How do you simplify #(25-a^2) / (a^2 +a -30)#?

3 Answers
Hihi
Mar 13, 2018

#(5+a)/(-a-6)#

Explanation:

Given the following identities:

#x^2-y^2=(x+y)(x-y)#

#x^2+(a+b)x+ab=(x+a)(x+b)#

Therefore,

#(25-a^2)/(a^2+a-30)#

#=[(5+a)(5-a)]/[(a+6)(a-5)]#

#=[(5+a)(5-a)]/[(a+6)(-1)(5-a)]#

#=(5+a)/(-a-6)#

Note: for the second identity, it is rather more on factorization than a real identity. Hence, more practice could yield faster calculation speed and accuracy.

sjc
Mar 13, 2018

#-(a+5)/(a+6)#

Explanation:

#(1) " "#Factorise top and bottom

#(a)" "#the top with difference of squares

#(b)" "#the bottom with usual quadratic form

#(2)" "#then cancel down

#(25-a^2)/(a^2+a-30)#

#=((5-a)(5+a))/((a+6)(a-5)#

#=(-cancel((a-5))(5+a))/((a+6)cancel((a-5))#

#=-(a+5)/(a+6)#

F. Javier B.
Mar 13, 2018

Using notable identityties. See below

Explanation:

First: we note that #25-a^2=(5+a)(5-a)#

Secondly we look for zeros on denominator in order to factorize it

#a=(-b+-sqrt(b^2-4ac))/(2a)=(-1+-sqrt(1^2-4·(-30)·17))/2#

#a=(-1+-sqrt121)/2#. This give us two solutions (roots) of denominator expresion #a=-6# and #a=5#. For this reason we can express denominator as #(a+6)(a-5).

Summarizing all results, we have

#((5+a)(5-a))/((a+6)(a-5))=-((5+a)(cancel(a-5)))/((a+6)(cancel(a-5)))=-(5+a)/(a+6)#

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