Ma15 - Partial fractions with linear factors | OCR A-Level Mathematics B (MEI)

Replace an algebraic fraction by simpler fractions whose numerators are constants. Handle distinct and squared linear factors, then use supplied rules and a short mathematical passage to see what the decomposition can reveal.

Splitting two distinct linear factors

How can a hundred positive fractions add to less than one? The first four terms of $1/(n(n+1))$ are $1/2,1/6,1/12,1/20$ , and their sum is $4/5$ . A different way of writing each fraction will explain the pattern without adding a hundred denominators.

You already know

To add fractions, use a common denominator. A fraction excludes denominator zeros; a polynomial identity obtained after clearing the denominator can still hold at those inputs.

Partial fractions reverse addition: we know the combined numerator and want to recover the simpler pieces.

Partial-fraction decomposition

A partial-fraction decomposition writes a rational expression as a sum of simpler fractions, with the same values at its allowed inputs. Each distinct linear factor receives a constant numerator. For $(x+1)(x+3)$ , start with $A/(x+1)+B/(x+3)$ .

Choose constants by making the recombined numerator identical to the starting numerator.

Worked example 1: two distinct factors

Question. Express $(4x+6)/((x+1)(x+3))$ in partial fractions.

Plan. Clear the denominator, then substitute $-1,-3$ into the polynomial identity.

\begin{aligned} 4x+6&\equiv A(x+3)+B(x+1) && \text{recombine numerators}\\ x=-1:\quad2&=2A\quad\Rightarrow\quad A=1 && \text{isolate }A\\ x=-3:\quad-6&=-2B\quad\Rightarrow\quad B=3 && \text{isolate }B \end{aligned}

Why this step? Those inputs remove one unknown at a time. They are used in the polynomial identity, not in the undefined fractions.

Answer. $1/(x+1)+3/(x+3)$ , for $x\ne-1,-3$ .

Check. $(x+3)+3(x+1)=4x+6$ , so recombination establishes the identity everywhere it is defined.

Change one feature: replace the first factor by a non-monic linear factor. Its zero is fractional.

Worked example 2: keep a non-monic factor intact

Question. Express $(4x+7)/((2x+1)(x+3))$ in partial fractions.

Plan. Use constants over the factors as written and substitute $-1/2,-3$ .

\begin{aligned} 4x+7&\equiv A(x+3)+B(2x+1) && \text{clear denominator}\\ x=-\tfrac12:\quad5&=\tfrac52 A\quad\Rightarrow\quad A=2 && \text{first constant}\\ x=-3:\quad-5&=-5B\quad\Rightarrow\quad B=1 && \text{second constant} \end{aligned}

Why this step? The zero of $2x+1$ is $-1/2$ , not $-1$ . Keeping its coefficient avoids a hidden rescaling error.

Answer. $2/(2x+1)+1/(x+3)$ , with $x\ne-1/2,-3$ .

Check. $2(x+3)+(2x+1)=4x+7$ .

Coefficient comparison is another valid route. The last identity gives $A+2B=4$ and $3A+B=7$ . This decomposition method is not in the formula sheet: learn it.

Watch out

Wrong: $1/(2x+1)=1/(x+1/2)$ . Right: $1/(2x+1)=(1/2)/(x+1/2)$ . Changing a factor's scale changes the required numerator.

Keep the same factors and change the numerator.

Handling three distinct factors

With three different factors, include three unknown constants. The numerator identity now contains products of two factors.

Worked example 3: three-factor decomposition

Question. Express $12/((x-2)(x+1)(x+2))$ in partial fractions.

Plan. Clear all three factors and evaluate at each factor's zero.

\begin{aligned} 12&\equiv A(x+1)(x+2)+B(x-2)(x+2)+C(x-2)(x+1) && \text{clear denominator}\\ x=2:\quad12&=12A\quad\Rightarrow\quad A=1 && \text{first constant}\\ x=-1:\quad12&=-3B\quad\Rightarrow\quad B=-4 && \text{second constant}\\ x=-2:\quad12&=4C\quad\Rightarrow\quad C=3 && \text{third constant} \end{aligned}

Why this step? At a factor zero, exactly one of the three numerator products survives.

Answer. $1/(x-2)-4/(x+1)+3/(x+2)$ , for $x\ne2,-1,-2$ .

Check. $(x^2+3x+2)-4(x^2-4)+3(x^2-x-2)=12$ . The quadratic and linear coefficients cancel.

Change one feature: the starting numerator is now linear.

Worked example 4: a linear numerator over three factors

Question. Express $(2x+8)/((x-2)(x+1)(x+2))$ in partial fractions.

Plan. Use the same three-factor identity with $2x+8$ on its left.

\begin{aligned} x=2:\quad12&=12A\quad\Rightarrow\quad A=1 && \text{first constant}\\ x=-1:\quad6&=-3B\quad\Rightarrow\quad B=-2 && \text{second constant}\\ x=-2:\quad4&=4C\quad\Rightarrow\quad C=1 && \text{third constant} \end{aligned}

Why this step? The constants must be recomputed from the new numerator, even though the denominator is unchanged.

Answer. $1/(x-2)-2/(x+1)+1/(x+2)$ , with $x\ne2,-1,-2$ .

Check. $(x^2+3x+2)-2(x^2-4)+(x^2-x-2)=2x+8$ .

Ma15 uses constant or linear numerators and denominators with up to three linear factors, including squared factors. A quadratic or cubic factor that cannot be factorised into real linear factors requires a different form and is excluded here. These setup rules are not in the formula sheet: learn them.

Problem-solving

Use the coefficients that ought to vanish as a check. A linear numerator means the total quadratic coefficient must be zero, even when individual pieces contain large quadratic terms.

Change only the numerator's constant in the next question.

Including every power of a repeated factor

A squared factor contributes two possible fractions, not one: its first power and its square each need a constant numerator.

Repeated-factor form

For distinct factor zeros and non-zero linear coefficients,

\frac{px+q}{(ax+b)(cx+d)^2} =\frac A{ax+b}+\frac B{cx+d}+\frac C{(cx+d)^2}.

Retain all denominator-zero exclusions. A solved coefficient may be zero, but do not omit its possible term at the start.

Not in the formula sheet: learn it.

Substituting the repeated zero isolates the squared-denominator coefficient. It does not isolate the first-power coefficient, so compare a power afterwards.

Worked example 5: a repeated factor

Question. Express $(8-x)/((x+1)(x-2)^2)$ in partial fractions.

Plan. Include both powers of $x-2$ and use the quadratic coefficient last.

\begin{aligned} 8-x&\equiv A(x-2)^2+B(x+1)(x-2)+C(x+1) && \text{clear denominator}\\ x=-1:\quad9&=9A\quad\Rightarrow\quad A=1 && \text{isolate }A\\ x=2:\quad6&=3C\quad\Rightarrow\quad C=2 && \text{isolate }C\\ 0&=A+B\quad\Rightarrow\quad B=-1 && \text{compare }x^2 \end{aligned}

Why this step? The substitutions leave $B$ undetermined; the missing quadratic term supplies the needed equation.

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

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

Change one feature: the repeated factor becomes non-monic. Keep its coefficient when expanding its square.

Worked example 6: a non-monic repeated factor

Question. Express $(5-4x)/((x+1)(2x-1)^2)$ in partial fractions.

Plan. Use $-1,1/2$ , then compare the quadratic coefficient.

\begin{aligned} 5-4x&\equiv A(2x-1)^2+B(x+1)(2x-1)+C(x+1) && \text{complete identity}\\ x=-1:\quad9&=9A\quad\Rightarrow\quad A=1 && \text{first constant}\\ x=\tfrac12:\quad3&=\tfrac32 C\quad\Rightarrow\quad C=2 && \text{third constant}\\ 0&=4A+2B\quad\Rightarrow\quad B=-2 && \text{compare }x^2 \end{aligned}

Why this step? $(2x-1)^2$ contributes $4x^2$ , whereas $(x+1)(2x-1)$ contributes $2x^2$ .

Answer. $1/(x+1)-2/(2x-1)+2/(2x-1)^2$ , with $x\ne-1,1/2$ .

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

A fraction over $(2x-1)^2$ with a linear numerator can combine the last two terms, but splitting into constant-numerator pieces is what makes the later rules useful.

Watch out

Wrong: set up only $A/(x+1)+C/(2x-1)^2$ . Right: include $B/(2x-1)$ as well. The full identity needs that freedom.

Keep the non-monic repeated factor and change the numerator.

Using the split in a power series

A series application becomes easier once each denominator has just one linear factor. The expansion rules are supplied here; your task is to choose the right substitution and combine the terms.

Binomial expansion

For real $n$ and $|u|<1$ ,

(1+u)^n=1+nu+\frac{n(n-1)}{2!}u^2+\cdots.

A non-negative integer $n$ gives a terminating expansion valid for every real $u$ .

In the formula sheet.

The following special cases are derived from that general rule, not printed separately. We will use them for $|u|<1$ :

\frac1{1-u}=1+u+u^2+\cdots,\qquad \frac1{(1-u)^2}=1+2u+3u^2+\cdots.

Replacing $u$ by $-x$ expands $1/(1+x)$ . A truncated series gives the requested first terms, not an exact polynomial valid everywhere.

Worked example 7: expand a supplied split

Question. Expand $1/((1+x)(1-2x))$ through $x^2$ , using

\frac1{(1+x)(1-2x)}=\frac1{3(1+x)}+\frac2{3(1-2x)}.

Plan. Use $u=-x$ and $u=2x$ , then require both expansions to be valid.

\begin{aligned} &\tfrac13(1-x+x^2+\cdots)+\tfrac23(1+2x+4x^2+\cdots) && \text{substitute arguments}\\ &\qquad=1+x+3x^2+\cdots && \text{collect powers} \end{aligned}

Why this step? The sign of the linear term changes for $1/(1+x)$ , but its squared term stays positive.

Answer. $1+x+3x^2$ through $x^2$ , valid for $|x|<1/2$ .

Check. The combined numerator is $(1-2x)/3+2(1+x)/3=1$ . Multiplying the series by $1-x-2x^2$ gives zero coefficients of $x,x^2$ .

Change one feature: square $1-2x$ . The first-power and squared-factor expansions now both appear.

Worked example 8: expansion with a repeated factor

Question. Expand $1/((1+x)(1-2x)^2)$ through $x^2$ , using the supplied split

\frac1{9(1+x)}+\frac2{9(1-2x)}+\frac2{3(1-2x)^2}.

Plan. Apply the squared rule to the last fraction and collect coefficients exactly.

\begin{aligned} &\tfrac19(1-x+x^2+\cdots)+\tfrac29(1+2x+4x^2+\cdots) && \text{simple factors}\\ &\quad+\tfrac23(1+4x+12x^2+\cdots) && \text{squared factor}\\ &=1+3x+9x^2+\cdots && \text{collect powers} \end{aligned}

Why this step? The squared-factor coefficient is $3(2x)^2=12x^2$ , before multiplying by $2/3$ .

Answer. $1+3x+9x^2$ through $x^2$ , valid for $|x|<1/2$ .

Check. Recombining the split gives numerator $1$ . Its denominator is $1-3x+4x^3$ ; multiplying by the displayed series gives no $x$ or $x^2$ term.

The expression can be defined outside that interval, but the expansions cannot. A squared factor changes coefficients, rather than changing the boundary $|2x|<1$ .

Watch out

Wrong: use $|x|<1$ for every factor. Right: a factor $1-2x$ requires $|2x|<1$ . Intersect all conditions, keeping endpoints strict.

Change the scale in the repeated factor, using a supplied new split.

Using the split in an integration step

Another application is integration: once a rational expression is split, each piece has a simple antiderivative. An antiderivative differentiates back to its integrand.

Integration rules

On an interval where a differentiable $\mathrm f$ is non-zero,

\int\frac{\mathrm f'(x)}{\mathrm f(x)}\,\mathrm dx=\ln|\mathrm f(x)|+C,

and, for $n\ne-1$ ,

\int\mathrm f'(x)(\mathrm f(x))^n\,\mathrm dx=\frac{(\mathrm f(x))^{n+1}}{n+1}+C.

Use real-valued powers on their domain.

In the formula sheet.

The linear-factor cases are not printed separately. We supply them here, for $a\ne0$ and on an interval avoiding $ax+b=0$ :

\int\frac1{ax+b}\,\mathrm dx=\frac1a\ln|ax+b|+C, \qquad \int\frac1{(ax+b)^2}\,\mathrm dx=-\frac1{a(ax+b)}+C.

A first power gives a logarithm; a square gives a reciprocal.

Worked example 9: two logarithmic terms

Question. Integrate $(4x+1)/((x+1)(2x-1))$ , using its split $1/(x+1)+2/(2x-1)$ .

Plan. Apply the supplied logarithm rule with linear coefficients $1,2$ .

\begin{aligned} \int\left(\frac1{x+1}+\frac2{2x-1}\right)\mathrm dx &=\ln|x+1|+\frac22\ln|2x-1|+C && \text{include }1/a\\ &=\ln|x+1|+\ln|2x-1|+C && \text{simplify coefficient} \end{aligned}

Why this step? The outer numerator $2$ cancels the $1/2$ in the integration rule; neither may be ignored.

Answer. $\ln|x+1|+\ln|2x-1|+C$ , on an interval avoiding $-1,1/2$ .

Check. Differentiation gives $1/(x+1)+2/(2x-1)$ , whose numerator is $(2x-1)+2(x+1)=4x+1$ .

Change one feature: square the second linear factor, as in the earlier repeated-factor example.

Worked example 10: add a reciprocal term

Question. Integrate $(5-4x)/((x+1)(2x-1)^2)$ , using its earlier decomposition.

Plan. Integrate $1/(x+1)-2/(2x-1)+2/(2x-1)^2$ term by term.

\begin{aligned} \int\left(\frac1{x+1}-\frac2{2x-1}+\frac2{(2x-1)^2}\right)\mathrm dx &=\ln|x+1|-\ln|2x-1|\\ &\quad-\frac1{2x-1}+C && \text{apply both rules} \end{aligned}

Why this step? The last denominator is squared. Its integral is a reciprocal with a minus sign, not another logarithm.

Answer. $\ln|x+1|-\ln|2x-1|-1/(2x-1)+C$ , on an interval avoiding $-1,1/2$ .

Check. The reciprocal term differentiates to $2/(2x-1)^2$ . The logarithms give the other two fractions, whose recombination we already checked.

Use one arbitrary constant for the sum on each connected interval. Absolute values keep the logarithms real for either sign of a non-zero argument; they do not permit crossing a denominator zero.

Exam tip

Keep the exact coefficient from each supplied rule, and include the constant in the finished antiderivative. Differentiate to check the mathematics; a check does not replace the working requested by “Determine” or detailed reasoning.

Change the repeated term's numerator coefficient.

Reciprocal gaps in a rational model

A short mathematical article can define a new object using familiar algebra. Read the definitions first, keep their conditions, and identify which result you are being asked to establish rather than assume.

Reciprocal gaps

For a positive real input $t$ , define a reciprocal level $L(t)=1/t$ . Increase the input by a positive step $h$ . The decrease in level is

D_h(t)=L(t)-L(t+h),\qquad t>0,\quad h>0.

Combining the two fractions gives the rational model

D_h(t)=\frac{h}{t(t+h)}.

The subscript $h$ names the chosen step; it is not multiplication by $h$ . The model is defined for positive inputs and positive steps. Its algebraic fraction alone has the weaker restrictions $t\ne0,-h$ . When several decreases are added, an intermediate reciprocal level may appear with opposite signs and disappear. Only levels that actually match can cancel. The formula is exact for the defined reciprocal levels; it makes no claim about another system's behaviour without evidence that its levels follow $1/t$ .

The passage gives two ways of writing the same decrease. We can establish their equivalence by a numerator identity, then use the split to expose cancellation.

Worked example 11: read the unit-step model

Question. For $h=1$ , establish a partial-fraction form for $D_1(t)$ , then find $D_1(1)+D_1(2)+D_1(3)+D_1(4)$ .

Plan. Decompose $1/(t(t+1))$ , then write every reciprocal before cancelling. Call the requested sum $S$ .

\begin{aligned} 1&\equiv A(t+1)+Bt && \text{cleared identity}\\ A&=1,\quad B=-1 && \text{compare coefficients}\\ D_1(t)&=\frac1t-\frac1{t+1} && \text{partial fractions}\\ S&=1-\tfrac12+\tfrac12-\tfrac13+\tfrac13-\tfrac14+\tfrac14-\tfrac15 && \text{write four terms}\\ &=1-\tfrac15=\tfrac45 && \text{cancel matching levels} \end{aligned}

Why this step? The cancellation concerns equal reciprocal levels, not cancelling terms inside a single numerator.

Answer. $D_1(t)=1/t-1/(t+1)$ for $t>0$ ; the requested total is $4/5$ .

Check. Directly, $1/2+1/6+1/12+1/20=4/5$ . More surprisingly, the first $100$ positive terms add to exactly $1-1/101=100/101$ , still less than $1$ .

Change one feature: the step is $h=2$ . The matching levels are two positions apart, so a different set of boundary terms survives.

Worked example 12: change the step size

Question. For $h=2$ , establish a partial-fraction form for $D_2(t)$ , then find $D_2(1)+D_2(2)+D_2(3)+D_2(4)$ .

Plan. Decompose $2/(t(t+2))$ , then expand the four-term sum, called $S$ .

\begin{aligned} 2&\equiv A(t+2)+Bt\quad\Rightarrow\quad A=1,\ B=-1 && \text{compare coefficients}\\ D_2(t)&=\frac1t-\frac1{t+2} && \text{use the new step}\\ S&=1-\tfrac13+\tfrac12-\tfrac14+\tfrac13-\tfrac15+\tfrac14-\tfrac16 && \text{write every term}\\ &=1+\tfrac12-\tfrac15-\tfrac16=\tfrac{17}{15} && \text{cancel matching levels} \end{aligned}

Why this step? Adjacent terms no longer cancel completely. Keeping the terms visible prevents you from importing the unit-step answer.

Answer. $D_2(t)=1/t-1/(t+2)$ for $t>0$ ; the total is $17/15$ .

Check. Direct substitution gives $2/3+1/4+2/15+1/12=17/15$ . All four inputs satisfy the model's condition.

For a general positive step the same identity is $h\equiv(t+h)-t$ , so the two numerators are always $1,-1$ . This rational-model decomposition is not in the formula sheet: derive it from the given definition.

Watch out

Wrong: an input is acceptable whenever the simplified fraction is defined. Right: keep the passage's conditions too. At $h=2,t=-1$ , the fraction is defined, but the model requires $t>0$ .

Keep the two-unit step and shift the inputs in the sum.

Quick Recap

Give each distinct linear factor a constant numerator; a squared factor needs both powers.
Clear denominators and use the resulting polynomial identity to find and check the constants.
Supplied expansion or integration rules can then act on one simple fraction at a time, with their own validity conditions.
In a mathematical passage, distinguish a new definition, the result to establish and the input conditions; matching terms may reveal cancellation.

The hundred fractions add to less than one because each is $1/n-1/(n+1)$ : every intermediate reciprocal cancels, leaving exactly $100/101$ .

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Ma15 - Partial fractions with linear factors

Splitting two distinct linear factors

Handling three distinct factors

Including every power of a repeated factor

Using the split in a power series

Using the split in an integration step

Reciprocal gaps in a rational model

Reciprocal gaps

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