Question
Answer
$$(x-2)(x+2)(x+2)(x-2)=x^4-8x^2+16$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(x-2)(x+2)(x+2)(x-2)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(x^2+2x-2x-4)(x+2)(x-2) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}(x^2-4)(x+2)(x-2) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}(x^3+2x^2-4x-8)(x-2) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}x^4-8x^2+16\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{x-2}\right) $ by each term in $ \left( x+2\right) $. $$ \left( \color{blue}{x-2}\right) \cdot \left( x+2\right) = x^2+ \cancel{2x} -\cancel{2x}-4 $$ |
| ② | Combine like terms: $$ x^2+ \, \color{blue}{ \cancel{2x}} \, \, \color{blue}{ -\cancel{2x}} \,-4 = x^2-4 $$ |
| ③ | Multiply each term of $ \left( \color{blue}{x^2-4}\right) $ by each term in $ \left( x+2\right) $. $$ \left( \color{blue}{x^2-4}\right) \cdot \left( x+2\right) = x^3+2x^2-4x-8 $$ |
| ④ | Multiply each term of $ \left( \color{blue}{x^3+2x^2-4x-8}\right) $ by each term in $ \left( x-2\right) $. $$ \left( \color{blue}{x^3+2x^2-4x-8}\right) \cdot \left( x-2\right) = \\ = x^4 -\cancel{2x^3}+ \cancel{2x^3}-4x^2-4x^2+ \cancel{8x} -\cancel{8x}+16 $$ |
| ⑤ | Combine like terms: $$ x^4 \, \color{blue}{ -\cancel{2x^3}} \,+ \, \color{blue}{ \cancel{2x^3}} \, \color{green}{-4x^2} \color{green}{-4x^2} + \, \color{orange}{ \cancel{8x}} \, \, \color{orange}{ -\cancel{8x}} \,+16 = x^4 \color{green}{-8x^2} +16 $$ |
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