Question
Answer
$$\dfrac{2x^3+7x^2-9}{(x+3)^3(x-3)}=\dfrac{2x^3+7x^2-9}{x^4+6x^3-54x-81}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{2x^3+7x^2-9}{(x+3)^3(x-3)}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2x^3+7x^2-9}{(x^3+9x^2+27x+27)(x-3)} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{2x^3+7x^2-9}{x^4+6x^3-54x-81}\end{aligned} $$ | |
| ① | Find $ \left(x+3\right)^3 $ using formula $$ (A + B) = A^3 + 3A^2B + 3AB^2 + B^3 $$where $ A = x $ and $ B = 3 $. $$ \left(x+3\right)^3 = x^3+3 \cdot x^2 \cdot 3 + 3 \cdot x \cdot 3^2+3^3 = x^3+9x^2+27x+27 $$ |
| ② | Multiply each term of $ \left( \color{blue}{x^3+9x^2+27x+27}\right) $ by each term in $ \left( x-3\right) $. $$ \left( \color{blue}{x^3+9x^2+27x+27}\right) \cdot \left( x-3\right) = \\ = x^4-3x^3+9x^3 -\cancel{27x^2}+ \cancel{27x^2}-81x+27x-81 $$ |
| ③ | Combine like terms: $$ x^4 \color{blue}{-3x^3} + \color{blue}{9x^3} \, \color{red}{ -\cancel{27x^2}} \,+ \, \color{red}{ \cancel{27x^2}} \, \color{orange}{-81x} + \color{orange}{27x} -81 = x^4+ \color{blue}{6x^3} \color{orange}{-54x} -81 $$ |
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