Partial Fractions (Edexcel International A Level (IAL) Maths: Pure 4): Flashcards

Exam code: YMA01

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For $\frac{4 x + 1}{(x + 2) (x - 1) {(x - 3)}^{2}}$ , how many partial fractions are needed?
Four, not three.
The squared bracket contributes two factors, $(x - 3)$ and ${(x - 3)}^{2}$ , and each of them needs its own fraction.
Complete the split:
$\frac{5}{(x + 1) {(x - 4)}^{2}} \equiv \frac{A}{x + 1} + \frac{B}{______} + \frac{C}{______}$
$\frac{5}{(x + 1) {(x - 4)}^{2}} \equiv \frac{A}{x + 1} + \frac{B}{x - 4} + \frac{C}{{(x - 4)}^{2}}$
The three constants are then found in the usual way, by multiplying through and substituting values of $x$ .
Define squared linear factor.
A factor of the form ${(a x + b)}^{2}$ , that is a linear factor repeated.
The repetition is what makes it behave differently from two distinct linear factors.
True or False?
An $x^{2}$ in a denominator counts as a squared linear factor.
True.
A linear factor is $(a x + b)$ and $b$ is allowed to be zero, so $x$ is linear and $x^{2}$ is its square.
Such a denominator therefore needs fractions over both $x$ and $x^{2}$ .
Why is $\frac{A}{x + 1} + \frac{B}{{(x - 4)}^{2}}$ not a complete split of $\frac{5}{(x + 1) {(x - 4)}^{2}}$ ?
Because the term over $(x - 4)$ is missing.
That leaves only two constants to match a numerator which in general needs three, so the identity cannot be made to hold for every value of $x$ .

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