Question

Partial fraction decomposition (s-4)/(s^2+4)(s+1)

Answer 1

(s-4) A Bx+C
----------- = ---- + ------
(s^2+4)(s+1) s+1 s^2+4

Answer 2

s=x

Answer 3

(s-4)/(s^2 +4)(s+1) = As+B/(s^2+4) + C/(s+1)

Multiply both sides by the denominator

s-4 = (As+B)(s+1) + C(s^2+4)

now plug numbers in. Plug in s=-1

-5 = 0 + 5C
C = -1

s=0
s-4 = (As+B)(s+1) - (s^2+4)
-4 = B - 4

B = 0

s-4 = As(s+1) - (s^2 + 4)

plug in 1
-3 = 2A - 5
2 A = 2
A = 1

so A = 1
B = 0
C = -1

Back to the original...
As+B/(s^2+4) + C/(s+1)

s/(s^2+4) - 1/(s+1)

Answer 4

(s-4) = A(s^2+4) + (Bx+C)(s+1)

when s = -1 we get

-5 = A(s^2+4) + (Bx+C)0

-5 = A(5)

A = -1

Answer 5

lol amistre you keep writing x

Answer 6

(s-4) = -(s^2+4) + (Bx+C)(s+1)

(s-4) = -s^2-4 + Bx(s+1)+ C(s+1); this can get tricker :)

Answer 7

lol ... bummer

Answer 8

s-4= -s^2 -4 +B(s^2) +B(s) + C(s) + C

Answer 9

now we relate variables to coefficients i think

Answer 10

s^1 +s = (B)s^2 +(B+C)s +C

Answer 11

s^2 +s = (B)s^2 +(B+C)s +C

B = 1; C = 0

Answer 12

(s-4) -1 s
----------- = ----- + ------ ; maybe :)
(s^2+4)(s+1) s+1 s^2+4

Answer 13

A+B=0 => A=-B
B+C=1 => C=1-B
4A+C=-4 => 4(-B)+(1-B)=-4 => -5B+1=-4 B=1
C=1-1=0
A=-1

Answer 14

thats right! gj

Answer 15

\[= A/(s+1) + (Bs + C)/(s ^{2}+ 4).\]
We can clear this equation of fractions to get
s-4=\[ s - 4 = A(s ^{2}+4) + (Bs + C)(s+1).\]

Answer 16

\frac{top}{bottom} gives you a nice fraction

\[\frac{top}{bottom}\]

Answer 17

(A+ B)s^2 + (4A + B + C) s + (B + 4C) = s - 4
A + B = 0
4A + B + C = 1
B + 4C = -4

Answer 18

I may have miscalculated somewhere, so TIWAGOS. The method I described here will always work for partial fractions decompositions.