# Factorize the following algebraic expressions:(i) $25x^2-10x+1-36y^2$(ii) $a^2-b^2+2bc-c^2$(iii) $a^2+2ab+b^2-c^2$

Given:

The given expressions are:

(i) $25x^2-10x+1-36y^2$

(ii) $a^2-b^2+2bc-c^2$

(iii) $a^2+2ab+b^2-c^2$

To do:

We have to factorize the given algebraic expressions.

Solution:

Factorizing algebraic expressions:

Factorizing an algebraic expression implies 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.

(i) The given expression is $25x^2-10x+1-36y^2$.

$25x^2-10x+1-36y^2$ can be written as,

$25x^2-10x+1-36y^2=[(5x)^2-2(5x)(1)+(1)^2]-(6y)^2$                [Since $25x^2=(5x)^2, 10x=2(5x)(1)$ and $36y^2=(6y)^2$]

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,

$m=5x$ and $n=1$

Therefore,

$25x^2-10x+1-36y^2=[(5x)^2-2(5x)(1)+(1)^2]-(6y)^2$

$25x^2-10x+1-36y^2=(5x-1)^2-(6y)^2$

Now,

Using the formula $m^2-n^2=(m+n)(m-n)$, we can factorize $(5x-1)^2-(6y)^2$ as,

$25x^2-10x+1-36y^2=(5x-1)^2-(6y)^2$

$25x^2-10x+1-36y^2=(5x-1+6y)(5x-1-6y)$

Hence, the given expression can be factorized as $(5x+6y-1)(5x-6y-1)$.

(ii) The given expression is $a^2-b^2+2bc-c^2$.

$a^2-b^2+2bc-c^2$ can be written as,

$a^2-b^2+2bc-c^2=a^2-(b^2-2bc+c^2)$

$a^2-b^2+2bc-c^2=a^2-[b^2-2(b)(c)+(c)^2]$

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,

$m=b$ and $n=c$

Therefore,

$a^2-b^2+2bc-c^2=a^2-[b^2-2(b)(c)+(c)^2]$

$a^2-b^2+2bc-c^2=a^2-(b-c)^2$

Now,

Using the formula $m^2-n^2=(m+n)(m-n)$, we can factorize $a^2-(b-c)^2$ as,

$a^2-b^2+2bc-c^2=a^2-(b-c)^2$

$a^2-b^2+2bc-c^2=(a+b-c)(a-b+c)$

Hence, the given expression can be factorized as $(a+b-c)(a-b+c)$.

(iii) The given expression is $a^2+2ab+b^2-c^2$.

$a^2+2ab+b^2-c^2$ can be written as,

$a^2+2ab+b^2-c^2=[(a)^2+2(a)(b)+(b)^2]-(c)^2$

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,

$m=a$ and $n=b$

Therefore,

$a^2+2ab+b^2-c^2=[(a)^2+2(a)(b)+(b)^2]-(c)^2$

$a^2+2ab+b^2-c^2=(a+b)^2-(c)^2$

Now,

Using the formula $m^2-n^2=(m+n)(m-n)$, we can factorize $(a+b)^2-(c)^2$ as,

$a^2+2ab+b^2-c^2=(a+b)^2-(c)^2$

$a^2+2ab+b^2-c^2=(a+b+c)(a+b-c)$

Hence, the given expression can be factorized as $(a+b+c)(a+b-c)$.

Updated on: 10-Apr-2023

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