Factorize the following algebraic expressions:
(i) $a^2-8ab+16b^2-25c^2$
(ii) $x^2-y^2+6y-9$

Given:

The given expressions are:

(i) $a^2-8ab+16b^2-25c^2$

(ii) $x^2-y^2+6y-9$

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 $a^2-8ab+16b^2-25c^2$.

$a^2-8ab+16b^2-25c^2$ can be written as,

$a^2-8ab+16b^2-25c^2=[(a)^2-2(a)(4b)+(4b)^2]-(5c)^2$          [Since $8ab=2(a)(4b), 16b^2=(4b)^2$ and $25c^2=(5c)^2$]

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 $n=4b$ 

Therefore,

$a^2-8ab+16b^2-25c^2=[(a)^2-2(a)(4b)+(4b)^2]-(5c)^2$

$a^2-8ab+16b^2-25c^2=(a-4b)^2-(5c)^2$

Now,

Using the formula $m^2-n^2=(m+n)(m-n)$, we can factorize $(a-4b)^2-(5c)^2$ as,

$a^2-8ab+16b^2-25c^2=(a-4b)^2-(5c)^2$

$a^2-8ab+16b^2-25c^2=(a-4b+5c)(a-4b-5c)$

Hence, the given expression can be factorized as $(a-4b+5c)(a-4b-5c)$.

(ii) The given expression is $x^2-y^2+6y-9$.

$x^2-y^2+6y-9$ can be written as,

$x^2-y^2+6y-9=x^2-[(y)^2-2(y)(3)+(3)^2]$

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,

$m=y$ and $n=3$ 

Therefore,

$x^2-y^2+6y-9=x^2-[(y)^2-2(y)(3)+(3)^2]$

$x^2-y^2+6y-9=x^2-(y-3)^2$

Now,

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

$x^2-y^2+6y-9=x^2-(y-3)^2$

$x^2-y^2+6y-9=(x+y-3)(x-y+3)$

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

Updated on: 11-Apr-2023

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