Answer
$$0.5((x+1)^4-x^4)+0.5((x-1)^4-x^4)=0$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}0.5((x+1)^4-x^4)+0.5((x-1)^4-x^4)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} \htmlClass{explanationCircle explanationCircle7}{\textcircled {7}} } }}}0.5(x^4+4x^3+6x^2+4x+1-x^4)+0.5(x^4-4x^3+6x^2-4x+1-x^4) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle8}{\textcircled {8}} } }}}0.5(4x^3+6x^2+4x+1)+0.5(-4x^3+6x^2-4x+1) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle9}{\textcircled {9}} } }}}0x^3+0x^2+0x+0+0x^3+0x^2+0x+0 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle10}{\textcircled {10}} } }}}0\end{aligned} $$ | |
| ① | $$ (x+1)^4 = (x+1)^2 \cdot (x+1)^2 $$ |
| ② | Find $ \left(x+1\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}{ 1 }$. $$ \begin{aligned}\left(x+1\right)^2 = \color{blue}{x^2} +2 \cdot x \cdot 1 + \color{red}{1^2} = x^2+2x+1\end{aligned} $$ |
| ③ | Multiply each term of $ \left( \color{blue}{x^2+2x+1}\right) $ by each term in $ \left( x^2+2x+1\right) $. $$ \left( \color{blue}{x^2+2x+1}\right) \cdot \left( x^2+2x+1\right) = x^4+2x^3+x^2+2x^3+4x^2+2x+x^2+2x+1 $$ |
| ④ | Combine like terms: $$ x^4+ \color{blue}{2x^3} + \color{red}{x^2} + \color{blue}{2x^3} + \color{green}{4x^2} + \color{orange}{2x} + \color{green}{x^2} + \color{orange}{2x} +1 = x^4+ \color{blue}{4x^3} + \color{green}{6x^2} + \color{orange}{4x} +1 $$$$ (x-1)^4 = (x-1)^2 \cdot (x-1)^2 $$ |
| ⑤ | Find $ \left(x-1\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}{ 1 }$. $$ \begin{aligned}\left(x-1\right)^2 = \color{blue}{x^2} -2 \cdot x \cdot 1 + \color{red}{1^2} = x^2-2x+1\end{aligned} $$ |
| ⑥ | Multiply each term of $ \left( \color{blue}{x^2-2x+1}\right) $ by each term in $ \left( x^2-2x+1\right) $. $$ \left( \color{blue}{x^2-2x+1}\right) \cdot \left( x^2-2x+1\right) = x^4-2x^3+x^2-2x^3+4x^2-2x+x^2-2x+1 $$ |
| ⑦ | Combine like terms: $$ x^4 \color{blue}{-2x^3} + \color{red}{x^2} \color{blue}{-2x^3} + \color{green}{4x^2} \color{orange}{-2x} + \color{green}{x^2} \color{orange}{-2x} +1 = x^4 \color{blue}{-4x^3} + \color{green}{6x^2} \color{orange}{-4x} +1 $$ |
| ⑧ | Combine like terms: $$ \, \color{blue}{ \cancel{x^4}} \,+4x^3+6x^2+4x+1 \, \color{blue}{ -\cancel{x^4}} \, = 4x^3+6x^2+4x+1 $$Combine like terms: $$ \, \color{blue}{ \cancel{x^4}} \,-4x^3+6x^2-4x+1 \, \color{blue}{ -\cancel{x^4}} \, = -4x^3+6x^2-4x+1 $$ |
| ⑨ | Multiply $ \color{blue}{0} $ by $ \left( 4x^3+6x^2+4x+1\right) $ $$ \color{blue}{0} \cdot \left( 4x^3+6x^2+4x+1\right) = 0x^30x^20x0 $$Multiply $ \color{blue}{0} $ by $ \left( -4x^3+6x^2-4x+1\right) $ $$ \color{blue}{0} \cdot \left( -4x^3+6x^2-4x+1\right) = 0x^30x^20x0 $$ |
| ⑩ | Combine like terms: $$ \, \color{blue}{ \cancel{0x^3}} \, \, \color{green}{ \cancel{0x^2}} \, \, \color{blue}{ \cancel{0x}} \, \, \color{green}{ \cancel{0}} \, \, \color{blue}{ \cancel{0x^3}} \, \, \color{green}{ \cancel{0x^2}} \, \, \color{blue}{ \cancel{0x}} \, \, \color{green}{ \cancel{0}} \, = \color{green}{0} $$ |
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