# Partial Fractions ```Partial Fractions
fractions
To add fractions we rewrite the fractions with a common denominator then add the
numerators.
Example
Find the sum of
3
2
+
3x + 4 x − 5
The common denominator of
b
b g
b gb g
Rewrite each fraction with a
denominator of 3x + 4 x − 5
b
gb g
3
2
and
is 3x + 4 x − 5
3x + 4
x −5
3 x −5
3
=
3x + 4
3x + 4 x − 5
gb g
and
b
g
2 3x + 4
2
=
x − 5 3x + 4 x − 5
b
gb g
b g
b g b g b g
b gb g b gb g b gb g
expand all brackets and
simplify
3 x −5
2 3x + 4
3 x − 5 + 2 3x + 4
+
=
3x + 4 x − 5
3x + 4 x − 5
3x + 4 x − 5
=
9x − 7
3x − 11x − 20
2
3
2
9x − 7
+
= 2
3x + 4 x − 5 3x − 11x − 20
Therefore
The reverse of this process is to split a fraction into partial fractions. In the above example
9x − 7
3
2
=
+
3x − 11x − 20 3x + 4 x − 5
2
Algebraic fraction
Partial fractions
If the degree of the numerator of the algebraic fraction is greater than that of the
denominator, divide the denominator into the numerator then express the remaining
fractional part as partial fractions.
Page 1 of 8
The form of the numerator in the partial fractions depends only on the type of the
factors in the denominator of the original fraction, as indicated below.
Each distinct linear factor eg. (x− a) has a
corresponding partial fraction of the form
A
( x − a)
Each repeated linear factor eg. (x− a)2 has a
corresponding partial fractions of the form
A
B
+
A and B constants.
( x − a ) ( x − a )2
where A is a constant.
.
2
Each quadratic factor eg. ax + bx + c has a
corresponding partial fraction of the form.
Ax + B
ax + bx + c
A and B constants.
2
Examples The following fractions have been written in partial fraction form (without evaluating
the constants in the numerators).
(1)
(2)
3x + 1
A
B
=
+
x −1 x + 2
x −1
x+2
b gb g b g b g
3x + 1
linear factors only in the denominator
A
B
C
+
+
x +1
x+3
x+3
b x + 1gb x + 3g b g b g b g
2
=
2
repeated linear factor in the denominator
gives rise to a partial fraction for both
b x − 3g and b x − 3g
(3)
d
3x + 1
i
x x2 − 2x + 3
=
A
Bx + C
+ 2
x
x − 2x + 3
d
i
2
In general the numerator of a partial fraction is a polynomial of degree one less than
the factor in the denominator.
Note the special case of repeated factors, example (2) above.
Exercise 1
Write the following in partial fraction form, but do not calculate the numerical values
for the constants in the numerator. If possible, factorise the quadratic factor first.
x+6
5
(a)
(b)
2x − 3 x + 4
x x + 4 x 2 − 3x + 2
b
(c)
gb g
2x
d x + 3ib x + 1g
2
b gd
(d)
dx
i
x2 − x + 3
2
ib g
− 3x + 7 x + 2
2
Page 2 of 8
Linear
factors
Example
The next task is to find values for the constants in the numerator. The first example will
contain linear factors in the denominator only.
Write
x −5
as the sum of partial fractions.
x + 2x − 3
2
factorise the denominator
write the general expression for the
partial fractions
add the fractions on the right side
the equation holds for all values of x
(except x = 1 and x =− 3) so we can
equate the numerators
expand brackets and collect like
terms
equate coefficients of powers of x on
both sides
solve these simultaneous equations
to give
substitute for A and B
the partial fractions are
therefore
x −5
x −5
=
x −1 x + 3
x + 2x − 3
b gb g
2
x −5
A
B
=
+
x −1 x + 3
x −1
x+3
b gb g b g b g
b g b g
b g b g b gb g
A x + 3 + B x −1
A
B
+
=
x −1
x+3
x −1 x + 3
b g b g
b gb g b gb g
x − 5 = Ab x + 3g + Bb x − 1g
A x + 3 + B x −1
x −5
=
x −1 x + 3
x −1 x + 3
b
g b
x − 5 = A + B x + 3A − B
g
A+ B =1
(coefficients of x)
3 A − B = −5
(constant terms)
A = −1 and
B=2
−1
2
and
x −1
x+3
x −5
2
1
=
−
x + 2x − 3 x + 3 x − 1
2
Page 3 of 8
Solving the simultaneous equations that result from equating coefficients can
sometimes be quite lengthy. An alternative method is to equate the numerators and,
before expanding the brackets, substitute a value of x into both sides of the equation so
that only one variable remains. Repeat this to find other variables. This method will
not necessarily find all variables, but will often make calculations easier.
Example
Using the previous example, after equating numerators we had
b g b g
x − 5 = A x + 3 + B x −1
substituting x = 1 will eliminate B and substituting x = −3 will eliminate A.
if x = 1 then
if x = −3 then
1 − 5 = A(1 + 3) + B(1− 1)
−3 −5 = A(−3 + 3) + B(−3 − 1)
solving gives
A=−1
solving gives
B=2
Repeated
linear
factors
In this example the denominator contains a repeated linear factor.
Example
Express
5x 2 + 3 x + 1
in terms of partial fractions.
x 3 − 3x − 2
factorise the
denominator
write the general
expression for the
partial fractions
on the right side
5x 2 + 3 x + 1
5x 2 + 3 x + 1
=
2
x 3 − 3x − 2
x +1 x − 2
b gb g
5x 2 + 3 x + 1
A
B
C
=
+
+
3
2
x−2
x +1
x − 3x − 2
x +1
b g b g b g
b g
b gb g b g
b g b g b g
b gb g
5x + 3x + 1 Ab x + 1g + Bb x + 1gb x − 2g + Cb x − 2g
∴
=
x − 3x − 2
b x − 2gb x + 1g
2
A x +1 + B x +1 x − 2 + C x − 2
A
B
C
+
+
=
2
2
x−2
x +1
x +1
x − 2 x +1
2
2
3
equate numerators
2
b g
2
b gb g b g
5x 2 + 3x + 1 = A x + 1 + B x + 1 x − 2 + C x − 2
Page 4 of 8
b g b gb g b g
b g b gb g b g
b g b gb g b g
to find A substitute x = 2 5x 2 + 3x + 1 = A x + 1 2 + B x + 1 x − 2 + C x − 2
in the equation
(the terms involving B and
2
2
C will equal zero and we 5 &times; 2 + 3 &times; 2 + 1 = A 2 + 1 + B 2 + 1 2 − 2 + C 2 − 2
2
can solve for A)
= A 3 + B 3 0 +C 0
27 = 9 A
A=3
b g b gb g b g
b g b g b gb g b g
2
2
to find C substitute x = −1 5x + 3x + 1 = A x + 1 + B x + 1 x − 2 + C x − 2
( the terms involving A
and B will equal zero) 5 &times; −1 2 + 3 &times; −1 + 1 = A 0 2 + B 0 −3 + C −3
3 = −3C
b g
C = −1
b g b gb g b g
b g b gb g b g
to find B substitute the
2
2
values already found (A=1 5x + 3x + 1 = A x + 1 + B x + 1 x − 2 + C x − 2
and C=−1) and a value
2
for x (substituting x=0
0 + 0 + 1 = 3 1 + B 1 −2 − 1 −2
keeps the arithmetic
1 = −2 B + 5
simple)
B=2
therefore A = 3,
substitute for A, B, and C
the partial fractions are
the solution is
3
,
( x − 2)
B = 2, and C = − 1
2
1
and
2
( x + 1)
( x + 1)
5x 2 + 3 x + 1
3
2
1
=
+
−
3
2
x−2
x +1
x − 3x − 2
x +1
b g b g b g
Page 5 of 8
or higher
factor
Example
The numerator for a quadratic factor has the form Ax+B. In general if the
denominator is of degree n then the numerator of the partial fraction is a polynomial
of degree n−1.
Express
5x 2 − 6 x + 5
in terms of partial fractions.
x 3 − 2 x 2 + 3x − 2
factorise the denominator
5x 2 − 6 x + 5
5x 2 − 6 x + 5
=
x 3 − 2 x 2 + 3x − 2
x − 1 x2 − x + 2
b gd
write the general expression
for the partial fractions
right side
i
5x 2 − 6 x + 5
A
Bx + C
=
+ 2
3
2
x −1
x − 2 x + 3x − 2
x − x+2
b g d
i
d
i b
b gd
gb g
i
d
i b
b gd
gb g
i
A x 2 − x + 2 + Bx + C x − 1
A
Bx + C
+ 2
=
x −1
x − x+2
x − 1 x2 − x + 2
b g d
i
A x 2 − x + 2 + Bx + C x − 1
5x 2 − 6 x + 5
=
x 3 − 2 x 2 + 3x − 2
x − 1 x2 − x + 2
equate numerators
expand the brackets on the
right-hand side of the
equation
equate coefficients of
powers of x
d
i b
gb g
5x 2 − 6 x + 5 = A x 2 − x + 2 + Bx + C x − 1
5 x 2 − 6 x + 5 = Ax 2 − Ax + 2 A + Bx 2 − Bx + Cx − C
= ( A + B ) x2 + ( C − A − B ) x + ( 2 A − C )
A+ B = 5
C − A − B = −6
2A − C = 5
solve the simultaneous
equations to give A = 2, B = 3 , and C =−1
substitute for A, B and C
the solution is
2
3x − 1
5x2 − 6 x + 5
=
+ 2
3
2
x − 2 x + 3 x − 2 ( x − 1) x − x + 2
Page 6 of 8
Exercise 2
Express the following as partial fractions.
(a)
(c)
(e)
2x
x−3 x+2
b gb g
x2 + 2x + 4
b x + 1gd x
2
(b)
i
+ 3x + 3
(d)
x+2
x − 5x + 6
2
3x
b1 − xg
2
x2 − 2x + 2
x3 + x2 + x
Page 7 of 8
Exercise 1
A
B
(a)
+
2x − 3 x + 4
(b)
A
B
C
D
+
+
+
x x + 4 x −1 x − 2
A
Bx + C
+ 2
x +1 x + 3
(d)
Exercise 2
6
4
(a)
+
5 ( x − 3) 5 ( x + 2 )
(b)
5
4
−
x−3 x−2
(e)
2
x+4
− 2
x x + x +1
(c)
(d)
3
(1 − x )
2
−
3
(1 − x )
A
B
Cx + D
+
+ 2
2
( x + 2 ) ( x + 2 ) x − 3x + 7
(c)
3
2x + 5
− 2
( x + 1) x + 3x + 3