naivlingr

2020-11-08

Need to calculate:The factorization of the polynomial, $3{x}^{2}-2x+3x-2$.

Tuthornt

Formula used:
The factors of a polynomial can be find by taking a common factor and this method is called factor by grouping,
$ab+ac+bd+cd=a\left(b+c\right)+d\left(b+c\right)$
$=\left(a+d\right)\left(b+c\right)$
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 $3{x}^{2}—2x+3x—2$.
This is a four term polynomial, factorization of this polynomial can be find by factor by grouping as,
$3{x}^{2}—2x+3x—2=\left(3{x}^{2}—2x\right)+\left(3x—2\right)$
$=x\left(3x-2\right)+1\left(3x-2\right)$
As, the binomial $\left(3x—2\right)$ is the common factor of the polynomial.
The polynomial can be factorized as,
$3{x}^{2}—2x+3x-2=x\left(3x—2\right)+1\left(3x-2\right)$
$=\left(3x-2\right)\left(x+1\right)$
Therefore, the factorization of the polynomial is $\left(3x—2\right)\left(x+1\right)$.

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