babeeb0oL

2021-02-14

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

SchulzD

Skilled2021-02-15Added 83 answers

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$6{z}^{3}+3{z}^{2}+2z+1$ .

This is a four term polynomial, factorization of this polynomial can be found by factor by grouping as,

$6{z}^{3}+3{z}^{2}+2z+1=(6{z}^{3}+3{z}^{2})+(2z+1)$

$=3{z}^{2}(2z+1)+1(2z+1)$

As,$(2z+1)$ is the common factor of the polynomial,

The polynomial can be factorized as,

$6{z}^{3}+3{z}^{2}+2z+1=3{z}^{2}(2z+1)+1(2z+1)$

$=2z+1)(3{z}^{2}+1)$

Therefore, the factorization of the polynomial is$(2z+1)(3{z}^{2}+1)$ .

The factors of a polynomial can be found by taking a common factor and this method is called factor by grouping,

Or,

Calculation:

Consider the polynomial

This is a four term polynomial, factorization of this polynomial can be found by factor by grouping as,

As,

The polynomial can be factorized as,

Therefore, the factorization of the polynomial is