Question
Answer
$$(x^2-3x+2)^2=x^4-6x^3+13x^2-12x+4$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(x^2-3x+2)^2& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}x^4-6x^3+13x^2-12x+4\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{x^2-3x+2}\right) $ by each term in $ \left( x^2-3x+2\right) $. $$ \left( \color{blue}{x^2-3x+2}\right) \cdot \left( x^2-3x+2\right) = x^4-3x^3+2x^2-3x^3+9x^2-6x+2x^2-6x+4 $$ |
| ② | Combine like terms: $$ x^4 \color{blue}{-3x^3} + \color{red}{2x^2} \color{blue}{-3x^3} + \color{green}{9x^2} \color{orange}{-6x} + \color{green}{2x^2} \color{orange}{-6x} +4 = \\ = x^4 \color{blue}{-6x^3} + \color{green}{13x^2} \color{orange}{-12x} +4 $$ |
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