P4 2.1 - Partial fractions | Edexcel A-Level International Mathematics

A single algebraic fraction can hide a much simpler sum. Learn how to recover that sum, including repeated factors and fractions needing division first, then check that the result is an identity on the original domain.

Distinct linear denominators

How can $\frac{3x+5}{(x+1)(x+2)}$ become two fractions with constant numerators? You already know how to add fractions; partial fractions reverse that process.

For example, adding $2/(x+1)$ and $1/(x+2)$ gives numerator $2(x+2)+(x+1)=3x+5$ . The denominators tell us the shape of the answer before we find its constants.

Partial-fraction decomposition

A partial-fraction decomposition rewrites a rational expression as a sum of simpler fractions, with the same value wherever the original expression is defined. For distinct linear factors, use one constant numerator above each factor.

Multiply by the full denominator to obtain a polynomial identity. That identity holds for all $x$ , so values making individual factors zero are useful for finding constants, even though those values remain excluded from the original fraction.

Worked example 1: two distinct factors

Question. Decompose $\frac{3x+5}{(x+1)(x+2)}$ .

Plan. Set up two fractions, clear denominators, then choose values isolating each constant.

\frac{3x+5}{(x+1)(x+2)}=\frac A{x+1}+\frac B{x+2}

3x+5=A(x+2)+B(x+1).

At $x=-1$ , $2=A$ ; at $x=-2$ , $-1=-B$ , so $B=1$ .

Answer. $\frac2{x+1}+\frac1{x+2}$ , with $x\ne-1, -2$ .

Check. Recombining gives numerator $2(x+2)+(x+1)=3x+5$ , proving equality throughout the allowed domain.

Change one feature: there are now three distinct factors. Add a third fraction and use three isolating values.

Worked example 2: three distinct factors

Question. Decompose $\frac{2x^2+5x-1}{(x-1)(x+1)(x+2)}$ .

Plan. Clear the three-factor denominator, then substitute its three roots.

2x^2+5x-1=A(x+1)(x+2)+B(x-1)(x+2)+C(x-1)(x+1).

The substitutions give

\begin{aligned} x=1 &: 6=6A &&\Rightarrow A=1\\ x=-1 &: -4=-2B &&\Rightarrow B=2\\ x=-2 &: -3=3C &&\Rightarrow C=-1. \end{aligned}

Answer. $\frac1{x-1}+\frac2{x+1}-\frac1{x+2}$ , for $x\ne1, -1, -2$ .

Check. The recombined numerator is

(x^2+3x+2)+2(x^2+x-2)-(x^2-1)=2x^2+5x-1.

A factor need not start with $x$ . For $\frac{4x-1}{(2x+1)(x-1)}=\frac A{2x+1}+\frac B{x-1}$ , the cleared identity is $4x-1=A(x-1)+B(2x+1)$ . Use $x=-1/2$ to get $-3=-3A/2$ , so $A=2$ ; use $x=1$ to get $3=3B$ , so $B=1$ . Recombining gives $2(x-1)+(2x+1)=4x-1$ , confirming the answer. In general, $ax+b$ is isolated with $x=-b/a$ , provided $a\ne0$ .

Keep every original denominator restriction in the result.

Watch out

Do not substitute a forbidden value into the original fraction. Substitute it into the polynomial identity after clearing denominators, where no division by zero remains.

For the next three-factor expression, set up all three simpler fractions before finding their constants.

Repeated linear factors

A repeated factor needs a fraction for each power of that factor. Otherwise the proposed form may lack enough flexibility to reproduce the numerator.

For a denominator $(x-a)^2$ , write $A/(x-a)+B/(x-a)^2$ . Combining those gives numerator $A(x-a)+B$ , which can match any linear numerator. A lone $B/(x-a)^2$ could match only a constant.

Repeated-factor form

For a proper fraction with denominator $(ax+b)(cx+d)^2$ , use

\frac A{ax+b}+\frac B{cx+d}+\frac C{(cx+d)^2},

where the two linear factors are distinct. Find the constants after clearing denominators.

Not in the formula booklet: learn the form.

An isolating value often finds one constant immediately. Compare coefficients or choose another convenient value for the others.

Worked example 3: a squared factor

Question. Decompose $\frac{3x+1}{(x-1)^2}$ .

Plan. Include both powers of $x-1$ and match the numerator.

3x+1=A(x-1)+B=Ax-A+B.

The coefficient of $x$ gives $A=3$ ; the constant term gives $-3+B=1$ , so $B=4$ .

Answer. $\frac3{x-1}+\frac4{(x-1)^2}$ , for $x\ne1$ .

Check. At $x=2$ , both expressions give $7$ . Recombining gives $3(x-1)+4=3x+1$ exactly.

Add one distinct linear factor. The squared factor still contributes two fractions.

Worked example 4: one distinct and one repeated factor

Question. Decompose $\frac{3x^2+9x-3}{(x-1)(x+2)^2}$ .

Plan. Use three constants; isolate two and obtain the third from the coefficient of $x^2$ .

3x^2+9x-3=A(x+2)^2+B(x-1)(x+2)+C(x-1).

At $x=1$ , $9=9A$ , so $A=1$ . At $x=-2$ , $-9=-3C$ , so $C=3$ . The coefficient of $x^2$ is $A+B=3$ , giving $B=2$ .

Answer. $\frac1{x-1}+\frac2{x+2}+\frac3{(x+2)^2}$ , for $x\ne1, -2$ .

Check. Expanding the proposed numerator gives $(x^2+4x+4)+2(x^2+x-2)+3x-3=3x^2+9x-3$ .

The power is part of the denominator, not a new constant to absorb.

Watch out

$A/(x+2)+B/(x+2)^2$ is the required form for a squared factor. $A/(x+2)+B/(x+2)$ just combines to one fraction and cannot supply both terms.

Include both powers of the repeated factor in this decomposition.

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P4 2.1 - Partial fractions

Distinct linear denominators

Repeated linear factors

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