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