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Factorize the algebraic expression $9z^2-x^2+4xy-4y^2$.
Given:
The given algebraic expression is $9z^2-x^2+4xy-4y^2$.
To do:
We have to factorize the expression $9z^2-x^2+4xy-4y^2$.
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.
$9z^2-x^2+4xy-4y^2$ can be written as,
$9z^2-x^2+4xy-4y^2=9z^2-(x^2-4xy+4y^2)$
$9z^2-x^2+4xy-4y^2=9z^2-[x^2-2(x)(2y)+(2y)^2]$ [Since $x^2=(x)^2, 4y^2=(2y)^2$ and $4xy=2(x)(2y)$]
Here, we can observe that the given expression is of the form $m^2-2mn+n^2$. So, by using the formula $(m-n)^2=m^2-2mn+n^2$, we can factorize the given expression.
Here,
$m=x$ and $n=2y$
Therefore,
$9z^2-x^2+4xy-4y^2=9z^2-[x^2-2(x)(2y)+(2y)^2]$
$9z^2-x^2+4xy-4y^2=9z^2-[(x-2y)^2]$
Now,
$9z^2-[(x-2y)^2]$ can be written as,
$9z^2-[(x-2y)^2]=(3z)^2-(x-2y)^2$ [Since $9z^2=(3z)^2$]
Using the formula $a^2-b^2=(a+b)(a-b)$, we can factorize $(3z)^2-(x-2y)^2$ as,
$9z^2-(x-2y)^2=(3z)^2-(x-2y)^2$
$9z^2-(x-2y)^2=(3z+x-2y)[3z-(x-2y)]$
$9z^2-(x-2y)^2=(x-2y+3z)(-x+2y+3z)$
Hence, the given expression can be factorized as $(x-2y+3z)(-x+2y+3z)$.
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