# Thread: Paratial fractions-need help

1. ## Paratial fractions-need help

I've worked some problems, but I keep on getting stuck. Can you please help me?

Write the partial fraction decomposition of the rational expression.

1. (7x^2-x-12)/(x)(x+1)(x-1)

Here is what I did:

(x)(x+1)(x-1)[(7x^2-x-12)/(x)(x+1)(x-1)] = (x)(x+1)(x-1)[(a)/(x) + (b)/(x+1) + (c)/(x-1)]

Then I find the terms that can cancel out and I get:

7x^2-x-12= a(x+1)(x-1) + b(x-1)x+ c(x+1)x
7x^2-x-12=ax^2-a+bx-b+bx+cx+c+cx
7x^2-x-12= ax^2-a+bx^2-b+cx^2+c

Then I get the first equation: a+b=7, this is where I get stuck.

3. (9x^2-x-14)/(x^3-x)

(x)(x+1)(x-1)[(9x^2-x-14)/(x)(x+1)(x-1)= (x)(x+1)(x-1)[ (a)/(x) + (b)/(x+1)+ (c)/(x-1)

9x^2-x-14= a(x+1)(x-1)+ b(x-1)x + c(x+1)x
9x^2-x-14= ax^2-a-bx^2-b+cx^2+c
9x^2-x-14= ax^2-bx^2+cx^2-a-b+c

The equations:
a-b+c=9
b=0
b+c=-1
c=-14

Write the partial fraction decomposition of the rational expression.

11. (6x+5)/((x-9)^2)

(x-9)^2[(6x+5)/((x-9)^2)]= (x-9)^2[(a)/(x-9) + (b)/((x-9)^2))

(6x+5)/(x-9)(a+b)
(6x+5)/ax+bx-9a-9b

the equations:

a+b=6
-9a-9b=5

To get a variable to cancel, I mulitplied the first equation by 9 and got:

9a+9b=54
-9a-9b=5

both 9a and 9b cancel out and I just 59.

13. (5x^2-5x+7)/((x-1)^3)

((x-1)^3)(5x^2-5x+7)/((x-1)^3)= ((x-1)^3))[ (a)/(x-1) + (b)/((x-1)^2)+ (c)/((x-1)^3)

(5x^2-5x+7)=((x-1)^2)(a) + (x-1)(b) +c
(5x^2-5x+7)=ax^2+a+bx-b+c

The equations I couldnt get on this problem.

15. (x+4)/(x^3-2x^2+x)- This one I dont understand.

Write the partial fraction decomposition of the rational expression.

17. (12x+3)/ (x^3-1) - This one I dont understand

20. (12x^2-12x+6)/(x-3)(x^2+4)

6(2x^2-2x+1)=(x-3)(x^2+4) [(a)/(x-3) + (bx+c)/(x^2+4)
6(2x^2-2x+1)= a(x^2+4) + bx+c(x-3)
6(2x^2-2x+1) = ax^2+4a+bx^2-3bx+cx-3c

The equations:
a+b=6
4a-3b (this is where I get stuck)

Write the form of the partial fraction decomposition of the rational expression. It is not necessary to solve for the constants.

26. (7x-3)/((x^2+x-7)^2)

I got: (ax+b)/(x^2+x-7) + (cx+d)/((x^2+x+7)^2)

29. (2x+1)/(x-6)((x^2+x-4)^2)

I got: (a)/(x-6) + (bx+c)/(x^2+x-4)

Write the partial fraction decomposition of the rational expression.

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

x^2+4x-1 = ax +b(x^2+3)+cx+b/((x^2+3)^2)
x^2+4x-1= ax+b(x^2+3)+cx+d
x^2+4x-1=ax^3+3a+bx^2+3b+cx+d
x^2+4x-1=ax^3+bx^2+ (3a+c)x + 3b+d

The equations:
a=1
b=0
3a+c=4
3b+d=-1

When I come up with the fraction I get: (x+4)/(x^2+3) + (-4x - 12)/((x^2+3)^2))

33. (4x^3+ 4x^2)/((x^2+5)^2))

I did: (4x^3+ 4x^2)= x^2+5 (ax+b) + cx+d
(4x^3+ 4x^2)= ax^3+bx^2+ 5a+5b+cx+d

The equations:
a=4
b=4
5a+c=0
5b+d=0

The final answer I got: (4x+4)/(x^2+1) + (-20x-20)/((x^2+5)^2)

2. Originally Posted by soly_sol
I've worked some problems, but I keep on getting stuck. Can you please help me?

Write the partial fraction decomposition of the rational expression.

1. (7x^2-x-12)/(x)(x+1)(x-1)
Put

$\displaystyle \frac{7x^2-x-12}{x(x+1)(x-1)} = \frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1}$

.......... $\displaystyle =\frac{A(x^2-1)+Bx(x+1)+Cx(x-1)}{x(x-1)(x+1)}$$\displaystyle =\frac{x^2(A+B+C)+x(B-C)+(-A)}{x(x-1)(x+1)}$

So equating coefficients in the denominators we are left with the system of equations:

$\displaystyle A+B+C = 7$

$\displaystyle B-C=-1$

$\displaystyle -A=-12$

to solve.

RonL

3. Hello, soly_sol!

1) Write the partial fraction decomposition of: .$\displaystyle \frac{7x^2-x-12}{x(x+1)(x-1)}$
Why don't you try this method?

We have: .$\displaystyle \frac{7x^2-x-12}{x(x+1)(x-1)} \;=\;\frac{A}{x} + \frac{B}{x+1} + \frac{C}{x-1}$

Then: .$\displaystyle 7x^2 -x-12 \;=\;(x+1)(x-1)A + x(x-1)B + x(x+1)C$

Now substitute some "good" values for $\displaystyle x.$

$\displaystyle \begin{array}{ccccccc}\text{Let }x = 0: & -12 & = & (1)(\text{-}1)A + 0\!\cdot\!B + 0\!\cdot\!C & \Rightarrow & A \:= \:12 \\ \text{Let }x = \text{-}1: &-4 & = & 0\!\cdot\!A + (\text{-}1)(2)B + 0\!\cdot\!C & \Rightarrow & B \:= \:\text{-}2 \\ \text{Let }x = 1: & -6 & = & 0\!\cdot\!A + 0\!\cdot\!B + 1\!\cdot\!2\!\cdot\!C & \Rightarrow & C \:= \:\text{-}3 \end{array}$

Therefore: .$\displaystyle \frac{7x^2-x-12}{x(x+1)(x-1)} \;=\;\frac{12}{x} - \frac{2}{x+1} - \frac{3}{x-1}$