# Need to calculate:The factorization of the polynomial, 2x^{4}+6x^{2}+5x^{2}+15.

Need to calculate:The factorization of the polynomial, $2{x}^{4}+6{x}^{2}+5{x}^{2}+15$.
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Formula used:
The factors of a polynomial can be find by taking a common factor and this method is called factor by grouping,
$ab+ac+bd+cd=a\left(b+c\right)+d\left(b+c\right)$
$=\left(a+d\left(b+c\right)$
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 provided polynomial,
$2{x}^{4}+6{x}^{2}+5{x}^{2}+15$
Since, this is a four term polynomial, factorization of this polynomial can be find with the help of factor by grouping as,
$2{x}^{4}+6{x}^{2}+5{x}^{2}+15=\left(2{x}^{4}+6{x}^{2}\right)+\left(5{x}^{2}+15\right)$
$=2{x}^{2}\left({x}^{2}+3\right)+5\left({x}^{2}+3\right)$
As, $\left(x2+3\right)$ is the common factor of the polynomial,
Therefore, the polynomial can be factorized as,
$2{x}^{4}+6{x}^{2}+5{x}^{2}+15=2{x}^{2}\left({x}^{2}+3\right)+5\left({x}^{2}+3\right)$
$=\left({x}^{2}+3\right)\left(2{x}^{2}+5\right)$
Hence, the factorization of the polynomial $2{x}^{4}+6{x}^{2}+5{x}^{2}+15$ is $\left({x}^{2}+3\right)\left(2{x}^{2}+5\right)$.