SOLUTION: Find the partial fraction decomposition of the rational expression below.
(-x-37)/(x^2-3x-28)
a.-(4/x-7)-(3/x+4)
b.(3/x-7)+(3/x+4)
c.-(3/x-7)+(3/x+4)
d.(4/x-7)+(3/x+4)
e.
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Question 471722: Find the partial fraction decomposition of the rational expression below.
(-x-37)/(x^2-3x-28)
a.-(4/x-7)-(3/x+4)
b.(3/x-7)+(3/x+4)
c.-(3/x-7)+(3/x+4)
d.(4/x-7)+(3/x+4)
e.-(4/x-7)+(3/x+4)
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Find the partial fraction decomposition of the rational expression below.
(-x-37)/(x^2-3x-28)
----
[-(x+37)]/[(x-7)(x+4)]
----
= A/(x-7) + B/(x+4)
----
= [A(x+4) + B(x-7)]/[x^2-3x-28]
----
Equate the numerators:
(A+B)x + (4A-7B) = -x-37
---
Then:
A + B = -1
4A-7B = -37
----
Modify:
4A + 4B = -4
4A - 7B = -37
----
11B = 33
B = 3
---
Since A+B = -1
A = -4
----
Solution:
-4/(x-7) + 3/(x+4)
------------------------
Cheers,
Stan H.
========================
a.-(4/x-7)-(3/x+4)
b.(3/x-7)+(3/x+4)
c.-(3/x-7)+(3/x+4)
d.(4/x-7)+(3/x+4)
e.-(4/x-7)+(3/x+4)
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