SOLUTION: I'm having a rough day! Here's another mind boggler that the book doesn't explain the steps to. Decompose into partial fractions: 5x + 7/ x^2 + 2x -3

Question 12135: I'm having a rough day! Here's another mind boggler that the book doesn't explain the steps to. Decompose into partial fractions:
5x + 7/ x^2 + 2x -3
Answer by rapaljer(4671) (Show Source): You can put this solution on YOUR website!

Multiply both sides of this equation by the LCD, which is (x+3)(x-1).
The result will be:
5x + 7 = A(x-1) + B(x+3)

You can substitute any two values of x into this equation, and obtain two equations and two unknowns. However, the easiest way to do it is to let x=1 and let x= -3, since these particular values of x will "zero out" half of the problem and give immediate results:
If x= 1
5x + 7 = A(x-1) + B(x+3)
5 + 7 = A(0) + B(1+3)
12=4B, so B= 3

If x = -3,
5x + 7 = A(x-1) + B(x+3)
-15+7 = A(-3-1) + B(0)
-8 = -4A, so A = 2

Therefore the partial fractions would be:

R^2 at SCC

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