Answer
$$(2x-3)(3x^2+2x-1)=6x^3-5x^2-8x+3$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(2x-3)(3x^2+2x-1)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}6x^3+4x^2-2x-9x^2-6x+3 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}6x^3-5x^2-8x+3\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{2x-3}\right) $ by each term in $ \left( 3x^2+2x-1\right) $. $$ \left( \color{blue}{2x-3}\right) \cdot \left( 3x^2+2x-1\right) = 6x^3+4x^2-2x-9x^2-6x+3 $$ |
| ② | Combine like terms: $$ 6x^3+ \color{blue}{4x^2} \color{red}{-2x} \color{blue}{-9x^2} \color{red}{-6x} +3 = 6x^3 \color{blue}{-5x^2} \color{red}{-8x} +3 $$ |
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