Answer
$$(3y-2)(y^2+y+8)=3y^3+y^2+22y-16$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(3y-2)(y^2+y+8)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}3y^3+3y^2+24y-2y^2-2y-16 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}3y^3+y^2+22y-16\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{3y-2}\right) $ by each term in $ \left( y^2+y+8\right) $. $$ \left( \color{blue}{3y-2}\right) \cdot \left( y^2+y+8\right) = 3y^3+3y^2+24y-2y^2-2y-16 $$ |
| ② | Combine like terms: $$ 3y^3+ \color{blue}{3y^2} + \color{red}{24y} \color{blue}{-2y^2} \color{red}{-2y} -16 = 3y^3+ \color{blue}{y^2} + \color{red}{22y} -16 $$ |
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