Need to calculate:The factorization of 6z3+3z2+2z+1.

Formula used:
The factors of a polynomial can be found by taking a common factor and this method is called factor by grouping,
$ab+ac+bd+cd=a(b+c)+d(b+c)$
$=(a+d)(b+c)$
Or,
$ab-ac+bd-cd=a(b-c)+d(b-c)$
$=(a+d)(b-c)$
Calculation:
Consider the polynomial $6z^3+3z^2+2z+1$.
This is a four term polynomial, factorization of this polynomial can be found by factor by grouping as,
$6z^3+3z^2+2z+1=(6z^3+3z^2)+(2z+1)$
$=3z^2(2z+1)+1(2z+1)$
As, $(2z+1)$ is the common factor of the polynomial,
The polynomial can be factorized as,
$6z^3+3z^2+2z+1=3z^2(2z+1)+1(2z+1)$
$=2z+1)(3z^2+1)$
Therefore, the factorization of the polynomial is $(2z+1)(3z^2+1)$.