Answer
$$(6c-2)(4c+2)-(c+7)^2=23c^2-10c-53$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(6c-2)(4c+2)-(c+7)^2& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(6c-2)(4c+2)-(1c^2+14c+49) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}24c^2+12c-8c-4-(1c^2+14c+49) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}24c^2+4c-4-(1c^2+14c+49) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}24c^2+4c-4-c^2-14c-49 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}23c^2-10c-53\end{aligned} $$ | |
| ① | Find $ \left(c+7\right)^2 $ using formula. $$ (A + B)^2 = \color{blue}{A^2} + 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ c } $ and $ B = \color{red}{ 7 }$. $$ \begin{aligned}\left(c+7\right)^2 = \color{blue}{c^2} +2 \cdot c \cdot 7 + \color{red}{7^2} = c^2+14c+49\end{aligned} $$ |
| ② | Multiply each term of $ \left( \color{blue}{6c-2}\right) $ by each term in $ \left( 4c+2\right) $. $$ \left( \color{blue}{6c-2}\right) \cdot \left( 4c+2\right) = 24c^2+12c-8c-4 $$ |
| ③ | Combine like terms: $$ 24c^2+ \color{blue}{12c} \color{blue}{-8c} -4 = 24c^2+ \color{blue}{4c} -4 $$ |
| ④ | Remove the parentheses by changing the sign of each term within them. $$ - \left( c^2+14c+49 \right) = -c^2-14c-49 $$ |
| ⑤ | Combine like terms: $$ \color{blue}{24c^2} + \color{red}{4c} \color{green}{-4} \color{blue}{-c^2} \color{red}{-14c} \color{green}{-49} = \color{blue}{23c^2} \color{red}{-10c} \color{green}{-53} $$ |
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