Question
Answer
$$(7x+2)^4-24(3x-18)(4x+3)=2401x^4+2744x^3+888x^2+1736x+1312$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(7x+2)^4-24(3x-18)(4x+3)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}2401x^4+2744x^3+1176x^2+224x+16-24(3x-18)(4x+3) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}2401x^4+2744x^3+1176x^2+224x+16-(72x-432)(4x+3) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}2401x^4+2744x^3+1176x^2+224x+16-(288x^2+216x-1728x-1296) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle7}{\textcircled {7}} } }}}2401x^4+2744x^3+1176x^2+224x+16-(288x^2-1512x-1296) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle8}{\textcircled {8}} } }}}2401x^4+2744x^3+1176x^2+224x+16-288x^2+1512x+1296 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle9}{\textcircled {9}} } }}}2401x^4+2744x^3+888x^2+1736x+1312\end{aligned} $$ | |
| ① | $$ (7x+2)^4 = (7x+2)^2 \cdot (7x+2)^2 $$ |
| ② | Find $ \left(7x+2\right)^2 $ using formula. $$ (A + B)^2 = \color{blue}{A^2} + 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ 7x } $ and $ B = \color{red}{ 2 }$. $$ \begin{aligned}\left(7x+2\right)^2 = \color{blue}{\left( 7x \right)^2} +2 \cdot 7x \cdot 2 + \color{red}{2^2} = 49x^2+28x+4\end{aligned} $$ |
| ③ | Multiply each term of $ \left( \color{blue}{49x^2+28x+4}\right) $ by each term in $ \left( 49x^2+28x+4\right) $. $$ \left( \color{blue}{49x^2+28x+4}\right) \cdot \left( 49x^2+28x+4\right) = \\ = 2401x^4+1372x^3+196x^2+1372x^3+784x^2+112x+196x^2+112x+16 $$ |
| ④ | Combine like terms: $$ 2401x^4+ \color{blue}{1372x^3} + \color{red}{196x^2} + \color{blue}{1372x^3} + \color{green}{784x^2} + \color{orange}{112x} + \color{green}{196x^2} + \color{orange}{112x} +16 = \\ = 2401x^4+ \color{blue}{2744x^3} + \color{green}{1176x^2} + \color{orange}{224x} +16 $$ |
| ⑤ | Multiply $ \color{blue}{24} $ by $ \left( 3x-18\right) $ $$ \color{blue}{24} \cdot \left( 3x-18\right) = 72x-432 $$ |
| ⑥ | Multiply each term of $ \left( \color{blue}{72x-432}\right) $ by each term in $ \left( 4x+3\right) $. $$ \left( \color{blue}{72x-432}\right) \cdot \left( 4x+3\right) = 288x^2+216x-1728x-1296 $$ |
| ⑦ | Combine like terms: $$ 288x^2+ \color{blue}{216x} \color{blue}{-1728x} -1296 = 288x^2 \color{blue}{-1512x} -1296 $$ |
| ⑧ | Remove the parentheses by changing the sign of each term within them. $$ - \left( 288x^2-1512x-1296 \right) = -288x^2+1512x+1296 $$ |
| ⑨ | Combine like terms: $$ 2401x^4+2744x^3+ \color{blue}{1176x^2} + \color{red}{224x} + \color{green}{16} \color{blue}{-288x^2} + \color{red}{1512x} + \color{green}{1296} = \\ = 2401x^4+2744x^3+ \color{blue}{888x^2} + \color{red}{1736x} + \color{green}{1312} $$ |
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