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