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Factorize the following algebraic expressions:
(i) $49-x^2-y^2+2xy$
(ii) $a^2+4b^2-4ab-4c^2$
(iii) $x^2-y^2-4xz+4z^2$
Given:
The given expressions are:
(i) $49-x^2-y^2+2xy$
(ii) $a^2+4b^2-4ab-4c^2$
(iii) $x^2-y^2-4xz+4z^2$
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 $49-x^2-y^2+2xy$.
$49-x^2-y^2+2xy$ can be written as,
$49-x^2-y^2+2xy=49-(x^2+y^2-2xy)$
$49-x^2-y^2+2xy=7^2-[(x)^2-2(x)(y)+(y)^2]$ [Since $49=7^2$ and $2xy=2(x)(y)$]
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 $n=y$
Therefore,
$49-x^2-y^2+2xy=7^2-[(x)^2-2(x)(y)+(y)^2]$
$49-x^2-y^2+2xy=7^2-(x-y)^2$
Now,
Using the formula $m^2-n^2=(m+n)(m-n)$, we can factorize $7^2-(x-y)^2$ as,
$49-x^2-y^2+2xy=7^2-(x-y)^2$
$49-x^2-y^2+2xy=(7+x-y)(7-x+y)$
Hence, the given expression can be factorized as $(x-y+7)(-x+y+7)$.
(ii) The given expression is $a^2+4b^2-4ab-4c^2$.
$a^2+4b^2-4ab-4c^2$ can be written as,
$a^2+4b^2-4ab-4c^2=[(a)^2-2(a)(2b)+(2b)^2]-(2c)^2$ [Since $4b^2=(2b)^2, 4ab=2(a)(2b)$ and $4c^2=(2c)^2$]
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 $n=2b$
Therefore,
$a^2+4b^2-4ab-4c^2=[(a)^2-2(a)(2b)+(2b)^2]-(2c)^2$
$a^2+4b^2-4ab-4c^2=(a-2b)^2-(2c)^2$
Now,
Using the formula $m^2-n^2=(m+n)(m-n)$, we can factorize $(a-2b)^2-(2c)^2$ as,
$a^2+4b^2-4ab-4c^2=(a-2b)^2-(2c)^2$
$a^2+4b^2-4ab-4c^2=(a-2b+2c)(a-2b-2c)$
Hence, the given expression can be factorized as $(a-2b+2c)(a-2b-2c)$.
(iii) The given expression is $x^2-y^2-4xz+4z^2$.
$x^2-y^2-4xz+4z^2$ can be written as,
$x^2-y^2-4xz+4z^2=[(x)^2-2(x)(2z)+(2z)^2]-(y)^2$ [Since $4xz=2(x)(2z)$ and $4z^2=(2z)^2$]
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 $n=2z$
Therefore,
$x^2-y^2-4xz+4z^2=[(x)^2-2(x)(2z)+(2z)^2]-(y)^2$
$x^2-y^2-4xz+4z^2=(x-2z)^2-(y)^2$
Now,
Using the formula $m^2-n^2=(m+n)(m-n)$, we can factorize $(x-2z)^2-(y)^2$ as,
$x^2-y^2-4xz+4z^2=(x-2z)^2-(y)^2$
$x^2-y^2-4xz+4z^2=(x-2z+y)(x-2z-y)$
Hence, the given expression can be factorized as $(x+y-2z)(x-y-2z)$.
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