Flashcards · 8 cards

Partial fractions

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Rule
How do you split $\dfrac{2x - 3}{(x - 1)(x + 1)(x + 2)}$ into partial fractions?
Answer
Write $\dfrac{A}{x - 1} + \dfrac{B}{x + 1} + \dfrac{C}{x + 2}$ , multiply through by the bottom, then make each bracket zero in turn to find $A$ , $B$ and $C$ .
Partial fractionsMake each bracket zero in turn to find A and B
From the lesson: Partial fractions with linear factors
Rule
What if the top's degree is at least the bottom's?
Answer
Divide first: a whole part plus a proper fraction. Then split the proper fraction.
Divide firstWhole part, plus a proper fraction to split
From the lesson: Partial fractions with linear factors
Know it
Where is $\dfrac{2x - 3}{(x^2 - 1)(x + 2)}$ undefined?
Answer
Where the bottom is zero. Factorise it fully, $(x - 1)(x + 1)(x + 2)$ : so $x = 1$ , $-1$ or $-2$ .
From the lesson: Partial fractions with linear factors
Know it
How do you check your partial fractions?
Answer
Put a simple value such as $x = 0$ into both the original fraction and your answer: they must agree.
From the lesson: Partial fractions with linear factors
Rule
Which fractions do you write for a repeated bracket such as $(x + 2)^3$ ?
Answer
One for each power: $\dfrac{A}{x + 2} + \dfrac{B}{(x + 2)^2} + \dfrac{C}{(x + 2)^3}$ .
Which fractions to writeOne fraction for each power of a repeated bracket; Bx + C over a quadratic
From the lesson: Repeated and quadratic factors
Know it
Which fraction goes over a quadratic factor that won't factorise, such as $x^2 + 2$ ?
Answer
$\dfrac{Bx + C}{x^2 + 2}$ : a linear top, not just a number.
From the lesson: Repeated and quadratic factors
Know it
How many constants should the partial fractions have?
Answer
As many as the degree of the bottom: a cubic bottom always needs three.
From the lesson: Repeated and quadratic factors
Which method?
WAEC 2018 · Paper 2 · Q3
Using the substitution $u = x - 2$ , write $\dfrac{x^3 + 5}{(x - 2)^4}$ as an expression in terms of $u$ .
Using the answer in (a), express $\dfrac{x^3 + 5}{(x - 2)^4}$ in partial fractions.
Why substitute $u = x - 2$ ?
Answer
It turns the fraction into $\dfrac{(u + 2)^3 + 5}{u^4}$ . Expand the top and divide each term by $u^4$ : the partial fractions appear at once.
From the lesson: Repeated and quadratic factors Try the question