Question
Answer
$$3x^2(6-x)^2+2x(6-x)^3=x^4-108x^2+432x$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}3x^2(6-x)^2+2x(6-x)^3& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}3x^2(36-12x+x^2)+2x(216-108x+18x^2-x^3) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}108x^2-36x^3+3x^4+432x-216x^2+36x^3-2x^4 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}x^4-108x^2+432x\end{aligned} $$ | |
| ① | Find $ \left(6-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}{ 6 } $ and $ B = \color{red}{ x }$. $$ \begin{aligned}\left(6-x\right)^2 = \color{blue}{6^2} -2 \cdot 6 \cdot x + \color{red}{x^2} = 36-12x+x^2\end{aligned} $$Find $ \left(6-x\right)^3 $ using formula $$ (A - B) = A^3 - 3A^2B + 3AB^2 - B^3 $$where $ A = 6 $ and $ B = x $. $$ \left(6-x\right)^3 = 6^3-3 \cdot 6^2 \cdot x + 3 \cdot 6 \cdot x^2-x^3 = 216-108x+18x^2-x^3 $$ |
| ② | Multiply $ \color{blue}{3x^2} $ by $ \left( 36-12x+x^2\right) $ $$ \color{blue}{3x^2} \cdot \left( 36-12x+x^2\right) = 108x^2-36x^3+3x^4 $$Multiply $ \color{blue}{2x} $ by $ \left( 216-108x+18x^2-x^3\right) $ $$ \color{blue}{2x} \cdot \left( 216-108x+18x^2-x^3\right) = 432x-216x^2+36x^3-2x^4 $$ |
| ③ | Combine like terms: $$ \color{blue}{108x^2} \, \color{red}{ -\cancel{36x^3}} \,+ \color{orange}{3x^4} +432x \color{blue}{-216x^2} + \, \color{red}{ \cancel{36x^3}} \, \color{orange}{-2x^4} = \color{orange}{x^4} \color{blue}{-108x^2} +432x $$ |
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