geduiwelh

2020-12-02

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

SoosteethicU

Skilled2020-12-03Added 102 answers

Formula used:

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

$ab+ac+bd+cd=a(b+c)+d(b+c)$

$=(a+d)(b+c)$

Or,

$ab-ac+bd-cd=a(b-c)+d(b-c)$

$=(a+d)(b-c)$

Calculation:

Consider the polynomial$2{x}^{3}+6{x}^{2}+x+3$ .

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

$2{x}^{3}+6{x}^{2}+x+3=(2{x}^{3}+6{x}^{2})+(x+3)$

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

As,$(x+3)$ is the common factor of the polynomial,

The polynomial can be factorized as,

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

Therefore, the factorization of the polynomial$2{x}^{3}+6{x}^{2}+x+3$ is $(x+3)(2{x}^{2}+1)$ .

The factors of a polynomial can be find 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 find by factor by grouping as,

As,

The polynomial can be factorized as,

Therefore, the factorization of the polynomial

$\frac{20b}{{\left(4{b}^{3}\right)}^{3}}$

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