# Need to calculate:The factorization of 2x^{3}-8x^{2}-9x+36.

Need to calculate:The factorization of $2{x}^{3}-8{x}^{2}-9x+36$.
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Theodore Schwartz
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 $2{x}^{3}-8{x}^{2}-9x+36$.
This is a four term polynomial, factorization of this polynomial can be find by factor by grouping as,
$2{x}^{3}-8{x}^{2}-9x+36=\left(2{x}^{3}-8{x}^{2}\right)-\left(9x-36\right)$
$=2{x}^{2}\left(x-4\right)-9\left(x-4\right)$
As, $\left(x-4\right)$ is the common factor of the polynomial,
The polynomial can be factorized as,
$2{x}^{2}\left(x-4\right)-9\left(x-4\right)=\left(x-4\right)\left(2{x}^{2}-9\right)$
Therefore, the factorization of the polynomial $2{x}^{3}-8{x}^{2}-9x+36$ is $\left(x-4\right)\left(2{x}^{2}-9\right)$.