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Factorize the following algebraic expressions:
(i) $a^2+4ab+3b^2$
(ii) $96-4x-x^2$
(iii) $a^4+3a^2+4$
Given:
The given expressions are:
(i) $a^2+4ab+3b^2$.
(ii) $96-4x-x^2$
(iii) $a^4+3a^2+4$
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 $a^2+4ab+3b^2$.
$a^2+4ab+3b^2$ can be written as,
By splitting and grouping the terms we can factorize the given expression.
$a^2+4ab+3b^2=a^2+ab+3ab+3b^2$ [Since $4ab=ab+3ab$]
Therefore,
$a^2+4ab+3b^2=a^2+ab+3ab+3b^2$
$a^2+4ab+3b^2=a(a+b)+3b(a+b)$
$a^2+4ab+3b^2=(a+b)(a+3b)$
Hence, the given expression can be factorized as $(a+b)(a+3b)$.
(ii) The given expression is $96-4x-x^2$.
By splitting and grouping the terms we can factorize the given expression.
$96-4x-x^2$ can be written as,
$96-4x-x^2=-(x^2+4x-96)$
$96-4x-x^2=-(x^2+12x-8x-96)$ (Since $4x=12x-8x$)
Therefore,
$96-4x-x^2=-(x^2+12x-8x-96)$
$96-4x-x^2=-[x(x+12)-8(x+12)]$
$96-4x-x^2=-(x+12)(x-8)$
$96-4x-x^2=(x+12)(8-x)$ [Since $-(x-8)=(-x+8)=(8-x)$]
Hence, the given expression can be factorized as $(x+12)(8-x)$.
(iii) The given expression is $a^4+3a^2+4$.
By splitting and grouping the terms we can factorize the given expression.
$a^4+3a^2+4$ can be written as,
$a^4+3a^2+4=a^4+4a^2-a^2+4$ (Since $3a^2=4a^2-a^2$)
$a^4+3a^2+4=(a^2)^2+2(a^2)(2)+2^2-a^2$
$a^4+3a^2+4=(a^2+2)^2-a^2$ [Since $(m+n)^2=m^2+2mn+n^2$]
Therefore, by using the formula $m^2-n^2=(m+n)(m-n)$, we can factorize the given expression.
$a^4+3a^2+4=(a^2+2)^2-a^2$
$a^4+3a^2+4=(a^2+2+a)(a^2+2-a)$
Hence, the given expression can be factorized as $(a^2+a+2)(a^2-a+2)$.
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