# 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 $(x-2y+3z)(-x+2y+3z)$.

Updated on: 10-Apr-2023

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