Need to calculate:The factorization of x^{3}-7x^{2}+4x-28.

Need to calculate:The factorization of ${x}^{3}-7{x}^{2}+4x-28$.
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Formula used:
The factors of a polynomial can be found 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 ${x}^{3}—7{x}^{2}+4x-28$.
This is a four term polynomial, factorization of this polynomial can be find by factor by grouping as,
${x}^{3}-7{x}^{2}+4x-28=\left({x}^{3}-7{x}^{2}\right)+\left(4x-28\right)$
$={x}^{2}\left(x-7\right)+4\left(x-7\right)$
As, $\left(x-7\right)$ is the common factor of the polynomial,
The polynomial can be factorized as,
${x}^{3}-7{x}^{2}+4x-28={x}^{2}\left(x-7\right)+4\left(x-7\right)$
$=\left(x-7\right)\left({x}^{2}+4\right)$
Therefore, the factorization of the polynomial is $\left(x—7\right)\left({x}^{2}+4\right)$.