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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 $(a+b+c)(a+b-c)$.
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