# Factorize the following algebraic expressions:(i) $25-p^2-q^2-2pq$(ii) $x^2+9y^2-6xy-25a^2$(iii) $49-a^2+8ab-16b^2$

Given:

The given expressions are:

(i) $25-p^2-q^2-2pq$

(ii) $x^2+9y^2-6xy-25a^2$

(iii) $49-a^2+8ab-16b^2$

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 $25-p^2-q^2-2pq$.

$25-p^2-q^2-2pq$ can be written as,

$25-p^2-q^2-2pq=25-[p^2+2pq+q^2]$

$25-p^2-q^2-2pq=5^2-[(p)^2+2(p)(q)+(q)^2]$             [Since $25=5^2$ and $2pq=2(p)(q)$]

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,

$m=p$ and $n=q$

Therefore,

$25-p^2-q^2-2pq=5^2-[(p)^2+2(p)(q)+(q)^2]$

$25-p^2-q^2-2pq=5^2-(p+q)^2$

Now,

Using the formula $m^2-n^2=(m+n)(m-n)$, we can factorize $5^2-(p+q)^2$ as,

$25-p^2-q^2-2pq=5^2-(p+q)^2$

$25-p^2-q^2-2pq=(5+p+q)(5-p-q)$

Hence, the given expression can be factorized as $(p+q+5)(5-p-q)$.

(ii) The given expression is $x^2+9y^2-6xy-25a^2$.

$x^2+9y^2-6xy-25a^2$ can be written as,

$x^2+9y^2-6xy-25a^2=[(x)^2-2(x)(3y)+(3y)^2]-(5a)^2$      [Since $6xy=2(x)(3y), 9y^2=(3y)^2$ and $25a^2=(5a)^2$]

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 $n=3y$

Therefore,

$x^2+9y^2-6xy-25a^2=[(x)^2-2(x)(3y)+(3y)^2]-(5a)^2$

$x^2+9y^2-6xy-25a^2=(x-3y)^2-(5a)^2$

Now,

Using the formula $m^2-n^2=(m+n)(m-n)$, we can factorize $(x-3y)^2-(5a)^2$ as,

$x^2+9y^2-6xy-25a^2=(x-3y)^2-(5a)^2$

$x^2+9y^2-6xy-25a^2=(x-3y+5a)(x-3y-5a)$

Hence, the given expression can be factorized as $(x-3y+5a)(x-3y-5a)$.

(iii) The given expression is $49-a^2+8ab-16b^2$.

$49-a^2+8ab-16b^2$ can be written as,

$49-a^2+8ab-16b^2=7^2-[(a)^2-2(a)(4b)+(4b)^2]$             [Since $49=(7)^2, 8ab=2(a)(4b)$ and $16b^2=(4b)^2$]

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,

$49-a^2+8ab-16b^2=7^2-[(a)^2-2(a)(4b)+(4b)^2]$

$49-a^2+8ab-16b^2=7^2-(a-4b)^2$

Now,

Using the formula $m^2-n^2=(m+n)(m-n)$, we can factorize $7^2-(a-4b)^2$ as,

$49-a^2+8ab-16b^2=7^2-(a-4b)^2$

$49-a^2+8ab-16b^2=(7+a-4b)(7-a+4b)$

Hence, the given expression can be factorized as $(7+a-4b)(7-a+4b)$.

Updated on: 11-Apr-2023

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