1. How to Factorise This?

$\displaystyle x^5+x^3+x$

Answer: $\displaystyle x(1-x+x^2)(1+x+x^2)$

2. Originally Posted by cloud5
$\displaystyle x^5+x^3+x$

Answer: $\displaystyle x(1-x+x^2)(1+x+x^2)$
\displaystyle \begin{aligned} x^5+x^3+x &= x(x^4 + x^2 + 1) \\ &= x(x^4\;{\color{red}+\;2x^2} + 1\;{\color{red}-\;x^2}) \\ &= x[(x^2 + 1)^2 - x^2] \\ &= x[(x^2 + 1) + x][(x^2 + 1) - x] \\ &= x(x^2 + x + 1)(x^2 - x + 1) \end{aligned}

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3. Originally Posted by yeongil
\displaystyle \begin{aligned} x^5+x^3+x &= x(x^4 + x^2 + 1) \\ &= x(x^4\;{\color{red}+\;2x^2} + 1\;{\color{red}-\;x^2}) \\ &= x[(x^2 + 1)^2 - x^2] \\ &= x[(x^2 + 1) + x][(x^2 + 1) - x] \\ &= x(x^2 + x + 1)(x^2 - x + 1) \end{aligned}

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How did you get the red colored numbers? I don't understand...

4. Originally Posted by cloud5
How did you get the red colored numbers? I don't understand...
\displaystyle \begin{aligned} x^5+x^3+x &= x(x^4 + {\color{blue}x^2} + 1) \\ &= x(x^4\;{\color{red}+\;2x^2} + 1\;{\color{red}-\;x^2}) \\ \end{aligned}
...
Do you see the x-squared term in the previous step that I colored in blue? I replaced that with $\displaystyle 2x^2 - x^2$, which is in red and separated. After all, $\displaystyle x^2 = 2x^2 - x^2$.

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