Answer
$$(x-2)^2(x^2+4x-3)=x^4-15x^2+28x-12$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(x-2)^2(x^2+4x-3)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(x^2-4x+4)(x^2+4x-3) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}x^4-15x^2+28x-12\end{aligned} $$ | |
| 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 }$. $$ \begin{aligned}\left(x-2\right)^2 = \color{blue}{x^2} -2 \cdot x \cdot 2 + \color{red}{2^2} = x^2-4x+4\end{aligned} $$ |
| ② | Multiply each term of $ \left( \color{blue}{x^2-4x+4}\right) $ by each term in $ \left( x^2+4x-3\right) $. $$ \left( \color{blue}{x^2-4x+4}\right) \cdot \left( x^2+4x-3\right) = \\ = x^4+ \cancel{4x^3}-3x^2 -\cancel{4x^3}-16x^2+12x+4x^2+16x-12 $$ |
| ③ | Combine like terms: $$ x^4+ \, \color{blue}{ \cancel{4x^3}} \, \color{green}{-3x^2} \, \color{blue}{ -\cancel{4x^3}} \, \color{orange}{-16x^2} + \color{blue}{12x} + \color{orange}{4x^2} + \color{blue}{16x} -12 = x^4 \color{orange}{-15x^2} + \color{blue}{28x} -12 $$ |
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