Answer
$$(3x-2y)^4=81x^4-216x^3y+216x^2y^2-96xy^3+16y^4$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(3x-2y)^4& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}81x^4-216x^3y+216x^2y^2-96xy^3+16y^4\end{aligned} $$ | |
| ① | $$ (3x-2y)^4 = (3x-2y)^2 \cdot (3x-2y)^2 $$ |
| ② | Find $ \left(3x-2y\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}{ 2y }$. $$ \begin{aligned}\left(3x-2y\right)^2 = \color{blue}{\left( 3x \right)^2} -2 \cdot 3x \cdot 2y + \color{red}{\left( 2y \right)^2} = 9x^2-12xy+4y^2\end{aligned} $$ |
| ③ | Multiply each term of $ \left( \color{blue}{9x^2-12xy+4y^2}\right) $ by each term in $ \left( 9x^2-12xy+4y^2\right) $. $$ \left( \color{blue}{9x^2-12xy+4y^2}\right) \cdot \left( 9x^2-12xy+4y^2\right) = \\ = 81x^4-108x^3y+36x^2y^2-108x^3y+144x^2y^2-48xy^3+36x^2y^2-48xy^3+16y^4 $$ |
| ④ | Combine like terms: $$ 81x^4 \color{blue}{-108x^3y} + \color{red}{36x^2y^2} \color{blue}{-108x^3y} + \color{green}{144x^2y^2} \color{orange}{-48xy^3} + \color{green}{36x^2y^2} \color{orange}{-48xy^3} +16y^4 = \\ = 81x^4 \color{blue}{-216x^3y} + \color{green}{216x^2y^2} \color{orange}{-96xy^3} +16y^4 $$ |
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