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