WAEC 2018 · Paper 2 · Q3

Partial fractions

(a)
Using the substitution $u = x - 2$ , write $\dfrac{x^3 + 5}{(x - 2)^4}$ as an expression in terms of $u$ .
(b)
Using the answer in (a), express $\dfrac{x^3 + 5}{(x - 2)^4}$ in partial fractions.

Worked solution (try it first)

(a)

$u = x - 2$ , so $x = u + 2$ .
Expand: $x^3 = (u + 2)^3 = u^3 + 6u^2 + 12u + 8$ .
Add 5: $x^3 + 5 = u^3 + 6u^2 + 12u + 13$ .
The bottom is $u^4$ .
So the fraction is $\dfrac{u^3 + 6u^2 + 12u + 13}{u^4} = \dfrac1u + \dfrac{6}{u^2} + \dfrac{12}{u^3} + \dfrac{13}{u^4}$ .

(b)

Put $u = x - 2$ back: $\dfrac{1}{x - 2} + \dfrac{6}{(x - 2)^2} + \dfrac{12}{(x - 2)^3} + \dfrac{13}{(x - 2)^4}$ .

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