proper fractions with repeated linear factors complx

[Mike_Gannon1]

Hi.
I thought I had this. Simple partial fraction equations I can do. But I keep getting stuck once the denominators coefficient becomes more complex.
Im self studying and its frustrating me no end as I always hit the same issue on the second step, where equating coefficients doesn't work.

Question:Express this equation as a partial fraction.
5x²+23x+24
(2x +3)(x+2)²

I try solving it by starting with
A(2x+3) + B(x+2)²
and set x to -2 is
20-46+24=-2 and A(3-4) + B(0)
so A =2
Then equate coefficient.
5x² = Bx² so B=5
23x = 4Bx + 4x so B = 19/4
24 = 4A + 6 so B =9/2

and thats not right. What am I missing. Is there another step?

 

[Ishuda]

Hi.
I thought I had this. Simple partial fraction equations I can do. But I keep getting stuck once the denominators coefficient becomes more complex.
Im self studying and its frustrating me no end as I always hit the same issue on the second step, where equating coefficients doesn't work.

Question:Express this equation as a partial fraction.
5x²+23x+24
(2x +3)(x+2)²

I try solving it by starting with
A(2x+3) + B(x+2)²
and set x to -2 is
20-46+24=-2 and A(3-4) + B(0)
so A =2
Then equate coefficient.
5x² = Bx² so B=5
23x = 4Bx + 4x so B = 19/4
24 = 4A + 6 so B =9/2

and thats not right. What am I missing. Is there another step?

Because of the 2 exponent in the denominator [the (x+2)²] you will need three terms
\(\displaystyle \dfrac{5x^2+23x+24}{(2x+3)(x+2)^2}=\dfrac{A}{2x+3} + \dfrac{B}{x+2} + \dfrac{Cx+D}{(x+2)^2}\)
It might help to note that 5x²+23x+24 = (5x+8)(x+3)

You might want to look at
http://www.mathsisfun.com/algebra/partial-fractions.html

 

[Mike_Gannon1]

Thanks Ishuda. Looks like the problem is I was trying to solve it in two stages and treat the repeated linear as an independent equation to be solved. So If I treat it as one equation then I should be able to solve it. I will have another go tomorrow. Thanks for your help.

Mike

 

[Mike_Gannon1]

Thank you Isuda,

That worked perfectly!!