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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 $(7+a-4b)(7-a+4b)$.
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