Partial Fractions

Question 1

What 2 methods are used in impartial fractions?

Answer 1

Substitution and Comparing Coefficients

Question 2

6x-2/(x-3)(x+1) = A/(x-3) + B/(x+1) NEXT STEP

Answer 2

A(x+1)/(x-3)(x+1) + B(x-3)/(x-3)(x+1)

Question 3

6x-2/(x-3)(x+1) = A/(x-3) + B/(x+1)
A(x+1)/(x-3)(x+1) + B(x-3)/(x-3)(x+1) NEXT STEP

Answer 3

6x-2 = A(x+1) + B(x-3)

Question 4

6x-2/(x-3)(x+1) = A/(x-3) + B/(x+1)
A(x+1)/(x-3)(x+1) + B(x-3)/(x-3)(x+1)
6x-2 = A(x+1) + B(x-3) NEXT STEP (substitution)

Answer 4

x = 3
16 = 4A + 0B
4 = A
x = -1
-8 = 0A + -4B
B = 2
4/(x-3) + 2/(x+1)

Question 5

6x-2/(x-3)(x+1) = A/(x-3) + B/(x+1)
A(x+1)/(x-3)(x+1) + B(x-3)/(x-3)(x+1)
6x-2 = A(x+1) + B(x-3) NEXT STEP (comparing coefficients)

Answer 5

6x-2 = Ax+A + Bx-3B

 6 = A+B -   -2 = A-3B ------------------
 8 = 4B B = 2 6 = A+2 A = 4 4/(x-3) + 2/(x=1)

Question 6

What do you do if the Partial Fraction has a Repeated Factor? (11x²+14x+5/(x+1)²(2x+1))

Answer 6

You split it with an extra, C, like so:
A/(x+1)² + B/(x+1) + C/(2x+1), then
A(2x+1)/(x+1)²(2x+1) + B(x+1)(2x+1)/(x+1)²(2x+1) + C(x+1)²/(x+1)²(2x+1)