Answer
$$(3x-4)(2x^2-5x-5)=6x^3-23x^2+5x+20$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(3x-4)(2x^2-5x-5)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}6x^3-15x^2-15x-8x^2+20x+20 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}6x^3-23x^2+5x+20\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{3x-4}\right) $ by each term in $ \left( 2x^2-5x-5\right) $. $$ \left( \color{blue}{3x-4}\right) \cdot \left( 2x^2-5x-5\right) = 6x^3-15x^2-15x-8x^2+20x+20 $$ |
| ② | Combine like terms: $$ 6x^3 \color{blue}{-15x^2} \color{red}{-15x} \color{blue}{-8x^2} + \color{red}{20x} +20 = 6x^3 \color{blue}{-23x^2} + \color{red}{5x} +20 $$ |
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