Question
Answer
$$2(5y-1)(y-3)=10y^2-32y+6$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}2(5y-1)(y-3)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(10y-2)(y-3) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}10y^2-30y-2y+6 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}10y^2-32y+6\end{aligned} $$ | |
| ① | Multiply $ \color{blue}{2} $ by $ \left( 5y-1\right) $ $$ \color{blue}{2} \cdot \left( 5y-1\right) = 10y-2 $$ |
| ② | Multiply each term of $ \left( \color{blue}{10y-2}\right) $ by each term in $ \left( y-3\right) $. $$ \left( \color{blue}{10y-2}\right) \cdot \left( y-3\right) = 10y^2-30y-2y+6 $$ |
| ③ | Combine like terms: $$ 10y^2 \color{blue}{-30y} \color{blue}{-2y} +6 = 10y^2 \color{blue}{-32y} +6 $$ |
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