Question
Answer
$$(y^2-2x)^4=y^8-8xy^6+24x^2y^4-32x^3y^2+16x^4$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(y^2-2x)^4& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}y^8-8xy^6+24x^2y^4-32x^3y^2+16x^4\end{aligned} $$ | |
| ① | $$ (y^2-2x)^4 = (y^2-2x)^2 \cdot (y^2-2x)^2 $$ |
| ② | Find $ \left(y^2-2x\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^2 } $ and $ B = \color{red}{ 2x }$. $$ \begin{aligned}\left(y^2-2x\right)^2 = \color{blue}{\left( y^2 \right)^2} -2 \cdot y^2 \cdot 2x + \color{red}{\left( 2x \right)^2} = y^4-4xy^2+4x^2\end{aligned} $$ |
| ③ | Multiply each term of $ \left( \color{blue}{y^4-4xy^2+4x^2}\right) $ by each term in $ \left( y^4-4xy^2+4x^2\right) $. $$ \left( \color{blue}{y^4-4xy^2+4x^2}\right) \cdot \left( y^4-4xy^2+4x^2\right) = \\ = y^8-4xy^6+4x^2y^4-4xy^6+16x^2y^4-16x^3y^2+4x^2y^4-16x^3y^2+16x^4 $$ |
| ④ | Combine like terms: $$ y^8 \color{blue}{-4xy^6} + \color{red}{4x^2y^4} \color{blue}{-4xy^6} + \color{green}{16x^2y^4} \color{orange}{-16x^3y^2} + \color{green}{4x^2y^4} \color{orange}{-16x^3y^2} +16x^4 = \\ = y^8 \color{blue}{-8xy^6} + \color{green}{24x^2y^4} \color{orange}{-32x^3y^2} +16x^4 $$ |
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