Question
Answer
$$\dfrac{x^5+9x^4+31x^3+51x^2+40x+12}{(x^2-1)(x^2-4)}=\dfrac{x^5+9x^4+31x^3+51x^2+40x+12}{x^4-5x^2+4}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{x^5+9x^4+31x^3+51x^2+40x+12}{(x^2-1)(x^2-4)}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{x^5+9x^4+31x^3+51x^2+40x+12}{x^4-4x^2-x^2+4} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{x^5+9x^4+31x^3+51x^2+40x+12}{x^4-5x^2+4}\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{x^2-1}\right) $ by each term in $ \left( x^2-4\right) $. $$ \left( \color{blue}{x^2-1}\right) \cdot \left( x^2-4\right) = x^4-4x^2-x^2+4 $$ |
| ② | Simplify denominator $$ x^4 \color{blue}{-4x^2} \color{blue}{-x^2} +4 = x^4 \color{blue}{-5x^2} +4 $$ |
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