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