Question
Answer
$$(8x+2)^4-2(x-1)(xxxx+3)=-2x^5+4098x^4+4096x^3+1536x^2+250x+22$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(8x+2)^4-2(x-1)(xxxx+3)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(8x+2)^4-2(x-1)(x^4+3) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}4096x^4+4096x^3+1536x^2+256x+16-2(x-1)(x^4+3) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}4096x^4+4096x^3+1536x^2+256x+16-(2x-2)(x^4+3) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle7}{\textcircled {7}} } }}}4096x^4+4096x^3+1536x^2+256x+16-(2x^5+6x-2x^4-6) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle8}{\textcircled {8}} } }}}4096x^4+4096x^3+1536x^2+256x+16-2x^5-6x+2x^4+6 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle9}{\textcircled {9}} } }}}-2x^5+4098x^4+4096x^3+1536x^2+250x+22\end{aligned} $$ | |
| ① | $$ x x x x = x^{1 + 1 + 1 + 1} = x^4 $$ |
| ② | $$ (8x+2)^4 = (8x+2)^2 \cdot (8x+2)^2 $$ |
| ③ | Find $ \left(8x+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}{ 8x } $ and $ B = \color{red}{ 2 }$. $$ \begin{aligned}\left(8x+2\right)^2 = \color{blue}{\left( 8x \right)^2} +2 \cdot 8x \cdot 2 + \color{red}{2^2} = 64x^2+32x+4\end{aligned} $$ |
| ④ | Multiply each term of $ \left( \color{blue}{64x^2+32x+4}\right) $ by each term in $ \left( 64x^2+32x+4\right) $. $$ \left( \color{blue}{64x^2+32x+4}\right) \cdot \left( 64x^2+32x+4\right) = \\ = 4096x^4+2048x^3+256x^2+2048x^3+1024x^2+128x+256x^2+128x+16 $$ |
| ⑤ | Combine like terms: $$ 4096x^4+ \color{blue}{2048x^3} + \color{red}{256x^2} + \color{blue}{2048x^3} + \color{green}{1024x^2} + \color{orange}{128x} + \color{green}{256x^2} + \color{orange}{128x} +16 = \\ = 4096x^4+ \color{blue}{4096x^3} + \color{green}{1536x^2} + \color{orange}{256x} +16 $$ |
| ⑥ | Multiply $ \color{blue}{2} $ by $ \left( x-1\right) $ $$ \color{blue}{2} \cdot \left( x-1\right) = 2x-2 $$ |
| ⑦ | Multiply each term of $ \left( \color{blue}{2x-2}\right) $ by each term in $ \left( x^4+3\right) $. $$ \left( \color{blue}{2x-2}\right) \cdot \left( x^4+3\right) = 2x^5+6x-2x^4-6 $$ |
| ⑧ | Remove the parentheses by changing the sign of each term within them. $$ - \left( 2x^5+6x-2x^4-6 \right) = -2x^5-6x+2x^4+6 $$ |
| ⑨ | Combine like terms: $$ \color{blue}{4096x^4} +4096x^3+1536x^2+ \color{red}{256x} + \color{green}{16} -2x^5 \color{red}{-6x} + \color{blue}{2x^4} + \color{green}{6} = \\ = -2x^5+ \color{blue}{4098x^4} +4096x^3+1536x^2+ \color{red}{250x} + \color{green}{22} $$ |
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