2 a (x + 1)(x + 2) = x2 + 3x + 2
b (x − 3)(x + 4) = x2 + x − 12
c (x − 2)(x − 3) = x2 − 5x + 6
d (x − 3)2 = x2 − 6x + 9
e (2x + 1)(x + 2) = 2x2 + 5x + 2
f (x + 1)(x + 2)(x + 3) = x3 + 6x2 + 11x + 6
3 a b a
n
a b
...
2 a (x + 1)(x + 2) = x2 + 3x + 2
b (x − 3)(x + 4) = x2 + x − 12
c (x − 2)(x − 3) = x2 − 5x + 6
d (x − 3)2 = x2 − 6x + 9
e (2x + 1)(x + 2) = 2x2 + 5x + 2
f (x + 1)(x + 2)(x + 3) = x3 + 6x2 + 11x + 6
3 a b a
n
a b
n
a b
n
a b
1 2 3
...
( ) + = n n n + 1 2 n n 2 3 3
+
+
− − − +
a b a
n
a b
n
a b
n
a b
1 2 3
...
( ) + = n n n + 1 2 n n 2 3 3
+
+
− − − +
x3 term = = 6
3 3 ! ! ! 23 3 x x 160 3 so the coeffi cient is 160
Using the factor theorem:
Substitute x = − 2 into denominator
8 + 12 − 8 − 12 = 0
So (x + 2) is a factor.
Using the factor theorem:
Substitute x = 3 into denominator
−27 + 27 + 12 − 12 = 0
So (3 − x) is a factor as coefficient of x3 is negative.
Express the expression as partial fractions.
The two different methods to split a rational
function with linear factors in its denominator
into partial fractions are substitution and equating
coefficients.
For partial fractions without repeated terms,
the substitution method relies only on basic
mathematical operations without the need for
any algebraic manipulation, whereas the equating
coefficients method results in the solving of
simultaneous equations.
For partial fractions with repeated terms, the
substitution method relies on a combination of
basic mathematical operations and the solving
of simultaneous equations, whereas the equating
coefficients method results in the solving of
simultaneous equations with three unknowns
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