How do you simplify #(x^2-6x+9)/(81-x^4)#?

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Answer 1 · 2016-06-06T07:40:24

#(x^2-6x+9)/(81-x^4)=(3-x)/((9+x^2)(3+x))#

To simplify #(x^2-6x+9)/(81-x^4)#, first factorize polynomials in numerator and denominator.

#x^2-6x+9# is complete square of the type #(a-b)^2=a^2-2ab+b^2#, as #x^2-6x+9=(x)^2-2xx x xx 3 +3^2#. Hence,

#x^2-6x+9=(x-3)^2#.

This can also be written as #9-6x+x^2=(3-x)^2#

For factorizing #81-x^4#, we use the identity #(a^2-b^2)=(a+b)(a-b)#

Hence #81-x^4=(9)^2-*x^2)^2=(9+x^2)(9+x^2)=(9+x^2)(3^2-x^2)#

= #(9+x^2)(3+x)(3-x)#

Hence #(x^2-6x+9)/(81-x^4)=(3-x)^2/((9+x^2)(3+x)(3-x))#

= #(3-x)/((9+x^2)(3+x))#