Answer
$$(14+22m)^2-4(m^2+1)\cdot153=-128m^2+616m-416$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(14+22m)^2-4(m^2+1)\cdot153& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}196+616m+484m^2-4(m^2+1)\cdot153 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}196+616m+484m^2-(4m^2+4)\cdot153 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}196+616m+484m^2-(612m^2+612) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}196+616m+484m^2-612m^2-612 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}-128m^2+616m-416\end{aligned} $$ | |
| ① | Find $ \left(14+22m\right)^2 $ using formula. $$ (A + B)^2 = \color{blue}{A^2} + 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ 14 } $ and $ B = \color{red}{ 22m }$. $$ \begin{aligned}\left(14+22m\right)^2 = \color{blue}{14^2} +2 \cdot 14 \cdot 22m + \color{red}{\left( 22m \right)^2} = 196+616m+484m^2\end{aligned} $$ |
| ② | Multiply $ \color{blue}{4} $ by $ \left( m^2+1\right) $ $$ \color{blue}{4} \cdot \left( m^2+1\right) = 4m^2+4 $$ |
| ③ | $$ \left( \color{blue}{4m^2+4}\right) \cdot 153 = 612m^2+612 $$ |
| ④ | Remove the parentheses by changing the sign of each term within them. $$ - \left( 612m^2+612 \right) = -612m^2-612 $$ |
| ⑤ | Combine like terms: $$ \color{blue}{196} +616m+ \color{red}{484m^2} \color{red}{-612m^2} \color{blue}{-612} = \color{red}{-128m^2} +616m \color{blue}{-416} $$ |
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