Inyalan0

2021-12-10

Factor each polynomial completely. If the polynomial cannot be factored, say it is prime.
${x}^{4}+{x}^{3}+x+1$

Medicim6

Step 1
Given:
Let $p={x}^{4}+{x}^{3}+x+1$
To Find: To factorise the given polynomial
Step 2
Solution:
$p={x}^{4}+{x}^{3}+x+1$
$p={x}^{3}\left(x+1\right)+1\left(x+1\right)$
$p=\left({x}^{3}+1\right)\left(x+1\right)$
$p=\left({x}^{3}+{1}^{3}\right)\left(x+1\right)$
We know that ${a}^{3}+{b}^{3}=\left(a+b\right)\left({a}^{2}-ab+{b}^{2}\right)$
Hence $p=\left(x+1\right)\left[{x}^{2}-\left(x\right)\left(1\right)+{1}^{2}\right]\left(x+1\right)$
$p=\left(x+1\right)\left({x}^{2}-x+1\right)\left(x+1\right)$
$p=\left(x+1\right)\left(x+1\right)\left({x}^{2}-x+1\right)$
Hence ${x}^{4}+{x}^{3}+x+1=\left(x+1\right)\left(x+1\right)\left({x}^{2}-x+1\right)$

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