Question
Answer
$$(1-4x)^4=256x^4-256x^3+96x^2-16x+1$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(1-4x)^4& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}256x^4-256x^3+96x^2-16x+1\end{aligned} $$ | |
| ① | $$ (1-4x)^4 = (1-4x)^2 \cdot (1-4x)^2 $$ |
| ② | Find $ \left(1-4x\right)^2 $ using formula. $$ (A - B)^2 = \color{blue}{A^2} - 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ 1 } $ and $ B = \color{red}{ 4x }$. $$ \begin{aligned}\left(1-4x\right)^2 = \color{blue}{1^2} -2 \cdot 1 \cdot 4x + \color{red}{\left( 4x \right)^2} = 1-8x+16x^2\end{aligned} $$ |
| ③ | Multiply each term of $ \left( \color{blue}{1-8x+16x^2}\right) $ by each term in $ \left( 1-8x+16x^2\right) $. $$ \left( \color{blue}{1-8x+16x^2}\right) \cdot \left( 1-8x+16x^2\right) = 1-8x+16x^2-8x+64x^2-128x^3+16x^2-128x^3+256x^4 $$ |
| ④ | Combine like terms: $$ 1 \color{blue}{-8x} + \color{red}{16x^2} \color{blue}{-8x} + \color{green}{64x^2} \color{orange}{-128x^3} + \color{green}{16x^2} \color{orange}{-128x^3} +256x^4 = \\ = 256x^4 \color{orange}{-256x^3} + \color{green}{96x^2} \color{blue}{-16x} +1 $$ |
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