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Partial fractions

Adding fractions is easy and everyone learns it forwards. This lesson runs it backwards, splitting one awkward algebraic fraction into simple pieces with unseen constants, found by substituting cunning values of x. The payoff lands later, when integration and series expansions want the pieces, not the whole.

Year 12-13EDEXCEL 9MA0 2.10

Builds on Polynomials and the factor theorem.

IN THIS TOPIC

  • Split fractions with distinct linear factors, two or three of them.
  • Handle a repeated linear factor with the two-fraction template.
  • Choose substitution values that make constants fall out one at a time.

WHAT YOU PROBABLY THINK

A squared bracket in the denominator needs just one fraction.

Un-adding fractions

Adding 2/(x + 1) and 3/(x + 2) over a common denominator gives (5x + 7)/((x + 1)(x + 2)), and partial fractions is that computation reversed. Write the split with unknown constants, multiply up, and choose values of x that silence one bracket at a time.

The rational function 5x plus 7 over bracket x plus 1 bracket x plus 2 drawn with its two partial fractions 2 over x plus 1 and 3 over x plus 2: at every x the two simple curves stack to the complicated onethe whole fraction2/(x + 1)3/(x + 2)two simple pieces stack to the awkward whole
FIG. 1The decomposition is a fact about heights: at every x, the curves of 2/(x + 1) and 3/(x + 2) stack to the curve of the whole fraction.

WORKED EXAMPLE

Two brackets, two constants

Express (5x + 7)/((x + 1)(x + 2)) in partial fractions.

Set (5x + 7)/((x + 1)(x + 2)) = A/(x + 1) + B/(x + 2) and multiply up: 5x + 7 = A(x + 2) + B(x + 1).

Substitute x = −1 to kill the B bracket: 2 = A. Substitute x = −2 to kill the A bracket: −3 = −B, so B = 3.

So the fraction is 2/(x + 1) + 3/(x + 2).

Any spare value of x checks the answer; x = 0 gives 7/2 on the left and 2 + 3/2 on the right, agreeing.

YOUR TURN

Three brackets, three constants

Express (6x2 + 5x − 2)/(x(x − 1)(2x + 1)) in partial fractions, before opening the working.

Show the working

The template is A/x + B/(x − 1) + C/(2x + 1), and multiplying up gives 6x2 + 5x − 2 = A(x − 1)(2x + 1) + Bx(2x + 1) + Cx(x − 1).

x = 0 gives −2 = −A, so A = 2. x = 1 gives 9 = 3B, so B = 3. x = −½ gives −3 = ¾C, so C = −4.

The fraction is 2/x + 3/(x − 1) − 4/(2x + 1).

The value x = −½ silences two brackets at once, which is why awkward-looking roots of brackets are the best substitutions on offer.

The repeated factor

A squared bracket breaks the one-fraction-per-bracket pattern, and the opening lie with it. The denominator (cx + d)2 can shelter two distinct simple pieces, one over the bracket and one over its square, and both are needed in general.

The two decomposition templates: distinct linear brackets take one unknown constant each, while a squared bracket takes two fractions, one over the bracket and one over its square(ax + b)(cx + d)A/(ax + b) + B/(cx + d)(ax + b)(cx + d)²A/(ax + b) + B/(cx + d)+ C/(cx + d)²distinct brackets: one constant eacha squared bracket needs two fractions
FIG. 2The two templates. Distinct brackets take one constant each; a squared bracket takes a fraction over itself and another over its square.

TRY IT UNSEEN

A squared bracket, in full

Express (x + 4)/((x + 1)(x − 2)2) in partial fractions.

Show the working

Template: A/(x + 1) + B/(x − 2) + C/(x − 2)2, so x + 4 = A(x − 2)2 + B(x + 1)(x − 2) + C(x + 1).

x = −1 gives 3 = 9A, so A = ⅓. x = 2 gives 6 = 3C, so C = 2.

No value kills only B, so compare a coefficient: the constant terms give 4 = 4A − 2B + C, and with A and C known, B = −⅓.

The fraction is ⅓/(x + 1) − ⅓/(x − 2) + 2/(x − 2)2, and a spare substitution such as x = 1 confirms both sides give 5.

Substitution reaches the constants at the brackets' roots; the leftover constant always needs a compared coefficient or one more substitution.

THE EXAM BIT

  • Write the template first, one fraction per distinct bracket and two for a squared one; the template line carries its own mark.
  • Multiply up and substitute each bracket's root to pick constants off one at a time.
  • When a constant survives every substitution, compare coefficients, the constant term being usually quickest.
  • Check with a spare value of x; ten seconds of arithmetic protects every mark downstream.
  • Keep the constants as exact fractions; thirds and quarters are normal here and decimals are marked down.

CHECK YOURSELF

Express (7x − 1)/((x − 1)(x + 3)) in partial fractions.

Show a hint

Two brackets, two constants, two substitutions.

Show the answer

Template and multiply up: 7x − 1 = A(x + 3) + B(x − 1).

x = 1 gives 6 = 4A, so A = 3/2. x = −3 gives −22 = −4B, so B = 11/2.

The fraction is 3/(2(x − 1)) + 11/(2(x + 3)), and x = 0 checks: 1/3 = −3/2 + 11/6 holds.

One fraction per distinct bracket; a squared bracket takes the bracket and its square.

Substitute the roots to harvest constants; compare a coefficient for whatever survives.

CHECK YOUR PROGRESS

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  • Split fractions with distinct linear factors, two or three of them.
  • Handle a repeated linear factor with the two-fraction template.
  • Choose substitution values that make constants fall out one at a time.

No animated video for this topic yet; these notes stand alone.