# Solve the factorization of x^{12}+x^{7}+x^{5}+1

Solve the factorization of ${x}^{12}+{x}^{7}+{x}^{5}+1$
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Formula used:
The factors of a polynomial can be find by taking a common factor and this method is called factor by grouping,
$ab+ac+bd+cd=a\left(b+c\right)+d\left(b+c\right)$
$=\left(a+d\right)\left(b+c\right)$
Or,
$ab-ac+bd-cd=a\left(b-c\right)+d\left(b-c\right)$
$=\left(a+d\right)\left(b-c\right)$
Calculation:
Consider the polynomial ${x}^{12}+{x}^{7}+{x}^{5}+1$.
This is a four term polynomial, factorization of this polynomial can be find by factor by grouping as,
${x}^{12}+{x}^{7}+{x}^{5}+1=\left({x}^{12}+{x}^{7}\right)\left({x}^{5}+1\right)$
$={x}^{7}\left({x}^{5}+1\right)+1\left({x}^{5}+1\right)$
As, $\left({x}^{5}+1\right)$ is the common factor of the polynomial,
The polynomial can be factorized as,
${x}^{7}\left({x}^{5}+1\right)+1\left({x}^{5}+1\right)=\left({x}^{5}+1\right)\left({x}^{7}+1\right)$
Therefore, the factorization of the polynomial ${x}^{12}+{x}^{7}+{x}^{5}+1$ is $\left({x}^{5}+1\right)\left({x}^{7}+1\right)$.