Question
Answer
$$(3x+8)(2x^2-3x+3)=6x^3+7x^2-15x+24$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(3x+8)(2x^2-3x+3)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}6x^3-9x^2+9x+16x^2-24x+24 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}6x^3+7x^2-15x+24\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{3x+8}\right) $ by each term in $ \left( 2x^2-3x+3\right) $. $$ \left( \color{blue}{3x+8}\right) \cdot \left( 2x^2-3x+3\right) = 6x^3-9x^2+9x+16x^2-24x+24 $$ |
| ② | Combine like terms: $$ 6x^3 \color{blue}{-9x^2} + \color{red}{9x} + \color{blue}{16x^2} \color{red}{-24x} +24 = 6x^3+ \color{blue}{7x^2} \color{red}{-15x} +24 $$ |
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