Question
Answer
$$4(x^2-2y+4xy+2y^2-3x+5xy^2)=20xy^2+4x^2+16xy+8y^2-12x-8y$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}4(x^2-2y+4xy+2y^2-3x+5xy^2)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}4x^2-8y+16xy+8y^2-12x+20xy^2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}20xy^2+4x^2+16xy+8y^2-12x-8y\end{aligned} $$ | |
| ① | Multiply $ \color{blue}{4} $ by $ \left( x^2-2y+4xy+2y^2-3x+5xy^2\right) $ $$ \color{blue}{4} \cdot \left( x^2-2y+4xy+2y^2-3x+5xy^2\right) = 4x^2-8y+16xy+8y^2-12x+20xy^2 $$ |
| ② | Combine like terms: $$ 20xy^2+4x^2+16xy+8y^2-12x-8y = 20xy^2+4x^2+16xy+8y^2-12x-8y $$ |
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