^3+88x^2+32x} $ to $ \dfrac{5x-10}{7x+4} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{8x^2+8x}$. $$ \begin{aligned} \frac{40x^3-40x^2-80x}{56x^3+88x^2+32x} & =\frac{ \left( 5x-10 \right) \cdot \color{blue}{ \left( 8x^2+8x \right) }}{ \left( 7x+4 \right) \cdot \color{blue}{ \left( 8x^2+8x \right) }} = \\[1ex] &= \frac{5x-10}{7x+4} \end{aligned} $$ |
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