Question
Answer
$$(4-r)(-(8-r))=-r^2+12r-32$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(4-r)(-(8-r))& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(4-r)\cdot(-8+r) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}-32+4r+8r-r^2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}-r^2+12r-32\end{aligned} $$ | |
| ① | Remove the parentheses by changing the sign of each term within them. $$ - \left(8-r \right) = -8+r $$ |
| ② | Multiply each term of $ \left( \color{blue}{4-r}\right) $ by each term in $ \left( -8+r\right) $. $$ \left( \color{blue}{4-r}\right) \cdot \left( -8+r\right) = -32+4r+8r-r^2 $$ |
| ③ | Combine like terms: $$ -32+ \color{blue}{4r} + \color{blue}{8r} -r^2 = -r^2+ \color{blue}{12r} -32 $$ |
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