# Factorize the following algebraic expressions:(i) $16-a^6+4a^3b^3-4b^6$(ii) $a^2-2ab+b^2-c^2$(iii) $x^2+2x+1-9y^2$

Given:

The given expressions are:

(i) $16-a^6+4a^3b^3-4b^6$.

(ii) $a^2-2ab+b^2-c^2$

(iii) $x^2+2x+1-9y^2$

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 $16-a^6+4a^3b^3-4b^6$.

$16-a^6+4a^3b^3-4b^6$ can be written as,

$16-a^6+4a^3b^3-4b^6=16-[(a^3)^2-2(a^3)(2b^3)+(2b^3)^2$

$16-a^6+4a^3b^3-4b^6=4^2-[(a^3)^2-2(a^3)(2b^3)+(2b^3)^2]$             [Since $16=4^2, a^6=(a^3)^2, 4b^6=(2b^3)^2$ and $4a^3b^3=2(a^3)(2b^3)$]

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,

$m=a^3$ and $n=2b^3$

Therefore,

$16-a^6+4a^3b^3-4b^6=4^2-[(a^3)^2-2(a^3)(2b^3)+(2b^2)^3]$

$16-a^6+4a^3b^3-4b^6=4^2-(a^3-2b^3)^2$

Now,

Using the formula $m^2-n^2=(m+n)(m-n)$, we can factorize $4^2-(a^3-2b^3)^2$ as,

$16-a^6+4a^3b^3-4b^6=4^2-(a^3-2b^3)^2$

$16-a^6+4a^3b^3-4b^6=(4+a^3-2b^3)(4-a^3+2b^3)$

Hence, the given expression can be factorized as $(4+a^3-2b^3)(4-a^3+2b^3)$.

(ii) 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 $(a-b+c)(a-b-c)$.

(iii) The given expression is $x^2+2x+1-9y^2$.

$x^2+2x+1-9y^2$ can be written as,

$x^2+2x+1-9y^2=[(x)^2+2(x)(1)+(1)^2]-(3y)^2$             [Since $9y^2=(3y)^2$]

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,

$m=x$ and $n=1$

Therefore,

$x^2+2x+1-9y^2=[(x)^2+2(x)(1)+(1)^2]-(3y)^2$

$x^2+2x+1-9y^2=(x+1)^2-(3y)^2$

Now,

Using the formula $m^2-n^2=(m+n)(m-n)$, we can factorize $(x+1)^2-(3y)^2$ as,

$x^2+2x+1-9y^2=(x+1)^2-(3y)^2$

$x^2+2x+1-9y^2=(x+1+3y)(x+1-3y)$

Hence, the given expression can be factorized as $(x+1+3y)(x+1-3y)$.

Updated on: 10-Apr-2023

53 Views