Question
Answer
$$(y-2)^4=y^4-8y^3+24y^2-32y+16$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(y-2)^4& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}y^4-8y^3+24y^2-32y+16\end{aligned} $$ | |
| ① | $$ (y-2)^4 = (y-2)^2 \cdot (y-2)^2 $$ |
| ② | Find $ \left(y-2\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}{ 2 }$. $$ \begin{aligned}\left(y-2\right)^2 = \color{blue}{y^2} -2 \cdot y \cdot 2 + \color{red}{2^2} = y^2-4y+4\end{aligned} $$ |
| ③ | Multiply each term of $ \left( \color{blue}{y^2-4y+4}\right) $ by each term in $ \left( y^2-4y+4\right) $. $$ \left( \color{blue}{y^2-4y+4}\right) \cdot \left( y^2-4y+4\right) = y^4-4y^3+4y^2-4y^3+16y^2-16y+4y^2-16y+16 $$ |
| ④ | Combine like terms: $$ y^4 \color{blue}{-4y^3} + \color{red}{4y^2} \color{blue}{-4y^3} + \color{green}{16y^2} \color{orange}{-16y} + \color{green}{4y^2} \color{orange}{-16y} +16 = \\ = y^4 \color{blue}{-8y^3} + \color{green}{24y^2} \color{orange}{-32y} +16 $$ |
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