The form of a partial fraction decomposition

[rowdy3]

For the following problem provide the "form" of the partial fraction decomposition for the given fractional expression. You do not have to solve the undetermined coefficients.

4. 2x^2 - 3x + 8 / x^3 + 9x
I took an x out and it's no x(x^2+9) My answer is A/X + BX+C/x^2+9

5. x- 7 / x^4 - 16
I did (x^2+4)(x^2-4). I factor (x^2 - 4) into (x-2)(x+2). My answer is AX+B/x^2+4 + C/(x+2) + D/(x-2)

6. x^2 - 4x + 6 / (x+3)^2 (x^2+1)^2.
My answer is A/(x+3) + B/(x+3) + CX+D/(x^2+1) + EX+F/(x^2+10^2)
Here's a scan of the problems.
http://pic20.picturetrail.com/VOL1370/5671323/23539305/392720956.jpg

 

[rock.freak667]

For the following problem provide the "form" of the partial fraction decomposition for the given fractional expression. You do not have to solve the undetermined coefficients.

4. 2x^2 - 3x + 8 / x^3 + 9x
I took an x out and it's no x(x^2+9) My answer is A/X + BX+C/x^2+9

5. x- 7 / x^4 - 16
I did (x^2+4)(x^2-4). I factor (x^2 - 4) into (x-2)(x+2). My answer is AX+B/x^2+4 + C/(x+2) + D/(x-2)

These are correct.

x^2 - 4x + 6 / (x+3)^2 (x^2+1)^2.

When you have repeated roots, you need to reflect that in your decomposition.

So if you had x/(x-a)² the decomposition would be A/(x-a²) + B/(x-a).

 

[Mark44]

Rockfreak.667 has already answered your question, but how you represent fractions deserves comment.

For the following problem provide the "form" of the partial fraction decomposition for the given fractional expression. You do not have to solve the undetermined coefficients.

4. 2x^2 - 3x + 8 / x^3 + 9x

When you write a fraction as a single line of text, use parentheses when the numerator or denominator contains multiple terms. As you have written this, someone could legitimately interpret the above as
2x^2 - 3x + (8/x^3) + 9x.

Since this isn't what you meant, you should have written this as
(2x^2 - 3x + 8) / (x^3 + 9x).

I took an x out and it's no x(x^2+9) My answer is A/X + BX+C/x^2+9

5. x- 7 / x^4 - 16

Write this as (x - 7)/(x^4 - 16).
And your answer should be written as A/x + (Bx + C)/(x^2 + 9). Similar for the expression below.

I did (x^2+4)(x^2-4). I factor (x^2 - 4) into (x-2)(x+2). My answer is AX+B/x^2+4 + C/(x+2) + D/(x-2)

6. x^2 - 4x + 6 / (x+3)^2 (x^2+1)^2.

Write this as ( x^2 - 4x + 6)/((x+3)^2 (x^2+1)^2).
Note the extra pair of parentheses used in the denominator. These are used to show that the denominator is (x + 3)^2 * (x^2 + 1)^2.

My answer is A/(x+3) + B/(x+3) + CX+D/(x^2+1) + EX+F/(x^2+10^2)
Here's a scan of the problems.
http://pic20.picturetrail.com/VOL1370/5671323/23539305/392720956.jpg

The 3rd and 4th terms should be written as (Cx + D)/(x^2 + 1) + (Ex + F)/(x^2 + 1)^2.