Question
Answer
$$(2x^3-3x^2+6x)(3x^2-x-1)=6x^5-11x^4+19x^3-3x^2-6x$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(2x^3-3x^2+6x)(3x^2-x-1)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}6x^5-11x^4+19x^3-3x^2-6x\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{2x^3-3x^2+6x}\right) $ by each term in $ \left( 3x^2-x-1\right) $. $$ \left( \color{blue}{2x^3-3x^2+6x}\right) \cdot \left( 3x^2-x-1\right) = 6x^5-2x^4-2x^3-9x^4+3x^3+3x^2+18x^3-6x^2-6x $$ |
| ② | Combine like terms: $$ 6x^5 \color{blue}{-2x^4} \color{red}{-2x^3} \color{blue}{-9x^4} + \color{green}{3x^3} + \color{orange}{3x^2} + \color{green}{18x^3} \color{orange}{-6x^2} -6x = \\ = 6x^5 \color{blue}{-11x^4} + \color{green}{19x^3} \color{orange}{-3x^2} -6x $$ |
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