Simplify these fractions as far as possible?

Question

Whats the method used to simplify these fractions.
1) 2x^2 + 7x +6
(X-5)(X+2)
2) 2x^2 + 9x-18
(x+6)(x+1)
3) 3x^2 - 7x+2
(3x-1)(x+2)
its the top equation over the bottom
i need to know how they are done.
Thanks

Answer 1

Factor the top numbers..

1) (2x+3)(x+2)
(x-5)(x+2)

Cancel the (x+2) on bottom and top and you get:

(2x+3)/(x-5)

2) (2x-3)(x+6)
(x+6)(x+1)

Cancel out the (x+6) and you get:

(2x-3)/(x+1)

3) (3x-1)(x-2)
(3x-1)(x+2)

Cancel out the (3x-1) and you get :

(x-2)/(x+2)

Answer 2

Simplify these fractions as far as possible?
Whats the method used to simplify these fractions.
1) 2x^2 + 7x +6
------------------- =
(X-5)(X+2)

The numerator can be written as 2x + 3 and x + 2 and thus x + 2 gets cancelled. The resultant fraction is (2x + 3) / (x - 5)


2) 2x^2 + 9x -18
-----------------------
(x+6)(x+1)

The numerator can be written as 2x - 3 and x + 6 and thus x + 6 gets cancelled. The fraction becomes 2x - 3 divided by x + 1

3) 3x^2 - 7x+2
--------------------
(3x-1)(x+2)

The numerator can be written as 3x - 1 and x - 2 and thus the 3x - 1 get cancelled. So, the fraction becomes (x - 2) / (x + 2)

Look for one of the expressions given in the denominator and see how the equation in the numerator can be factorised using the expressions in the denominator.