Answer
$$(10y+6)(3y+7)-(y+2)(y-4)=29y^2+90y+50$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(10y+6)(3y+7)-(y+2)(y-4)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}30y^2+70y+18y+42-(1y^2-4y+2y-8) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}30y^2+88y+42-(1y^2-2y-8) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}30y^2+88y+42-y^2+2y+8 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}29y^2+90y+50\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{10y+6}\right) $ by each term in $ \left( 3y+7\right) $. $$ \left( \color{blue}{10y+6}\right) \cdot \left( 3y+7\right) = 30y^2+70y+18y+42 $$Multiply each term of $ \left( \color{blue}{y+2}\right) $ by each term in $ \left( y-4\right) $. $$ \left( \color{blue}{y+2}\right) \cdot \left( y-4\right) = y^2-4y+2y-8 $$ |
| ② | Combine like terms: $$ 30y^2+ \color{blue}{70y} + \color{blue}{18y} +42 = 30y^2+ \color{blue}{88y} +42 $$Combine like terms: $$ y^2 \color{blue}{-4y} + \color{blue}{2y} -8 = y^2 \color{blue}{-2y} -8 $$ |
| ③ | Remove the parentheses by changing the sign of each term within them. $$ - \left( y^2-2y-8 \right) = -y^2+2y+8 $$ |
| ④ | Combine like terms: $$ \color{blue}{30y^2} + \color{red}{88y} + \color{green}{42} \color{blue}{-y^2} + \color{red}{2y} + \color{green}{8} = \color{blue}{29y^2} + \color{red}{90y} + \color{green}{50} $$ |
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