How do you simplify (25-a^2) / (a^2 +a -30)25a2a2+a30?

3 Answers
Mar 13, 2018

(5+a)/(-a-6)5+aa6

Explanation:

Given the following identities:

x^2-y^2=(x+y)(x-y)x2y2=(x+y)(xy)

x^2+(a+b)x+ab=(x+a)(x+b)x2+(a+b)x+ab=(x+a)(x+b)

Therefore,

(25-a^2)/(a^2+a-30)25a2a2+a30

=[(5+a)(5-a)]/[(a+6)(a-5)]=(5+a)(5a)(a+6)(a5)

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

=(5+a)/(-a-6)=5+aa6

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.

Mar 13, 2018

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

Explanation:

(1) " "(1) Factorise top and bottom

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

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

(2)" "(2) then cancel down

(25-a^2)/(a^2+a-30)25a2a2+a30

=((5-a)(5+a))/((a+6)(a-5)=(5a)(5+a)(a+6)(a5)

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

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

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)