Question
Answer
$$\dfrac{x^2+11}{(2-x)(3x+4)}=\dfrac{x^2+11}{-3x^2+2x+8}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{x^2+11}{(2-x)(3x+4)}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{x^2+11}{6x+8-3x^2-4x} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{x^2+11}{-3x^2+2x+8}\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{2-x}\right) $ by each term in $ \left( 3x+4\right) $. $$ \left( \color{blue}{2-x}\right) \cdot \left( 3x+4\right) = 6x+8-3x^2-4x $$ |
| ② | Simplify denominator $$ \color{blue}{6x} +8-3x^2 \color{blue}{-4x} = -3x^2+ \color{blue}{2x} +8 $$ |
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