Summary Partial fractions

Partial fractions
of demons
Degree 1
= :




:
diff factors
2 factor
squared
:
I ,
I 2 3 2×+2-3
¥2
2 " + ↳ + ←+
+ =

#
Simplify
= -
2
:
C3C + 1)(2+2) [✗ + , > ( ✗ + 1) 2
2×-1
3×+5



Resolve into
° + 1) (set 2)
x+iJ
partial fractions
I make it
equal A1B over [
the factors of demon .




3×+5
⇐ + 1) ( ✗ + 2)
=
A

Cat 1) +2+2
B
Solve for
t
A and B
z
=
¥+1 , >
2



t
2×-1 =
Afc + 1) TB
3×+5 = A (xt2) +
BLOCH) sub back into
original


= +

3- ⇐ =
Ec +
IT



3 factors :
same as 2 but use A ,B and Cover factors
,




£ zfz-tqf-z.ge
61×2-4×+43 -6pct -2×3+13><
- =
x (x
+25
6×2 -
24×+24-6=2+12×+132
☐ =
✗ bet 2) 2


25

I

É¥→= +1=-2 +
É
Boxset 2)
>
✗ +24 =
Abc +
2) + + ( (




É = +
Eg +
E- →
¥2 =
E- +

3- § +




Bottom
degree 2 of
:
?:
degree 1
= -
num -_
. : oc in num .




3=-4 2 3×2 -25×+28 +2×2+2
+
#
=
22+1 (→ +
1) (2-7)
☐ 5×2-25×+30


-
=
(x -
7) (x2 +
1)




5×2-25×+30
(x -
7) (x2 +
1)
= +
¥
5×2-25×+30 __ ( Ax + B) Ex 7) - +
CCSCZ +
1)



fc2.TK#=AxtB
?
22 + I
+

§