Question
Answer
$$(3-x)^4(1+\dfrac{2}{x})^7=(x^4-12x^3+54x^2-108x+81)(1+\dfrac{2}{x})^7$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(3-x)^4(1+\frac{2}{x})^7& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}(x^4-12x^3+54x^2-108x+81)(1+\frac{2}{x})^7\end{aligned} $$ | |
| ① | $$ (3-x)^4 = (3-x)^2 \cdot (3-x)^2 $$ |
| ② | Find $ \left(3-x\right)^2 $ using formula. $$ (A - B)^2 = \color{blue}{A^2} - 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ 3 } $ and $ B = \color{red}{ x }$. $$ \begin{aligned}\left(3-x\right)^2 = \color{blue}{3^2} -2 \cdot 3 \cdot x + \color{red}{x^2} = 9-6x+x^2\end{aligned} $$ |
| ③ | Multiply each term of $ \left( \color{blue}{9-6x+x^2}\right) $ by each term in $ \left( 9-6x+x^2\right) $. $$ \left( \color{blue}{9-6x+x^2}\right) \cdot \left( 9-6x+x^2\right) = 81-54x+9x^2-54x+36x^2-6x^3+9x^2-6x^3+x^4 $$ |
| ④ | Combine like terms: $$ 81 \color{blue}{-54x} + \color{red}{9x^2} \color{blue}{-54x} + \color{green}{36x^2} \color{orange}{-6x^3} + \color{green}{9x^2} \color{orange}{-6x^3} +x^4 = \\ = x^4 \color{orange}{-12x^3} + \color{green}{54x^2} \color{blue}{-108x} +81 $$ |
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