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