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