babeeb0oL

2021-02-14

Need to calculate:The factorization of $6{z}^{3}+3{z}^{2}+2z+1$.

SchulzD

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 $6{z}^{3}+3{z}^{2}+2z+1$.
This is a four term polynomial, factorization of this polynomial can be found by factor by grouping as,
$6{z}^{3}+3{z}^{2}+2z+1=\left(6{z}^{3}+3{z}^{2}\right)+\left(2z+1\right)$
$=3{z}^{2}\left(2z+1\right)+1\left(2z+1\right)$
As, $\left(2z+1\right)$ is the common factor of the polynomial,
The polynomial can be factorized as,
$6{z}^{3}+3{z}^{2}+2z+1=3{z}^{2}\left(2z+1\right)+1\left(2z+1\right)$
$=2z+1\right)\left(3{z}^{2}+1\right)$
Therefore, the factorization of the polynomial is $\left(2z+1\right)\left(3{z}^{2}+1\right)$.

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