# Need to calculate: The factorization of the polynomial ax^{2}+ay+bx^{2}+by.

Need to calculate: The factorization of the polynomial $a{x}^{2}+ay+b{x}^{2}+by$.
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Tasneem Almond
Formula used:
The factors of a polynomial can be found by taking a common factor and this method is called factor by grouping,
$ab+ac+bd+cd=a\left(b+c\right)+d\left(b+c\right)$
$=\left(a+d\right)\left(b+c\right)$
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, $a{x}^{2}+ay+b{x}^{2}+by$.
This is a four terms polynomial, factorization of this polynomial can be found by factor by grouping as,
$a{x}^{2}+ay+b{x}^{2}+by=\left(a{x}^{2}+ay\right)+\left(b{x}^{2}+by\right)$
$=a\left({x}^{2}+y\right)+b\left({x}^{2}+y\right)$
As, $\left({x}^{2}+y\right)$ is the common factor of the polynomial factor it out as follows,
$a{x}^{2}+ay+b{x}^{2}+by=a\left({x}^{2}+y\right)+b\left({x}^{2}+y\right)$
$=\left({x}^{2}+y\right)\left(a+b\right)$
The factorization of the polynomial $a{x}^{2}+ay+b{x}^{2}+$ by is $\left({x}^{2}+y\right)\left(a+b\right)$.
Check the result as follows,
$\left({x}^{2}+y\right)\left(a+b\right)={x}^{2}\ast a+{x}^{2}\ast b+y\ast a+y\ast b$
$=a{x}^{2}+b{x}^{2}+ay+by$
$=a{x}^{2}+ay+b{x}^{2}+by$
Thus, the factorization of the polynomial $a{x}^{2}+ay+b{x}^{2}+$ by is $\left({x}^{2}+y\right)\left(a+b\right)$.