termegolz6

2022-07-14

Factor. ${x}^{5}+x+1$

Danica Ray

Expert

We can add and subtract $\phantom{\rule{thickmathspace}{0ex}}+{x}^{2}-{x}^{2}=0$ without changing the expression:
$\begin{array}{rl}{x}^{5}+x+1& =\left({x}^{5}-{x}^{2}\right)+\left({x}^{2}+x+1\right)\\ \\ & ={x}^{2}\left({x}^{3}-1\right)+\left({x}^{2}+x+1\right)\\ \\ & ={x}^{2}\left(x-1\right)\mathbf{\left(}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{x}\mathbf{+}\mathbf{1}\mathbf{\right)}+\mathbf{\left(}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{x}\mathbf{+}\mathbf{1}\mathbf{\right)}\end{array}$
Now factor out the common factor...which gives us
$\left({x}^{2}\left(x-1\right)+1\right)\left({x}^{2}+x+1\right)=\left({x}^{3}-{x}^{2}+1\right)\left({x}^{2}+x+1\right)$

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