SOLUTION: 1. (a) resolve into partial fractions 2x/[(x-2)(x+5)] (b) resolve into partial fractions 1/(x^2-x)

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: 1. (a) resolve into partial fractions 2x/[(x-2)(x+5)] (b) resolve into partial fractions 1/(x^2-x)      Log On


   

Question 1146014: 1. (a) resolve into partial fractions 2x/[(x-2)(x+5)]
(b) resolve into partial fractions 1/(x^2-x)
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39674) About Me  (Show Source):
You can put this solution on YOUR website!
(a)

%282x%29%2F%28%28x-2%29%28x%2B5%29%29=A%2F%28x-2%29%2BB%2F%28x%2B5%29



2x=A%28x%2B5%29%2BB%28x-2%29

2x=Ax%2B5A%2BBx-2B

2x=%28A%2BB%29x%2B%285A-2B%29

Follow the corresponding parts...
system%282=A%2BB%2C5A-2B=0%29-------solve this system for A and B.

system%282A%2B2B=4%2C5A-2B=0%29
ADD.
7A=4
highlight%28A=4%2F7%29

system%285A%2B5B=10%2C5A-2B=0%29
SUBTRACT.
7B=10
highlight%28B=10%2F7%29

Answer by MathTherapy(10579) About Me  (Show Source):
You can put this solution on YOUR website!
1. (a) resolve into partial fractions 2x/[(x-2)(x+5)]
   (b) resolve into partial fractions 1/(x^2-x)

1. (a) resolve into partial fractions 2x%2F%28x+-+2%29%28x+%2B+5%29
The other person’s solution: A = 2%2F3, and B = 4%2F3, is WRONG!!

2x%2F%28x+-+2%29%28x+%2B+5%29 = A%2F%28x+-+2%29+%2B+B%2F%28x+%2B+5%29
 ---- Multiplying by LCD, (x - 2)(x + 5)
           2x = A(x + 5) + B(x - 2) ---- Equating NUMERATORS, since denominators are the same
         2(2) = A(2 + 5) + B(2 - 2) ---- Substituting 2 for x to determine the value of A
            4 = 7A
                4%2F7+=+A
       2(- 5) = A(- 5 + 5) + B(- 5 - 2) ---- Substituting - 5 for x to determine the value of B
         - 10 = - 7B
        %28-+10%29%2F%28-+7%29+=+10%2F7+=+B
       (A, B) = (4%2F7, 10%2F7)

Therefore, highlight%282x%2F%28x+-+2%29%28x+%2B+5%29%29 = A%2F%28x+-+2%29+%2B+B%2F%28x+%2B+5%29 = highlight%28%284%2F7%29%2F%28x+-+2%29+%2B+%2810%2F7%29%2F%28x+%2B+5%29%29


1. (b) resolve into partial fractions 1%2F%28x%5E2+-+x%29

Use the same concept above, to decompose this PROPER FRACTION too. 

Before doing so though, we FACTORIZE the denominator in 1%2F%28x%5E2+-+x%29 to get: 1%2Fx%28x+-+1%29, and then: 1%2Fx%28x+-+1%29 = A%2Fx+%2B+B%2F%28x+-+1%29