Tuesday

December 1, 2015
Posted by **Sean** on Sunday, February 24, 2008 at 3:25pm.

I'm having a slight problem though, when either C, or C and D come in.

I understand

A/(something) + B/(something)

But I become confused when it's either

A/(something) + B/(something) + C/(something)

OR

A/(something) + B/(something) + (Cx + D)/(something)

Why is it that you get ONLY C, or Cx + D ?

I always get confused when it comes to that, so help would be greatly appreciated.

- Calculus (Partial Fractions) -
**Damon**, Sunday, February 24, 2008 at 4:23pmWhen you have a quadratic that does not have real roots in the denominator (complex roots only), you put Cx+D in the numerator.

like if your denominator were (x-1)(x+1)(x-2)

you might use

A/(x-1) + B/(x+1) +C/(x-2)

But if your denominator were:

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

You could deal with the (x-1) part just fine.

but what to do with the other part?

You can not write it as the sum of two fractions with numerators B and C because you can nor write

x^2-2x+2 as (x+p)(x+q)

so you have to resort to

(C + D x)/(x^2-2x+2)

- Calculus (Partial Fractions) -
**Reiny**, Sunday, February 24, 2008 at 4:42pmThis method has practical applications only if the denominator factors.

Since you don't give an example I will supply one

separate (5x^2+3x+4)/(x^3+x^2-2x)

(I started with known fractions and simplified, so that I would have a question that worked out)

The bottom factors to x(x+2)(x-1)

so let

(5x^2+3x+4)/(x^3+x^2-2x) = A/(x+2) + B/(x-1) + c/x

(5x^2+3x+4)/(x^3+x^2-2x)

= [Ax(x-1) + Bx(x+2) + C(x+2)(x-1)]/x(x+2)(x-1)

clearing the denominator we get:

5x^2+3x+4 = Ax(x-1) + Bx(x+2) + C(x+2)(x-1)

now let x=0, then -2C = 4, and C = -2

let x=1, then 3B=12, and B = 4

let x=-2, then 6A = 18 and A = 3

so my original fraction (5x^2+3x+4)/(x^3+x^2-2x) can

be split into

3/(x+2) + 4/(x-1) - 2/x

- Calculus (Partial Fractions) -
**Sean**, Monday, February 25, 2008 at 2:28pmReiny. You misunderstood. And I didn't word it right.

I meant where does (Cx + D)/Something

Pop up?

- Calculus (Partial Fractions) -
**Anonymous**, Monday, February 25, 2008 at 3:11pmOkay. Example.

If the Denominator is

x^4 - 2x^2 - 8 > (x-2) (x+2) (x^2+2)

Would I do A + B + C, or A + B + (Cx + D)?

- Calculus (Partial Fractions) -
**Sean**, Monday, February 25, 2008 at 3:15pmWow. I'm making this topic last a while.

I think- and this is just out loud- that you would do (Cx + D) Over the X^2 + 2?

Am I getting that right?

OR

You have Cx + D over something if the SOMETHING has a term of X to a power greater than one?