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.

Why does a quadratic factor need a linear numerator rather than a constant?

Because the numerator over any factor has to be one degree lower than the factor itself, and one degree below a quadratic is linear. The same rule is what gives a linear factor its single constant on top, so this is not a special case but the general pattern.

A LevelMathsCambridge (CIE)Pure 3FlashcardsAlgebra & FunctionsPartial Fractions

Partial Fractions (Cambridge (CIE) A Level Maths: Pure 3): Flashcards

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Define partial fractions.

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Cards in this collection (18)

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 $x$ .
Complete the correct form for this split:
$\frac{2 x^{2} + 8 x - 1}{(x + 3) (x^{2} + 5 x - 2)} \equiv \frac{A}{x + 3} + \frac{______}{x^{2} + 5 x - 2}$
The completed form is:
$\frac{2 x^{2} + 8 x - 1}{(x + 3) (x^{2} + 5 x - 2)} \equiv \frac{A}{x + 3} + \frac{B x + C}{x^{2} + 5 x - 2}$
The numerator sitting over the quadratic factor is linear, not a single constant.
Why does a quadratic factor need a linear numerator rather than a constant?
Because the numerator over any factor has to be one degree lower than the factor itself, and one degree below a quadratic is linear.
The same rule is what gives a linear factor its single constant on top, so this is not a special case but the general pattern.
Which quadratic factors need a linear numerator rather than being split further?
Only those that do not factorise, such as $x^{2} + 4$ or $2 x^{2} + 3$ .
On this course the quadratic factor usually appears with no $x$ term at all, in the form $c x^{2} + d$ .
With one linear and one quadratic factor, why does substituting roots not find every constant?
Because only the linear factor has a root you can substitute, and that single substitution removes only one of the unknowns.
The quadratic does not factorise, so it offers no convenient value of $x$ that makes it vanish, and the remaining constants have to be reached another way.
True or False?
If the linear factor is $x$ itself, you can still use $x = 0$ as one of your two extra values.
False.
Putting $x = 0$ is already the root of the factor $x$ , so it does the job of the first substitution and cannot then serve again as a second, independent equation.
You have to choose two other values of $x$ and solve the resulting pair simultaneously.
Which values of $x$ should you substitute to find all three constants?
Start with the root of the linear factor, then use $x = 0$ , then any small convenient value such as $x = 1$ .
The first removes one unknown outright, and the other two give equations that pin down the remaining pair.

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