Question
Answer
$$-(x+7)(x-7)+4(2x+1)(3x-4)=23x^2-20x+33$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}-(x+7)(x-7)+4(2x+1)(3x-4)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}-(x^2-7x+7x-49)+(8x+4)(3x-4) \xlongequal{ } \\[1 em] & \xlongequal{ }-(x^2-49)+(8x+4)(3x-4) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}-x^2+49+(8x+4)(3x-4) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}-x^2+49+24x^2-32x+12x-16 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}-x^2+49+24x^2-20x-16 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}23x^2-20x+33\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{x+7}\right) $ by each term in $ \left( x-7\right) $. $$ \left( \color{blue}{x+7}\right) \cdot \left( x-7\right) = x^2 -\cancel{7x}+ \cancel{7x}-49 $$Multiply $ \color{blue}{4} $ by $ \left( 2x+1\right) $ $$ \color{blue}{4} \cdot \left( 2x+1\right) = 8x+4 $$ |
| ② | Remove the parentheses by changing the sign of each term within them. $$ - \left(x^2-49 \right) = -x^2+49 $$ |
| ③ | Multiply each term of $ \left( \color{blue}{8x+4}\right) $ by each term in $ \left( 3x-4\right) $. $$ \left( \color{blue}{8x+4}\right) \cdot \left( 3x-4\right) = 24x^2-32x+12x-16 $$ |
| ④ | Combine like terms: $$ 24x^2 \color{blue}{-32x} + \color{blue}{12x} -16 = 24x^2 \color{blue}{-20x} -16 $$ |
| ⑤ | Combine like terms: $$ \color{blue}{-x^2} + \color{red}{49} + \color{blue}{24x^2} -20x \color{red}{-16} = \color{blue}{23x^2} -20x+ \color{red}{33} $$ |
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