Partial or whole?
( x + 1 ) ( x − 3 ) ( x + 4 ) 2 x + 1 = ( x + 1 ) A + ( x − 3 ) B + ( x + 4 ) C
The above rational fraction is expressed in the form of partial fractions. Evaluate A + B + C .
The answer is 0.
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1 solution
It's easier to solve it by partial fractions - cover up rule .
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If we need to find the values of A , B and C then yes, it's a better method. However, I'm extremely used to this method, and besides, for this particular question, I can find A + B + C with this method without knowing each individual value
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Fair point! Good work! =D =D
Ahahaha, you got me, I only realized halfway when I tried to solve for A , B and C . Anyway,
( x + 1 ) ( x − 3 ) ( x + 4 ) 2 x + 1 = x + 1 A + x − 3 B + x + 4 C = ( x + 1 ) ( x − 3 ) ( x + 4 ) A ( x − 3 ) ( x + 4 ) + B ( x + 1 ) ( x + 4 ) + C ( x + 1 ) ( x − 3 )
The denominators are the same, so we compare the numerator:
A ( x 2 + x − 1 2 ) + B ( x 2 + 5 x + 4 ) + C ( x 2 − 2 x − 3 ) = 2 x + 1 ( A + B + C ) x 2 + ( A + 5 B − 2 C ) x + ( − 1 2 A + 4 B − 3 C ) = 2 x + 1
By comparison:
( A + B + C ) x 2 = 0 x 2 A + B + C = 0