Asked by Anonymous
How do I write the partial fraction decomposition of 7/(x^2+13x+40)?
Answers
Answered by
Steve
the fractions will be
A/(x+8) + B/(x+5)
A(x+5) + B(x+8) = 7
(A+B)x + (5A+8B) = 0x + 7
A+B = 0
5A+8B = 7
3B = 7
B = 7/3
A = -7/3
7/(x^2+13x+40) = (7/3)/(x+5) - (7/3)/(x+8))
A/(x+8) + B/(x+5)
A(x+5) + B(x+8) = 7
(A+B)x + (5A+8B) = 0x + 7
A+B = 0
5A+8B = 7
3B = 7
B = 7/3
A = -7/3
7/(x^2+13x+40) = (7/3)/(x+5) - (7/3)/(x+8))
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