Answer
$$(-x-y)^4=x^4+4x^3y+6x^2y^2+4xy^3+y^4$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(-x-y)^4& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}x^4+4x^3y+6x^2y^2+4xy^3+y^4\end{aligned} $$ | |
| ① | $$ (-x-y)^4 = (-x-y)^2 \cdot (-x-y)^2 $$ |
| ② | Find $ \left(-x-y\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}{ x } $ and $ B = \color{red}{ y }$. $$ \begin{aligned}\left(-x-y\right)^2& \xlongequal{ S1 } \left(x+y\right)^2 \xlongequal{ S2 } \color{blue}{x^2} +2 \cdot x \cdot y + \color{red}{y^2} = \\[1 em] & = x^2+2xy+y^2\end{aligned} $$ |
| ③ | Multiply each term of $ \left( \color{blue}{x^2+2xy+y^2}\right) $ by each term in $ \left( x^2+2xy+y^2\right) $. $$ \left( \color{blue}{x^2+2xy+y^2}\right) \cdot \left( x^2+2xy+y^2\right) = \\ = x^4+2x^3y+x^2y^2+2x^3y+4x^2y^2+2xy^3+x^2y^2+2xy^3+y^4 $$ |
| ④ | Combine like terms: $$ x^4+ \color{blue}{2x^3y} + \color{red}{x^2y^2} + \color{blue}{2x^3y} + \color{green}{4x^2y^2} + \color{orange}{2xy^3} + \color{green}{x^2y^2} + \color{orange}{2xy^3} +y^4 = \\ = x^4+ \color{blue}{4x^3y} + \color{green}{6x^2y^2} + \color{orange}{4xy^3} +y^4 $$ |
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