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

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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) About Me  (Show Source):
You can put this solution on YOUR website!
+%285x+%2B+7%29%2F+%28x%5E2+%2B+2x+-3%29+
+%285x+%2B+7%29%2F+%28%28x+%2B+3%29%28x+-1%29%29+=+A%2F%28x%2B3%29+%2B+B%2F%28x-1%29

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:
+%285x+%2B+7%29%2F+%28%28x+%2B+3%29%28x+-1%29%29+=+A%2F%28x%2B3%29+%2B+B%2F%28x-1%29
+%285x+%2B+7%29%2F+%28%28x+%2B+3%29%28x+-1%29%29+=+2%2F%28x%2B3%29+%2B+3%2F%28x-1%29

R^2 at SCC