# Thread: Two quick factoring questions?

1. ## Two quick factoring questions?

Stars are to organize, I am very new here.

How do I simplify:

2y-1***1-4y^2
------ / ---------
2******4

How do I solve:

4*****5****3x
--- = ----- + ----
x-1***2x-2***4

2. question 1:

[ (2y-1)/2 ] / [ (1-4y^2)/4 ]

well you can see that,
1-4y^2
= 1^2 - (2y)^2
= (1 - 2y)(1 + 2y) =====> (x - a)(x + a) = x^2 - a^2

substituting back to the main question

[ (2y-1)/2 ] / [ (1-2y)(1+2y)/4 ]
= [ (2y-1)/2 ] / [ (-1)(2y-1)(1+2y)/4 ]
= [ 4.(2y-1)/2 ] / [ (-1)(2y-1)(1+2y) ] =====> cancel out the (2y-1) term on the numerator and denominator
= 2 / [ (-1)(1+2y) ]
= -2 / (1+2y)

Question 2:

4/(x-1) = 5/(2x-2) + 3x/4
4/(x-1) = 5/2(x-1) + 3x/4 ======> remove the 2 from (2x-2)
4/(x-1) = (5/2)/(x-1) + 3x/4
(4-5/2) / (x-1) = 3x/4
(3/2) / (x-1) = 3x/4
4(3/2) = (x-1)(3x)
6 = 3x^2 - 3x
3x^2 - 3x - 6 = 0 ======> rearranging
3( x^2 - x - 2) = 0
x^2 - x - 2 = 0
(x-2)(x+1) = 0 ========> factorize the 2nd order ODE
x-2 = 0, x+1 = 0
x=2, x= -1