SOLUTION: write the partial fraction decomompasitionof the rational function (2x+3)/x^2+3x+2?

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Question 382940: write the partial fraction decomompasitionof the rational function (2x+3)/x^2+3x+2?
Found 2 solutions by Fombitz, nyc_function:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
%282x%2B3%29%2F%28x%5E2%2B3x%2B2%29=%282x%2B3%29%2F%28%28x%2B2%29%28x%2B1%29%29
%282x%2B3%29%2F%28x%5E2%2B3x%2B2%29=A%2F%28x%2B2%29%2BB%2F%28x%2B1%29
.
.
2x%2B3=A%28x%2B1%29%2BB%28x%2B2%29
2x%2B3=Ax%2BBx%2BA%2B2B
2x%2B3=%28A%2BB%29x%2B%28A%2B2B%29
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.
1.A%2BB=2
2.A%2B2B=3
Subtract eq. 1 from eq. 2,
A%2B2B-A-B=3-2
B=1
Then from eq. 1,
A%2B1=2
A=1
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.
highlight%28%282x%2B3%29%2F%28x%5E2%2B3x%2B2%29=1%2F%28x%2B2%29%2B1%2F%28x%2B1%29%29

Answer by nyc_function(2741) About Me  (Show Source):
You can put this solution on YOUR website!
The function can be rewritten:
(2x + 3)/(x + 1)(x + 2); denominator has non repeating linear factors.
(2x + 3)/(x + 1)(x + 2) = A/(x + 1) + B/(x + 2)........multiply by (x + 1)(x + 2)
2x + 3 = A(x + 2) + B(x + 1) = (A + B)x + 2A + B
So A + B = 2 and 2A + B = 3==========> A = 1 and B = 1