B10 - Partial fractions | AQA A-Level Mathematics

Rewrite a rational expression as simpler fractions by choosing a complete denominator pattern and matching a polynomial identity. You will prepare expressions first, handle distinct and repeated linear factors, and check the result by recombining its terms.

Preparing a rational expression

Can you add $1/2+1/6+1/12+\cdots+1/10100$ , where the denominators are consecutive products $1\times2,2\times3,\ldots,100\times101$ , without finding a huge common denominator? The first three terms already have a useful form: $1/2=1-1/2$ , $1/6=1/2-1/3$ , and $1/12=1/3-1/4$ . Simpler fractions can make a complicated expression easier to use.

Rational expression and partial fractions

A rational expression is a ratio of polynomials. Its partial fractions express it as a sum of simpler rational terms. For example, $1/[x(x+1)]=1/x-1/(x+1)$ , for $x\ne0,-1$ . The equality is an identity on that domain, not an equation holding at only selected inputs.

A fraction is proper when its numerator has smaller degree than its denominator. Before decomposing, factor and cancel common factors, keeping every original excluded input. If the remaining fraction is improper, use algebraic division to write a polynomial plus a proper fraction. These preparation rules are not in the formula booklet: learn them.

You already know

Only factors can cancel. For example, $(x-1)(x+1)/[(x-1)(x+2)]$ simplifies to $(x+1)/(x+2)$ , but the original expression still excludes $x=1$ as well as $x=-2$ .

The simpler formula may be defined at a cancelled input; that does not put the input back into the original expression. Factorisation and division prepare the identity without changing what its original domain was.

Worked example 1: cancel and separate a constant

Question. Simplify $(x^2-1)/[(x-1)(x+2)]$ to a constant plus a proper fraction, stating the original excluded inputs.

Plan. Record denominator zeros, factor the numerator, then write the new numerator in terms of the denominator.

\begin{aligned} x&\ne1,-2 && \text{original denominator zeros}\\ \frac{x^2-1}{(x-1)(x+2)}&=\frac{(x-1)(x+1)}{(x-1)(x+2)} && \text{difference of squares}\\ &=\frac{x+1}{x+2} && \text{cancel a nonzero factor}\\ &=1-\frac1{x+2} && x+1=(x+2)-1 \end{aligned}

Why this step? Separating the whole denominator from the numerator leaves a proper remainder that can be handled directly.

Answer. $1-1/(x+2)$ , retaining $x\ne1,-2$ .

Check. Recombining gives $[(x+2)-1]/(x+2)=(x+1)/(x+2)$ . At $x=0$ , both original and final expressions give $\frac12$ .

Change only the numerator next. The cancelled factor remains, but the proper remainder changes sign.

Worked example 2: a different remainder

Question. Simplify $(x^2+2x-3)/[(x-1)(x+2)]$ to a constant plus a proper fraction, stating excluded inputs.

Plan. Factor the new numerator and use the same denominator-based division.

\begin{aligned} x^2+2x-3&=(x-1)(x+3) && \text{factorise}\\ \frac{x^2+2x-3}{(x-1)(x+2)}&=\frac{x+3}{x+2} && x\ne1,-2\\ &=1+\frac1{x+2} && x+3=(x+2)+1 \end{aligned}

Why this step? The original denominator decides the exclusions before cancellation; the changed numerator does not alter them.

Answer. $1+1/(x+2)$ , for $x\ne1,-2$ .

Check. At $x=0$ , the original fraction is $(-3)/(-2)=\frac32$ , matching $1+\frac12$ .

A proper fraction with three linear factors can have a quadratic original numerator. The partial-fraction numerators in the patterns below are constants; the degree of the original numerator is a different matter.

Watch out

Wrong: cancel the $x$ in $(x+1)/(x+2)$ . The $x$ is a term in a sum, not a common factor. Right: factor completely, cancel whole factors only, and retain original exclusions.

Change the numerator again and retain the same cancellation-and-division method.

Two distinct linear factors

For $1/[x(x+1)]$ , two separate factors suggest two simpler denominators. Recombining $A/x+B/(x+1)$ gives numerator $A(x+1)+Bx$ ; choosing $A=1$ , $B=-1$ produces the required constant numerator.

Two distinct linear factors

For distinct linear factors and a proper fraction,

\frac{px+q}{(ax+b)(cx+d)}\equiv\frac{A}{ax+b}+\frac{B}{cx+d}

Clearing the denominator gives

px+q\equiv A(cx+d)+B(ax+b)

The coefficients $A$ and $B$ are constants. Not in the formula booklet: learn it.

The cleared equation is a polynomial identity. Substitute convenient values, or compare coefficients of matching powers. Denominator roots are useful because each makes one term vanish. You substitute these roots into the cleared polynomial identity, which is defined there, rather than into the original undefined fractions.

Worked example 3: keep a non-monic factor intact

Question. Express $\dfrac{4x+3}{(2x-1)(x+2)}$ in partial fractions.

Plan. Use one constant over each factor, clear denominators, and substitute the two factor roots.

\begin{aligned} \frac{4x+3}{(2x-1)(x+2)}&=\frac{A}{2x-1}+\frac{B}{x+2} && \text{complete pattern}\\ 4x+3&\equiv A(x+2)+B(2x-1) && \text{clear denominators}\\ x=\frac12:\quad5&=\frac52A\quad\Longrightarrow\quad A=2 && \text{isolate }A\\ x=-2:\quad-5&=-5B\quad\Longrightarrow\quad B=1 && \text{isolate }B \end{aligned}

Why this step? Retaining $2x-1$ avoids losing its factor of $2$ when finding the corresponding numerator.

Answer. $\dfrac2{2x-1}+\dfrac1{x+2}$ , for $x\ne\frac12,-2$ .

Check. Recombination gives numerator $2(x+2)+(2x-1)=4x+3$ .

Change only the original numerator to a constant. The same pattern works, and a negative coefficient is perfectly possible.

Worked example 4: a constant original numerator

Question. Express $\dfrac5{(2x-1)(x+2)}$ in partial fractions.

Plan. Use the same two-factor pattern and evaluate its cleared identity at each factor root.

\begin{aligned} 5&\equiv A(x+2)+B(2x-1) && \text{clear denominators}\\ x=\frac12:\quad5&=\frac52A\quad\Longrightarrow\quad A=2 && \text{first root}\\ x=-2:\quad5&=-5B\quad\Longrightarrow\quad B=-1 && \text{second root} \end{aligned}

Why this step? The two terms' linear coefficients must cancel to leave a constant numerator; opposite signs achieve that here.

Answer. $\dfrac2{2x-1}-\dfrac1{x+2}$ , for $x\ne\frac12,-2$ .

Check. $2(x+2)-(2x-1)=5$ , exactly the original numerator.

The opening sum now collapses: $1/[n(n+1)]=1/n-1/(n+1)$ . Adding from $n=1$ to $100$ cancels every intermediate reciprocal, leaving $1-1/101=100/101$ . A hundred positive fractions have become one short subtraction. This identity is derived by recombination, not supplied in the formula booklet.

Watch out

Wrong: rewrite $2/(2x-1)$ as $2/(x-\frac12)$ . Right: $2/(2x-1)=1/(x-\frac12)$ . Factoring a constant from the denominator changes the numerator multiplier.

For AQA's ordinary “Express” task, a correct final decomposition can receive full credit. Writing the pattern and identity still makes your method clear and helps preserve method marks after an arithmetic slip.

Three distinct linear factors

Three different denominator factors need three coefficient constants. Clearing them produces quadratic products, so the original numerator may be quadratic even though each simpler numerator is constant.

Three distinct linear factors

For three distinct linear factors and a proper fraction,

\frac{px^2+qx+r}{(ax+b)(cx+d)(ex+f)} \equiv\frac{A}{ax+b}+\frac{B}{cx+d}+\frac{C}{ex+f}

Clear the whole denominator, then compare identities. Not in the formula booklet: learn it.

Substitution at the three different factor roots isolates the constants one at a time. Comparing the coefficients of $x^2$ , $x$ and $1$ is an equally valid method. Include the full pattern initially even if a coefficient later turns out to be zero.

Worked example 5: a quadratic original numerator

Question. Express $\dfrac{2x^2+3x+4}{x(x+1)(x+2)}$ in partial fractions.

Plan. Use denominators $x$ , $x+1$ , $x+2$ , then isolate each coefficient in the cleared identity.

2x^2+3x+4\equiv A(x+1)(x+2)+Bx(x+2)+Cx(x+1)

\begin{aligned} x=0:\quad4&=2A\quad\Longrightarrow\quad A=2 && \text{isolate }A\\ x=-1:\quad3&=-B\quad\Longrightarrow\quad B=-3 && \text{isolate }B\\ x=-2:\quad6&=2C\quad\Longrightarrow\quad C=3 && \text{isolate }C \end{aligned}

Why this step? Each factor root makes two products zero, leaving only one unknown coefficient.

Answer. $\dfrac2x-\dfrac3{x+1}+\dfrac3{x+2}$ , for $x\ne0,-1,-2$ .

Check. The recombined numerator is $2(x^2+3x+2)-3(x^2+2x)+3(x^2+x)=2x^2+3x+4$ .

Change only the constant term of the original numerator from $4$ to $6$ . All three coefficients may change because the identity must match at every input.

Worked example 6: change one numerator coefficient

Question. Express $\dfrac{2x^2+3x+6}{x(x+1)(x+2)}$ in partial fractions.

Plan. Keep the three-denominator pattern and substitute the same convenient inputs.

2x^2+3x+6\equiv A(x+1)(x+2)+Bx(x+2)+Cx(x+1)

\begin{aligned} x=0:\quad6&=2A\quad\Longrightarrow\quad A=3\\ x=-1:\quad5&=-B\quad\Longrightarrow\quad B=-5\\ x=-2:\quad8&=2C\quad\Longrightarrow\quad C=4 \end{aligned}

Each line isolates the coefficient whose denominator vanished before clearing.

Why this step? The factor roots are unchanged, so they remain the simplest choices even though the numerator has changed.

Answer. $\dfrac3x-\dfrac5{x+1}+\dfrac4{x+2}$ , for $x\ne0,-1,-2$ .

Check. $3(x+1)(x+2)-5x(x+2)+4x(x+1)=2x^2+3x+6$ .

An identity is stronger than a few numerical checks. Recombining verifies all coefficients together, so it is the best final check after solving for them.

Watch out

Wrong: use a linear numerator over each linear factor just because the original numerator is quadratic. Right: each distinct linear denominator has a constant numerator; the products created when clearing denominators supply the quadratic terms.

The next original numerator is constant. Keep the full three-term pattern.

Repeated linear factors

A squared factor needs two terms: one over its first power and one over its square. For example, $(3x+4)/(x+1)^2=3/(x+1)+1/(x+1)^2$ ; recombining gives $3(x+1)+1=3x+4$ .

A repeated linear factor

For distinct linear factors,

\frac{px^2+qx+r}{(ax+b)(cx+d)^2} \equiv\frac{A}{ax+b}+\frac{B}{cx+d}+\frac{C}{(cx+d)^2}

For a denominator containing only $(cx+d)^2$ , use just its first-power and squared terms. Not in the formula booklet: learn it.

Clearing the three-term pattern gives

px^2+qx+r\equiv A(cx+d)^2+B(ax+b)(cx+d)+C(ax+b)

The distinct-factor root isolates $A$ ; the repeated-factor root isolates $C$ . A further coefficient comparison or a third substituted input finds $B$ . The two root substitutions alone cannot determine all three constants.

Worked example 7: complete the repeated-factor pattern

Question. Express $\dfrac{2x^2+9x+8}{(x+1)(x+2)^2}$ in partial fractions.

Plan. Include both powers of $x+2$ , use the two roots, then compare quadratic coefficients.

2x^2+9x+8\equiv A(x+2)^2+B(x+1)(x+2)+C(x+1)

\begin{aligned} x=-1:\quad1&=A && \text{isolate the distinct-factor term}\\ x=-2:\quad-2&=-C\quad\Longrightarrow\quad C=2 && \text{isolate the squared-factor term}\\ 2&=A+B\quad\Longrightarrow\quad B=1 && \text{compare }x^2\text{ coefficients} \end{aligned}

Why this step? The repeated-factor root removes the term containing $B$ , so another piece of the identity is needed to find it.

Answer. $\dfrac1{x+1}+\dfrac1{x+2}+\dfrac2{(x+2)^2}$ , for $x\ne-1,-2$ .

Check. $(x+2)^2+(x+1)(x+2)+2(x+1)=2x^2+9x+8$ .

Change only the numerator next. The complete denominator pattern and the method of finding its constants remain the same.

Worked example 8: use the leading coefficient for the last constant

Question. Express $\dfrac{3x^2+13x+12}{(x+1)(x+2)^2}$ in partial fractions.

Plan. Isolate the first and third constants at the roots, then use the quadratic coefficient.

3x^2+13x+12\equiv A(x+2)^2+B(x+1)(x+2)+C(x+1)

\begin{aligned} x=-1:\quad2&=A && \text{first constant}\\ x=-2:\quad-2&=-C\quad\Longrightarrow\quad C=2 && \text{third constant}\\ 3&=A+B\quad\Longrightarrow\quad B=1 && \text{quadratic coefficient} \end{aligned}

Why this step? The $C$ -term is only linear, so it contributes no $x^2$ coefficient; the leading coefficient must equal $A+B$ .

Answer. $\dfrac2{x+1}+\dfrac1{x+2}+\dfrac2{(x+2)^2}$ , for $x\ne-1,-2$ .

Check. $2(x+2)^2+(x+1)(x+2)+2(x+1)=3x^2+13x+12$ .

For a denominator containing only $(x+1)^2$ , coefficient comparison is shorter: $3x+4\equiv B(x+1)+C$ gives $B=3$ , $C=1$ , proving the opening identity of this part. The same principle covers either two or three fractional terms. These decompositions later help with integration and binomial approximations.

Watch out

Wrong: use only $A/(x+1)+C/(x+2)^2$ . The general pattern also needs $B/(x+2)$ . Right: include every power from the first power up to the repeated power, even if a particular coefficient eventually equals zero.

Use the same denominator once more with a new numerator.

Quick Recap

Simplify first, divide if needed, and preserve the original excluded inputs.
Use one constant numerator for each distinct linear factor.
A squared factor needs both its first-power and squared terms.
Clear denominators, find constants from the polynomial identity, then recombine to check.

The opening hundred-term sum is $100/101$ , because its partial fractions cancel in pairs, leaving $1-1/101$ .

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B10 - Partial fractions

Preparing a rational expression

Two distinct linear factors

Three distinct linear factors

Repeated linear factors

Quick Recap

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