Define partial fractions .

Splitting a single algebraic fraction into a sum of simpler fractions , each having one factor of the original denominator underneath it. It is the reverse of adding fractions, where a common denominator is formed instead.

What is the first step in writing an expression as partial fractions?

Factorise the denominator , so that each factor can be given its own fraction. Until the denominator is written as a product there is nothing to split it into.

Where are partial fractions used later in the course?

In binomial expansions and in integration . Both are far easier to carry out on a sum of simple fractions than on a single complicated one.

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Partial Fractions (Edexcel A Level Maths: Pure): Flashcards

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Define partial fractions.

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Cards in this collection (12)

Define partial fractions.
Splitting a single algebraic fraction into a sum of simpler fractions, each having one factor of the original denominator underneath it.
It is the reverse of adding fractions, where a common denominator is formed instead.
What is the first step in writing an expression as partial fractions?
Factorise the denominator, so that each factor can be given its own fraction.
Until the denominator is written as a product there is nothing to split it into.
Complete the split into partial fractions:
$\frac{3 x - 2}{(5 x - 2) (x + 1)} \equiv \frac{A}{______} + \frac{B}{______}$
$\frac{3 x - 2}{(5 x - 2) (x + 1)} \equiv \frac{A}{5 x - 2} + \frac{B}{x + 1}$
Each linear factor of the denominator gets one fraction, with an unknown constant on top.
How do you find the unknown constants in a partial fraction split?
Multiply through by the original denominator to clear the fractions, then substitute values of $x$ that make one bracket zero.
Each substitution removes one unknown and leaves the other on its own.
What is the alternative to substituting values when finding partial fractions?
Comparing coefficients. The number of $x^{2}$ terms on each side must match, and likewise for the $x$ terms and the constants.
That produces simultaneous equations to solve for the unknowns.
True or False?
Partial fractions can be used on $\frac{3 x - 2}{5 x^{2} + 3 x - 2}$ .
True.
The denominator factorises to $(5 x - 2) (x + 1)$ , so it can be split.
A non-linear denominator is perfectly acceptable as long as it can be written as a product of linear factors.
Where are partial fractions used later in the course?
In binomial expansions and in integration.
Both are far easier to carry out on a sum of simple fractions than on a single complicated one.
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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