Answer
$$3x^2(-2x+1)^4=48x^6-96x^5+72x^4-24x^3+3x^2$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}3x^2(-2x+1)^4& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}3x^2(16x^4-32x^3+24x^2-8x+1) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}48x^6-96x^5+72x^4-24x^3+3x^2\end{aligned} $$ | |
| ① | $$ (-2x+1)^4 = (-2x+1)^2 \cdot (-2x+1)^2 $$ |
| ② | Find $ \left(-2x+1\right)^2 $ in two steps. S1: Change all signs inside bracket. S2: Apply formula $$ (A - B)^2 = \color{blue}{A^2} - 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ 2x } $ and $ B = \color{red}{ 1 }$. $$ \begin{aligned}\left(-2x+1\right)^2& \xlongequal{ S1 } \left(2x-1\right)^2 \xlongequal{ S2 } \color{blue}{\left( 2x \right)^2} -2 \cdot 2x \cdot 1 + \color{red}{1^2} = \\[1 em] & = 4x^2-4x+1\end{aligned} $$ |
| ③ | Multiply each term of $ \left( \color{blue}{4x^2-4x+1}\right) $ by each term in $ \left( 4x^2-4x+1\right) $. $$ \left( \color{blue}{4x^2-4x+1}\right) \cdot \left( 4x^2-4x+1\right) = 16x^4-16x^3+4x^2-16x^3+16x^2-4x+4x^2-4x+1 $$ |
| ④ | Combine like terms: $$ 16x^4 \color{blue}{-16x^3} + \color{red}{4x^2} \color{blue}{-16x^3} + \color{green}{16x^2} \color{orange}{-4x} + \color{green}{4x^2} \color{orange}{-4x} +1 = \\ = 16x^4 \color{blue}{-32x^3} + \color{green}{24x^2} \color{orange}{-8x} +1 $$ |
| ⑤ | Multiply $ \color{blue}{3x^2} $ by $ \left( 16x^4-32x^3+24x^2-8x+1\right) $ $$ \color{blue}{3x^2} \cdot \left( 16x^4-32x^3+24x^2-8x+1\right) = 48x^6-96x^5+72x^4-24x^3+3x^2 $$ |
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