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

Question

Eli Assouline · Accepted Answer

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

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.

Avery Alexander · Answer

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

Savannah Anderson · Answer

Using notable identityties. See below

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)#

Eva Abramson · Answer

undefined To simplify the expression (25-a^2) / (a^2 +a -30), we can factor the numerator and denominator. The numerator can be factored as the difference of squares: (5-a)(5+a). The denominator can be factored as (a+6)(a-5).

Therefore, the expression simplifies to (5-a)(5+a) / (a+6)(a-5).