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Factorize the algebraic expression $4(x+y)(3a-b)+6(x+y)(2b-3a)$.
Given:
The given algebraic expression is $4(x+y)(3a-b)+6(x+y)(2b-3a)$.
To do:
We have to factorize the expression $4(x+y)(3a-b)+6(x+y)(2b-3a)$.
Solution:
Factorizing algebraic expressions:
Factorizing an algebraic expression means writing the expression as a product of two or more factors. Factorization is the reverse of distribution.
An algebraic expression is factored completely when it is written as a product of prime factors.
Here, we can factorize the expression $4(x+y)(3a-b)+6(x+y)(2b-3a)$ by taking out the common factors. The highest common factor(HCF) of an algebraic expression is the highest factor that can be divided into each of the terms with no remainder.
The terms in the given expression are $4(x+y)(3a-b)$ and $6(x+y)(2b-3a)$.
We can observe that $(x+y)$ is common to both the terms.
Therefore, by taking $(x+y)$ as common, we get,
$4(x+y)(3a-b)+6(x+y)(2b-3a)=(x+y)[4(3a-b)+6(2b-3a)]$
Now, taking $2$ common in $[4(3a-b)+6(2b-3a)]$, we get,
$4(x+y)(3a-b)+6(x+y)(2b-3a)=(x+y)2[2(3a-b)+3(2b-3a)]$
$4(x+y)(3a-b)+6(x+y)(2b-3a)=2(x+y)[2(3a)-2(b)+3(2b)-3(3a)]$
$4(x+y)(3a-b)+6(x+y)(2b-3a)=2(x+y)[6a-2b+6b-9a]$
$4(x+y)(3a-b)+6(x+y)(2b-3a)=2(x+y)(-3a+4b)$
$4(x+y)(3a-b)+6(x+y)(2b-3a)=2(x+y)(4b-3a)$
Hence, the given expression can be factorized as $2(x+y)(4b-3a)$.
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