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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 $(x+1+3y)(x+1-3y)$.
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