Question
Answer
$$-(p+0.2)^2+12(p+0.2)-20=-p^2+12p-20$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}-(p+0.2)^2+12(p+0.2)-20& \xlongequal{ }-(1p^2+0p+0)+12(p+0.2)-20 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}-p^2+0p+0+12(p+0.2)-20 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}-p^2+0p+0+12p+0-20 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}-p^2+12p-20\end{aligned} $$ | |
| ① | Remove the parentheses by changing the sign of each term within them. $$ - \left(p^20p0 \right) = -p^20p0 $$ |
| ② | Multiply $ \color{blue}{12} $ by $ \left( p0\right) $ $$ \color{blue}{12} \cdot \left( p0\right) = 12p0 $$ |
| ③ | Combine like terms: $$ -p^2 \color{blue}{0p} \, \color{red}{ \cancel{0}} \,+ \color{blue}{12p} \, \color{red}{ \cancel{0}} \, = -p^2+ \color{blue}{12p} $$ |
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