# More polynomial division

Now I am having more troubles with more advanced problems.

I started with this problem:

[4y^2 - 7y - 12] / [(y)(y+2)(y-3)]

The problem is that I set it up the same way as I did in that thread and im able to solve for A, B, and C:
A= 3
B= 9/5
C= 1/5

so I integrate 3/y + (9/5)/(y+2) + (1/5)/(y-3)

and i get

3ln(y) + 9/5ln(y+2) + 1/5ln(y-3)

Beautiful right? Well im suppost to find the integral over the area from 1 to 2.

When I plug in 1 and 2 into the 1/5 ln (y-3) it yields a negetive number and you cant take the ln of a negetive number!!!!

Im sure im making the mistake in the polynomial division somewhere but i dont know. Im thinking there is a way to simplify the original question first before i solve for A B and C.... right? any insite would be wonderful!

Ok so I forgot to mention that I also tried to take the absolute values of the value inside the ln and it still did not work.

Oh ok. You didn't set up your equations right. With a denominator with three factors, it becomes (you can see this by yourself):

$$A(y)(y+2) + B(y)(y-3) + C(y+2)(y-3) = 4y^2 - 7y - 12$$

$$A(y^{2} + 2y) + B(y^{2} - 3y) + C(y^{2} - y - 6) = 4y^2 - 7y - 12$$

$$(A + B + C)y^{2} + (2A - 3B - C)y - 6C = 4y^2 - 7y - 12$$

If we want A, B and C to be constants, we compare each term with one another (equality of polynomials theorem). We find C = 2, and then we get $$2A - 3B - 2 = -7$$ and $$A + B + 2 = 4$$. Now we have two unknowns with two equations and we solve.

Last edited:
HallsofIvy