Question
Answer
$$2x^3y(3x^2-4y+2xy-8)=6x^5y+4x^4y^2-8x^3y^2-16x^3y$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}2x^3y(3x^2-4y+2xy-8)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}6x^5y-8x^3y^2+4x^4y^2-16x^3y \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}6x^5y+4x^4y^2-8x^3y^2-16x^3y\end{aligned} $$ | |
| ① | Multiply $ \color{blue}{2x^3y} $ by $ \left( 3x^2-4y+2xy-8\right) $ $$ \color{blue}{2x^3y} \cdot \left( 3x^2-4y+2xy-8\right) = 6x^5y-8x^3y^2+4x^4y^2-16x^3y $$ |
| ② | Combine like terms: $$ 6x^5y+4x^4y^2-8x^3y^2-16x^3y = 6x^5y+4x^4y^2-8x^3y^2-16x^3y $$ |
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