Ernstfalld

2021-03-06

Need to calculate: The factorization of the polynomial $a{x}^{2}+ay+b{x}^{2}+by$ .

Tasneem Almond

Skilled2021-03-07Added 91 answers

Formula used:

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

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

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

Or,

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

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

Calculation:

Consider the polynomial,$a{x}^{2}+ay+b{x}^{2}+by$ .

This is a four terms polynomial, factorization of this polynomial can be found by factor by grouping as,

$a{x}^{2}+ay+b{x}^{2}+by=(a{x}^{2}+ay)+(b{x}^{2}+by)$

$=a({x}^{2}+y)+b({x}^{2}+y)$

As,$({x}^{2}+y)$ is the common factor of the polynomial factor it out as follows,

$a{x}^{2}+ay+b{x}^{2}+by=a({x}^{2}+y)+b({x}^{2}+y)$

$=({x}^{2}+y)(a+b)$

The factorization of the polynomial$a{x}^{2}+ay+b{x}^{2}+$ by is $({x}^{2}+y)(a+b)$ .

Check the result as follows,

$({x}^{2}+y)(a+b)={x}^{2}\ast a+{x}^{2}\ast b+y\ast a+y\ast b$

$=a{x}^{2}+b{x}^{2}+ay+by$

$=a{x}^{2}+ay+b{x}^{2}+by$

Thus, the factorization of the polynomial$a{x}^{2}+ay+b{x}^{2}+$ by is $({x}^{2}+y)(a+b)$ .

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

Or,

Calculation:

Consider the polynomial,

This is a four terms polynomial, factorization of this polynomial can be found by factor by grouping as,

As,

The factorization of the polynomial

Check the result as follows,

Thus, the factorization of the polynomial