Define partial fractions .

Writing a single algebraic fraction as a sum of simpler fractions with smaller denominators. It is the reverse of adding fractions: instead of finding a common denominator, you split one back into its parts.

What are partial fractions actually used for?

Binomial expansions and integration . Both are straightforward on a simple fraction and awkward on a compound one, so splitting first is what makes them workable.

International A Level (IAL)MathsEdexcelPure 4FlashcardsAlgebra & FunctionsPartial Fractions

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

Exam code: YMA01

1/12

0Still learning

Know0

FrontPartial Fractions with Linear Denominators
Define partial fractions.

Was this flashcard helpful?

Cards in this collection (12)

Define partial fractions.
Writing a single algebraic fraction as a sum of simpler fractions with smaller denominators.
It is the reverse of adding fractions: instead of finding a common denominator, you split one back into its parts.
What form does a linear factor take?
$(a x + b)$ , with $x$ appearing to the first power only.
A denominator that is not itself linear can often be factorised into a product of such factors, and that is what makes the method work.
Complete the split into partial fractions:
$\frac{7 x + 1}{(x + 2) (x - 3)} \equiv \frac{A}{______} + \frac{B}{______}$
The completed split is:
$\frac{7 x + 1}{(x + 2) (x - 3)} \equiv \frac{A}{x + 2} + \frac{B}{x - 3}$
Each linear factor of the denominator gets a fraction of its own, with a constant on top.
You have written $\frac{7 x + 1}{(x + 2) (x - 3)} \equiv \frac{A}{x + 2} + \frac{B}{x - 3}$ . How do you find $A$ and $B$ ?
Multiply through by the whole denominator to clear the fractions, giving $7 x + 1 \equiv A (x - 3) + B (x + 2)$ .
Then substitute the values of $x$ that make each bracket zero, taking $x = 3$ and $x = - 2$ in turn.
Why do you substitute the values of $x$ that make a bracket zero?
Because each one kills off all but one of the unknown constants.
That leaves a single equation in a single unknown, which is far quicker than solving the constants simultaneously.
What is the alternative to substituting values of $x$ ?
Comparing coefficients on the two sides of the identity.
The two sides have to be equal for every value of $x$ , so the number of $x^{2}$ terms, of $x$ terms and of constants must match separately.
What are partial fractions actually used for?
Binomial expansions and integration.
Both are straightforward on a simple fraction and awkward on a compound one, so splitting first is what makes them workable.
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 $x$ .

Sign up to unlock flashcards

By signing up you agree to our Terms and Privacy Policy

Previous:Proof by ContradictionNext:General Binomial Expansion