Answer
$$(x+5)(3x-1)-2(x-4)^2=x^2+30x-37$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(x+5)(3x-1)-2(x-4)^2& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(x+5)(3x-1)-2(x^2-8x+16) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}3x^2-x+15x-5-(2x^2-16x+32) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}3x^2+14x-5-(2x^2-16x+32) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}3x^2+14x-5-2x^2+16x-32 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}x^2+30x-37\end{aligned} $$ | |
| ① | Find $ \left(x-4\right)^2 $ using formula. $$ (A - B)^2 = \color{blue}{A^2} - 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ x } $ and $ B = \color{red}{ 4 }$. $$ \begin{aligned}\left(x-4\right)^2 = \color{blue}{x^2} -2 \cdot x \cdot 4 + \color{red}{4^2} = x^2-8x+16\end{aligned} $$ |
| ② | Multiply each term of $ \left( \color{blue}{x+5}\right) $ by each term in $ \left( 3x-1\right) $. $$ \left( \color{blue}{x+5}\right) \cdot \left( 3x-1\right) = 3x^2-x+15x-5 $$Multiply $ \color{blue}{2} $ by $ \left( x^2-8x+16\right) $ $$ \color{blue}{2} \cdot \left( x^2-8x+16\right) = 2x^2-16x+32 $$ |
| ③ | Combine like terms: $$ 3x^2 \color{blue}{-x} + \color{blue}{15x} -5 = 3x^2+ \color{blue}{14x} -5 $$ |
| ④ | Remove the parentheses by changing the sign of each term within them. $$ - \left( 2x^2-16x+32 \right) = -2x^2+16x-32 $$ |
| ⑤ | Combine like terms: $$ \color{blue}{3x^2} + \color{red}{14x} \color{green}{-5} \color{blue}{-2x^2} + \color{red}{16x} \color{green}{-32} = \color{blue}{x^2} + \color{red}{30x} \color{green}{-37} $$ |
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