Study Companion: Programme F.8 — Partial Fractions

You already know how to add \frac2{x-1}+\frac3{x+2} into one fraction over a common denominator. Programme F.8 is that process run in reverse: given the single combined fraction, recover the simpler fractions it came from, which the book calls its partial fractions. The one idea underneath every case in the chapter — distinct simple factors, repeated factors, an irreducible quadratic factor, an unfactorised cubic — is that once the denominators are cleared, the two numerators are the same polynomial written two ways. That makes the equation between them an identity, true for every value of x, so any convenient value may be substituted, and like powers of x may be matched, to pin down one unknown at a time.

The book's words are used throughout: a denominator is broken into prime factors, a linear factor ax+b is a simple factor, and a quadratic factor that will not split is irreducible. Words and results that are ours rather than the book's are flagged where they first appear.

The rules, procedures and drill are in the companion reference sheet, which follows the book's topic order. This file tells the story in 5 units instead, because the programme's many templates are really 5 ideas.

How to use this companion

Each unit is an argument: a real situation and a question, the attempt a sensible person would make first, the picture that settles it, a check with numbers, and only then the rule. Read it once front to back. The Narration spine at the end of each unit summarises what the video shows.

Unit 01: Undoing an Addition

The problem

A survey drone flies 1200 m straight into a steady 5 m/s wind and then 1200 m back, at an airspeed of v m/s. Its flight planner stores the round trip as one formula,

so at v=15 m/s the round trip takes \frac{36\,000}{200}=180 s. How much of that is spent flying into the wind? The formula was made by adding the times of the two legs, and the question is how to take the sum apart again.

Strip away the drone and this is the book's opening question. Adding is routine:

Handed only the right-hand side, how do you recover the two fractions that were added to make it?

First attempt

Each simple factor of the denominator (the book's name for a linear factor) presumably came from one fraction, so write the split with unknown numerators,

and multiply both sides by (x-1)(x+2):

A and B are tangled together, but two ordinary values of x untangle them. At x=0: 1=2A-B. At x=2: 11=4A+B. Adding the two, 12=6A, so A=2 and then B=3. It works, but it costs a pair of simultaneous equations, and three factors would cost three. The values that would make life easy are x=1 and x=-2, since each one empties a bracket. Those are exactly the two values at which the original fraction does not exist. Is it legal to use them?

The picture

Read the cleared equation as a graph. Its left side, 5x+1, is a straight line. Its right side, A(x+2)+B(x-1), is a straight line too, whatever A and B are. The split was only claimed where the fraction exists, so all we know so far is that the two lines agree at every x except, possibly, x=1 and x=-2: the line y=5x+1 with two holes punched in it.

Two straight lines that share even two points are the same line, and these two share every point except the holes. So the right-hand line runs straight through both holes: it passes through (1,\ 6) and (-2,\ -9) as well. The equation holds for every x, the excluded values included. It is an identity, written with \equiv:

The same argument in symbols: the difference D(x)=5x+1-A(x+2)-B(x-1) is a polynomial of degree at most 1 that is zero at infinitely many values of x. By the factor theorem of F.3 Unit 04, each zero r of a polynomial gives it a factor (x-r), so a polynomial of degree n that is not identically zero has at most n zeros. D has more than one, so D is the zero polynomial, and it is zero at x=1 and x=-2 too. Nothing here depends on the degree being 1: two polynomials of degree at most n that agree at n+1 points are the same polynomial.

That licenses the key move: choose an x that empties one bracket.

The heights of the two holes, 6 and -9, are exactly the numbers the two substitutions use.

Check it with numbers

The substitutions give A=2 and B=3, the same pair the first attempt found the slow way. Check the split at an ordinary point, x=0, which neither substitution used:

That confirms the numbers. The line running through its holes is the reason they had to work.

The rule it gives

The book's procedure for a denominator that is a product of distinct simple factors:

1. Factorise the denominator into its prime factors.
2. Assume each simple factor gives one partial fraction with a constant numerator.
3. Add those fractions over the original denominator and equate the numerators. The result is an identity.
4. Substitute values of x that empty one bracket at a time.

with A=\frac{P(a)}{a-b} from x=a and B=\frac{P(b)}{b-a} from x=b.

The book says the assumption in step 2 is justified by succeeding in finding A and B. The picture proves more: the split always exists. Take A and B from those two formulas. Then A(x-b)+B(x-a) is a line that equals P(a) at x=a and P(b) at x=b, just as P(x) does, and P is a line too, since its degree is below 2. Two lines through the same two points coincide, so P(x)\equiv A(x-b)+B(x-a), and dividing by (x-a)(x-b) gives the split. The same argument works for any number of distinct simple factors. It also shows the answer is unique, because substituting each factor's root forces its constant. For repeated and quadratic factors the general existence theorem is stated without proof here (it belongs to a first course in algebra). Unit 04 proves the case of a single repeated factor; in every other case the template is confirmed by actually solving for its constants.

Beyond the book: substituting each factor's root is widely called the cover-up rule. To find A in \frac{5x+1}{(x-1)(x+2)}, cover up the (x-1) and evaluate what is left at x=1: \frac{5(1)+1}{1+2}=2. The book simply calls it substituting a suitable value of x.

Worked example

The book's first frame example: decompose \frac{8x-28}{x^2-6x+8}. Factorise the denominator first: x^2-6x+8=(x-2)(x-4), so

At x=4: 4=2B, so B=2. At x=2: -12=-2A, so A=6. So

Where this shows up

Back to the drone. Clearing the denominator of

gives 2400v\equiv A(v+5)+B(v-5). At v=5: 12\,000=10A, so A=1200. At v=-5: -12\,000=-10B, so B=1200:

Each piece is distance over ground speed: 1200 m at v-5 m/s into the wind, and 1200 m at v+5 m/s with it. At v=15 m/s the upwind leg takes \frac{1200}{10}=120 s and the return \frac{1200}{20}=60 s, which add to the 180 s the single formula gave. Two thirds of the flight is spent fighting the wind; in still air the round trip would take only \frac{2400}{15}=160 s. Notice which value of v found the upwind leg: v=5 m/s, the airspeed at which the drone hangs motionless against the wind and the formula for T has no value at all. The identity does not mind. That value is one of the picture's holes, filled in.

Narration spine. Two simple fractions combine into one over a common denominator, and the question runs it backwards with unknown numerators A and B. Clearing denominators gives one tangled equation, declared an identity that holds for every x. Substituting x=1 makes the B-term vanish and gives A=2; substituting x=-2 makes the A-term vanish and gives B=3. The split is checked at x=0, where both sides give -0.5, and the general rule and the book's \frac{8x-28}{x^2-6x+8} close the beat.

Unit 02: Improper Fractions — Divide First

The problem

A magnifying glass of focal length 10 cm throws a sharp image of a lamp u cm away onto a sheet of paper v cm behind the glass. The lens formula of F.5 Unit 02, \frac1{10}=\frac1u+\frac1v, solved for v, gives

so a lamp at 30 cm focuses at 15 cm. Carry the lamp further away and the image creeps in towards the glass. Where does it settle, and how far from that resting place is it for a given lamp distance?

The fraction \frac{10u}{u-10} has numerator and denominator of the same degree. So does the book's first example, \frac{x^2+3x-10}{x^2-2x-3}. Can the method of Unit 01 split it?

First attempt

Factorise the denominator, x^2-2x-3=(x+1)(x-3), assume \frac{x^2+3x-10}{(x+1)(x-3)}=\frac A{x+1}+\frac B{x-3}, and clear denominators:

Substituting still produces numbers without complaint: x=3 gives 8=4B, so B=2, and x=-1 gives -12=-4A, so A=3. Now test them at an ordinary point, x=0: the left side is -10, but the right side is 3(-3)+2(1)=-7. They disagree. The right side is only ever of degree 1, since nothing multiplies A or B by x^2, while the left side has an x^2 term that no choice of A and B can produce. The lens fails the same way: \frac{10u}{u-10}=\frac A{u-10} would need 10u\equiv A, a sloping line equal to a constant.

The picture

The template was built for fractions whose numerator has lower degree than the denominator. The book's rule when that fails: divide out first, by long division, exactly as in F.2 Unit 05:

The quotient is 1, with remainder 5x-7.

At x=10 this is ordinary arithmetic: 120=77\times1+43, and 5(10)-7=43. So x^2+3x-10=(x^2-2x-3)(1)+(5x-7), and dividing through by x^2-2x-3,

Now graph the fraction. The quotient 1 is the dotted line, and the remainder \frac{5x-7}{(x+1)(x-3)} is the gap between the curve and that line.

Far from the two vertical asymptotes the curve hugs the dotted line y=1: from below on the left, from above on the right. It crosses the line only once, at x=1.4, where the remainder's numerator 5x-7 is zero. The gap shrinks because the remainder's numerator has lower degree than its denominator:

So the fraction is a polynomial part, which rules far away, plus a remainder part, which rules near the asymptotes. A sum of constants over simple factors, \frac A{x+1}+\frac B{x-3}, dies away to 0 far out, so it can never follow a curve that settles at 1. That is why the first attempt failed: the template can only produce the remainder part, so it must be handed only the remainder.

Check it with numbers

Split the remainder as in Unit 01: 5x-7\equiv A(x-3)+B(x+1). At x=3: 8=4B, so B=2. At x=-1: -12=-4A, so A=3. So

At x=0 the left side is \frac{-10}{-3}=3.3333 and the right side is 1+3-0.6667=3.3333. \checkmark The numbers 3 and 2 from the first attempt were right all along. At a root of the denominator the quotient's contribution vanishes, so substituting found the remainder's constants; what the first attempt lacked was room for the quotient 1.

The rule it gives

The book's rule: the degree of the numerator must be less than the degree of the denominator. (Our shorthand, not the book's: such a fraction is proper, and otherwise it is improper.) If it is not, divide out first:

The division stops as soon as what remains has lower degree than Q, so the remainder R always meets the degree condition. The quotient S(x) has degree \deg P-\deg Q: a number when the degrees are equal, a linear polynomial when the numerator's degree is one more, and so on. Only the remainder fraction is split into partial fractions; the quotient is kept as it is.

Worked example

The book's Example 3: decompose \frac{3x^3-x^2-13x-13}{x^2-x-6}. The numerator's degree is one more than the denominator's, so the quotient is linear. Divide: 3x(x^2-x-6)=3x^3-3x^2-18x leaves 2x^2+5x-13, and then 2(x^2-x-6)=2x^2-2x-12 leaves 7x-1. So the quotient is 3x+2 and the remainder 7x-1:

Splitting the remainder, 7x-1\equiv A(x-3)+B(x+2): at x=3, 20=5B, so B=4; at x=-2, -15=-5A, so A=3. So

Far out, this curve hugs the sloping line y=3x+2 rather than a level one.

Where this shows up

Divide the lens formula the same way. Since 10u=10(u-10)+100,

The quotient, 10 cm, is the focal length: where the image of a very distant object settles. The remainder is how far beyond the focal length the image sits, and it shrinks as the lamp recedes: 5 cm for a lamp at 30 cm (so v=15 cm, as F.5 found), 1 cm for a lamp at 110 cm, and \frac{100}{490}=0.204 cm for a window 5 m away. This you can see for yourself. Stand a sheet of paper facing a window, and move a magnifying glass between them until a sharp, upside-down image of the window appears on the paper. The glass is then almost exactly its focal length from the paper: about 2 mm more, for a 10 cm glass and a window 5 m off. That is how a lens's focal length is measured. The remainder's vertical asymptote at u=10 cm is the blur that F.4 Unit 02 found when the glass is lifted to its focal length.

For a lens of focal length f the same division gives v=f+\frac{f^2}{u-f}, which is Newton's form of the lens equation, (u-f)(v-f)=f^2. It sets how far a camera must move its lens to focus. For a 50 mm lens, a subject at 2 m (2000 mm) needs the lens \frac{2500}{1950}=1.28 mm further from the sensor than for a distant scene, and a subject at 0.5 m needs \frac{2500}{450}=5.56 mm. The focusing mechanism is designed around that travel.

Narration spine. The Unit 01 template is tried directly on \frac{x^2+3x-10}{(x+1)(x-3)}; clearing denominators leaves a right side of degree one against an x^2 on the left, and the attempt is struck out. The fraction is plotted instead: far from its two asymptotes the curve hugs a dashed line at y=1, and the gap is the shrinking remainder. Long division gives quotient 1 and remainder 5x-7, the remainder splits as in Unit 01 into \frac3{x+1}+\frac2{x-3}, and the result is checked at x=0. The book's Example 3, with the quotient 3x+2, and the boxed rule close the beat.

Unit 03: Linear Factors, Including Non-Monic Ones

The problem

A full-wave rectifier turns a 50 Hz alternating voltage of peak 10 V into a pulsing voltage that never goes negative. Fourier analysis (beyond the book) shows that this output is a steady \frac{20}{\pi}=6.366 V plus ripple at 100, 200, 300,\ldots Hz, the kth ripple component having amplitude

How big are the first three components together, and what do all of them add up to? Adding a_1+a_2+a_3 is easy enough. Adding all of them is not, unless \frac1{(2k-1)(2k+1)} can be split, and its factors do not start with a bare k: their roots are \pm\frac12, not whole numbers.

The book's own example raises a second question. With three simple factors,

does substituting roots still work, and is there a method that does not depend on having a root for every unknown?

First attempt

Clearing the denominators of \frac A{x-2}+\frac B{x+4}+\frac C{x-1} gives

and substituting each root empties two of the three terms:

It is quick, and it works for any number of distinct simple factors. But it leans entirely on each unknown having a root of its own. Unit 04 meets a factor (3x+1)^3 that offers one root for three unknowns, and Unit 05 a quadratic factor with no rational root to substitute. There, substitution finds some of the constants but not all of them.

The picture

Expand every term of the right side and file each piece under its power of x:

The rows come from (x+4)(x-1)=x^2+3x-4, (x-2)(x-1)=x^2-3x+2 and (x-2)(x+4)=x^2+2x-8. Because the two sides are the same polynomial (Unit 01), each column must total the matching coefficient on the left. This is equating coefficients, and the book labels each equation by its power of x, with CT for the constant term:

No root was needed: every column gives one equation, whatever the factors are. Solve by eliminating one unknown. From [x^2], C=10-A-B. Substituting in [x]: 3A-3B+20-2A-2B=7, so A-5B=-13. Substituting in [\text{CT}]: -4A+2B-80+8A+8B=-42, so 4A+10B=38, that is 2A+5B=19. Adding A-5B=-13 and 2A+5B=19 gives 3A=6, so A=2. Then 5B=A+13=15, so B=3, and C=10-2-3=5.

The same three constants, by a second route. That is no coincidence. Substitution showed that whatever constants make the identity true, x=2 forces A=2, x=-4 forces B=3 and x=1 forces C=5: there is only one possible answer. Equating coefficients found constants that make every column total correct, so they make the identity true, and they can only be that one answer. The two methods read the same identity two ways, and uniqueness makes them agree.

Check it with numbers

At x=0 the left side is \frac{-42}{(-2)(4)(-1)}=\frac{-42}8=-5.25 and the right side is -1+0.75-5=-5.25. \checkmark The column totals check as well: 2+3+5=10, 6-9+10=7 and -8+6-40=-42.

The rule it gives

Equating coefficients (the book's method): expand the cleared identity, collect like powers of x, and set each coefficient on the right equal to the one on the left, one equation per power. It needs no roots, so it works for every kind of factor. Substitution stays the quicker route wherever a root isolates a constant: every simple factor that is not repeated, and the top rung of a repeated one (Unit 04). The two can be mixed freely, as Unit 05 does.

Simple factors ax+b ("non-monic" factors, in our words). A simple factor need not start with a bare x; the book's rule gives ax+b its own constant, \frac A{ax+b}. Its root is x=-\frac ba, so that is the value to substitute. For \frac{7x-5}{(2x-1)(3x-2)}, assume \frac A{2x-1}+\frac B{3x-2}, so 7x-5\equiv A(3x-2)+B(2x-1). At x=\frac12: -\frac32=-\frac12A, so A=3. At x=\frac23: -\frac13=\frac13B, so B=-1:

Nothing new is needed, because ax+b=a\left(x+\frac ba\right) is a factor with a bare x times a constant, and the unknown numerator absorbs the constant. In general, for distinct simple factors,

with each A_i found by substituting x=-\frac{b_i}{a_i} or by equating coefficients.

Worked example

Decompose \frac{9x^2-4x+1}{(2x+1)(x-1)(3x-2)}. Clearing denominators,

At x=-\frac12: \frac94+2+1=\frac{21}4 and A\left(-\frac32\right)\left(-\frac72\right)=\frac{21}4A, so A=1. At x=1: 6=B(3)(1), so B=2. At x=\frac23: 4-\frac83+1=\frac73 and C\left(\frac73\right)\left(-\frac13\right)=-\frac79C, so C=-3:

Check at x=0: the left side is \frac1{(1)(-1)(-2)}=0.5 and the right side 1-2+1.5=0.5. \checkmark

Where this shows up

Back to the rectifier. Clearing the denominator of \frac1{(2k-1)(2k+1)}=\frac A{2k-1}+\frac B{2k+1} gives 1\equiv A(2k+1)+B(2k-1). At k=\frac12: 1=2A, so A=\frac12. At k=-\frac12: 1=-2B, so B=-\frac12. So

Added up, the pieces cancel in pairs (beyond the book: a telescoping sum):

So the first n components total \frac{40}{\pi}\cdot\frac n{2n+1} V. For n=3 that is 12.732\times\frac37=5.457 V, matching 4.244+0.849+0.364=5.457 V added directly. As n grows, \frac n{2n+1} approaches \frac12, so all the ripple components together total \frac{40}{\pi}\cdot\frac12=6.366 V: exactly the steady level. They must. At the instants the output touches 0 V, every component is at its negative peak at once, and together they cancel the steady level. The split also says where to aim a smoothing filter: the 100 Hz component alone, 4.244 V, is two thirds of the total. (The values k=\pm\frac12 are harmonics that do not exist; as in Unit 01, the identity does not mind.)

Narration spine. Three simple factors: substitution at x=2, -4 and 1 in turn isolates one unknown at a time and gives A=2, B=3, C=5, but it needs a root for every unknown, which a repeated or quadratic factor will not supply. The identity is expanded instead and collected by powers of x; equating the [x^2], [x] and [\text{CT}] coefficients gives three equations whose solution is the same 2, 3, 5. A factor written ax+b works the same way: \frac{7x-5}{(2x-1)(3x-2)} is split by substituting the roots \frac12 and \frac23, and the rule for any simple factor closes the beat.

Unit 04: Repeated Factors Need a Ladder

The problem

A signal generator with an open-circuit output of 10 V and an internal resistance of 50 Ω drives a load of resistance R Ω. The load receives the power

Trying loads gives 0.444 W at R=25 Ω, 0.5 W at 50 Ω and 0.444 W again at 100 Ω. The best load looks like 50 Ω, but a table cannot prove that no value in between does better. The denominator is a repeated factor, (R+50)^2. The book's example has one cubed:

Does a repeated factor still get just one partial fraction, the way a simple factor did in Unit 01?

First attempt

One factor, one fraction:

The right side is a constant; the left is a full quadratic. At x=0 the identity would need A=6, and at x=1 it would need A=27. One unknown cannot match three independent coefficients.

The picture

Rename the repeated factor u=3x+1, so that x=\frac{u-1}3, and rewrite the numerator in powers of u:

Dividing by u^3 then splits it at once:

The numerator, written in powers of the repeated factor, is the answer. Its three "digits" in base u (the idea of F.2 Unit 05, a polynomial as a numeral in an unfixed base) become the numerators over u, u^2 and u^3. Every power of u below the third can appear in the numerator, so every power from u^1 to u^3 is needed underneath: a ladder of three rungs (our word, not the book's).

The same fact, seen through equating coefficients. Clear the denominators of \frac A{3x+1}+\frac B{(3x+1)^2}+\frac C{(3x+1)^3} and expand A(3x+1)^2+B(3x+1)+C column by column:

Each row starts one column further right than the row above it, so the rows form a staircase. The x^2 column involves only A, the x column brings in B, and the constant column brings in C. Reading left to right, each column adds exactly one new unknown with a nonzero multiplier (9, 3, 1), so each column fixes one unknown, and the equations have exactly one solution whatever the numerator is. Three coefficients, three rungs: exactly enough.

What if a rung is skipped? Drop the middle one and keep A and C: 18x^2+3x+6\equiv A(3x+1)^2+C=9Ax^2+6Ax+(A+C). The x^2 column demands 9A=18, so A=2; the x column demands 6A=3, so A=0.5. Two different values of A at once: no solution. In the staircase, removing a row leaves a column with no new unknown of its own, and its equation is almost always violated.

Check it with numbers

Solve down the staircase: 9A=18 gives A=2; 6A+3B=3 gives 12+3B=3, so B=-3; A+B+C=6 gives 2-3+C=6, so C=7. These are the digits 2, -3, 7 of the base-u numerator:

At x=0 (u=1) the left side is \frac61=6 and the right side 2-3+7=6. \checkmark Substituting the root x=-\frac13 still finds the top rung directly, 18\left(\frac19\right)-1+6=7=C, and that is the only constant it can reach. Each rung also rules at its own distance from the root: at u=0.1 the three terms are 20, -300 and 7000, while at u=10 they are 0.2, -0.03 and 0.007. No single term could follow the curve at both distances.

The rule it gives

The book's rule for a repeated simple factor:

and in general one term for every power from 1 to n, none skipped. The book states the rule without a reason. The reason (beyond the book) is the base-u rewrite above: with u=ax+b, a numerator of degree less than n is a polynomial of degree less than n in u, and dividing it by u^n gives the ladder. So for a single repeated factor the ladder always exists and is unique. When a repeated factor sits alongside other factors, each factor contributes its own terms, for example

That combined template is stated without proof here; solving for its constants confirms it each time.

Worked example

The book's squared-factor example: decompose \frac{35x-14}{(7x-2)^2}. Assume \frac A{7x-2}+\frac B{(7x-2)^2}, so 35x-14\equiv A(7x-2)+B. Equating the x coefficients: 35=7A, so A=5. Equating constant terms: -14=-2A+B=-10+B, so B=-4. (In base u=7x-2: 35x-14=5u-4, the same two digits.) So

Where this shows up

Split the power delivered to the load the same way: 100R\equiv A(R+50)+B. Equating the R coefficients gives A=100, and equating constant terms gives 0=50A+B, so B=-5000:

Write g=\frac1{R+50}, the total conductance of the circuit in siemens. Then P=100g-5000g^2, a quadratic in g, and completing the square (F.3 Unit 05) finishes the job:

The square is never negative, so P is never more than 0.5 W, and it reaches 0.5 W only when g=0.01 S, that is R+50=100, so R=50 Ω. This is the maximum power transfer rule: a load takes the most power when its resistance matches the source's internal resistance. Here it is proved with no calculus, because the two-rung ladder turned the fraction into a quadratic. It is a real design rule: an 8 Ω loudspeaker on this 50 Ω generator receives only \frac{100}{58}-\frac{5000}{3364}=1.7241-1.4863=0.238 W, under half of what is available. (A second example, beyond the book: in control engineering, a repeated factor in the transfer function of a critically damped system is what produces a term t\,e^{at} in its response, which no single unrepeated factor can produce.)

Narration spine. A single unknown over the cubed factor is tried and struck: a constant cannot match a quadratic numerator. Counting settles it: three coefficients, of x^2, x and 1, need three rungs, and the ladder \frac A{3x+1}+\frac B{(3x+1)^2}+\frac C{(3x+1)^3} is solved by matching x^2, x and the constant in turn, giving A=2, B=-3, C=7, checked at x=0. Dropping the middle rung demands A=2 and A=0.5 at once, and that attempt is struck too. The book's squared example, \frac{35x-14}{(7x-2)^2}, and the general ladder rule close the beat.

Unit 05: Quadratic Factors, and an Unfactorised Cubic

The problem

A river ferry runs a circuit: 8 km upstream against a 2 km/h current, 8 km back down with it, then 6 km up a tidal channel against a 3 km/h tide. At a speed through the water of v km/h, the operator's spreadsheet gives the circuit time as one fraction,

which at v=10 km/h is \frac{1696}{672}=2.524 h. How long is each leg? The denominator arrives as a single cubic, and it must be factorised before any template can be chosen: the book calls this case the peak of the programme. A cubic, once one factor has been found and divided out, leaves a quadratic that may or may not split further. The book's example of one that will not:

What numerator does a quadratic factor like 2x^2-6x+3 need?

First attempt

Treat the quadratic like a simple factor, with a constant numerator: \frac A{x-4}+\frac C{2x^2-6x+3}. Then

Substituting x=4 empties the C-term: 112-72-7=33=A(32-24+3)=11A, so A=3. But the x^2 column needs 7=2A, so A=3.5. Two different values of A: no solution. The constant over the quadratic is one unknown short.

The picture

First, confirm that the quadratic really will not split. The book's test for a quadratic ax^2+bx+c with whole-number coefficients: it has simple factors with rational coefficients only when b^2-4ac is a perfect square, because its roots \frac{-b\pm\sqrt{b^2-4ac}}{2a} are rational exactly when that square root is a whole number. Here b^2-4ac=36-24=12, not a perfect square, so 2x^2-6x+3 is an irreducible quadratic factor (the book's term). Its roots, x=\frac{3\pm\sqrt3}2\approx0.634 and 2.366, are real but irrational. Irreducible here means "no simple factors with rational coefficients", not "no real roots"; a negative b^2-4ac is only the special case with no real roots at all.

Now see why a constant numerator cannot work. Substituting x=4 is still sound, so A=3, and whatever sits over the quadratic must equal the original fraction minus \frac3{x-4}. Evaluate that leftover at two points placed symmetrically about x=1.5, the quadratic's line of symmetry. At x=0.5 the original fraction is \frac{57}7 and -\frac3{0.5-4}=\frac67, so the leftover is 9; at x=2.5 it is 11+2=13.

The quadratic is a parabola, symmetric about x=1.5, and it takes the same value, 0.5, at both points. So any \frac C{2x^2-6x+3} would give the same height, 2C, at both. The leftover is lopsided, 9 on one side and 13 on the other, and no constant can make it so. A numerator that changes with x can: Bx+C. In counting terms, the numerator over a quadratic may have any degree below 2, which leaves two coefficients to match, so the template needs two unknowns there: a linear numerator.

The capstone: an unfactorised cubic. When the whole denominator arrives as a cubic, such as x^3-4x^2+x+6, it is factorised first by the remainder theorem of F.3 Unit 04. Write it in nested form (F.3 Unit 03), f(x)=[(x-4)x+1]x+6, and try x=1,-1,2,-2,\ldots in turn. f(1)=[(1-4)(1)+1](1)+6=-2+6=4, not zero, so (x-1) is not a factor. f(-1)=[(-1-4)(-1)+1](-1)+6=-6+6=0, so (x+1) is a factor. Divide it out by long division (F.2 Unit 05):

The quotient is x^2-5x+6.

At x=10: 616\div11=56=100-50+6. The quotient x^2-5x+6 passes the perfect-square test (25-24=1) and splits as (x-2)(x-3). So

three distinct simple factors after all, and Unit 03's method finishes the job. Had the quotient failed the test, it would have stayed whole as an irreducible quadratic factor with a Bx+C numerator.

Check it with numbers

Quadratic factor. Clear the denominators with a linear numerator:

where (Bx+C)(x-4)=Bx^2+(C-4B)x-4C. Equating coefficients:

Substituting x=4 gave A=3 directly (the book reaches 11A=33 by elimination instead). Then [x^2] gives B=7-6=1, and [\text{CT}] gives -7=9-4C, so C=4. The [x] equation is left over as a check: -6(3)-4(1)+4=-18. \checkmark So

and the quadratic's term gives \frac{4.5}{0.5}=9 at x=0.5 and \frac{6.5}{0.5}=13 at x=2.5: the lopsided heights of the picture. At x=0 the left side is \frac{-7}{(-4)(3)}=0.5833 and the right side -0.75+1.3333=0.5833. \checkmark

Unfactorised cubic. With the denominator factorised, 8x^2-14x-10\equiv A(x-2)(x-3)+B(x+1)(x-3)+C(x+1)(x-2). At x=-1: 8+14-10=12=A(-3)(-4)=12A, so A=1. At x=2: 32-28-10=-6=B(3)(-1)=-3B, so B=2. At x=3: 72-42-10=20=C(4)(1)=4C, so C=5. (The book equates coefficients instead, A+B+C=8, 5A+2B+C=14 and 6A-3B-2C=-10, and reaches the same three numbers.) So

At x=0 the left side is \frac{-10}{6}=-1.6667 and the right side 1-1-1.6667=-1.6667. \checkmark

The rule it gives

The book's rule for an irreducible quadratic factor ax^2+bx+c (whole number coefficients, b^2-4ac not a perfect square): it gives a partial fraction with a linear numerator. With a simple factor px+q alongside,

Find A by substituting x=-\frac qp and the rest by equating coefficients, or everything by equating coefficients. (The book's frames write \frac{Bx+C}{\ldots} as here; its review summary writes the same rule as \frac{Ax+B}{ax^2+bx+c}.) The existence of this split in general is stated without proof; each time, solving for the constants confirms it.

A denominator handed over unfactorised is factorised first: nest it, try x=1,-1,2,-2,\ldots by the remainder theorem, and divide out each factor found, until every factor is visible. Only then is a template chosen, factor by factor.

Worked example

The book's first quadratic-factor example: decompose \frac{15x^2-x+2}{(x-5)(3x^2+4x-2)}. Test the quadratic: b^2-4ac=16+24=40, not a perfect square, so it is irreducible. Clear the denominators:

At x=5: 375-5+2=372=A(75+20-2)=93A, so A=4. Equating coefficients: [x^2] gives 15=3A+B, so B=3; [\text{CT}] gives 2=-2A-5C=-8-5C, so C=-2. Check with [x]: 4A-5B+C=16-15-2=-1. \checkmark

Where this shows up

Now the ferry. Nest the cubic, f(v)=[(v-3)v-4]v+12, and try values: f(1)=6, f(-1)=12, and f(2)=[(-1)(2)-4](2)+12=-12+12=0, so (v-2) is a factor. Dividing it out leaves v^2-v-6, which passes the perfect-square test (1+24=25) and splits as (v-3)(v+2). So the denominator is (v-2)(v+2)(v-3), and

At v=2: -32=-4A, so A=8. At v=-2: 160=20B, so B=8. At v=3: 30=5C, so C=6:

Each term is one leg: 8 km at v-2 km/h against the current, 8 km at v+2 km/h with it, and 6 km at v-3 km/h against the tide. At v=10 km/h the legs take 1 h, 0.667 h and 0.857 h, together the 2.524 h the single fraction gave. The split also shows what the combined fraction hides: the timetable fails outright for any boat that cannot beat 3 km/h through the water, because the tidal leg never ends. That is the root v=3 of the cubic.

(Beyond the book: when b^2-4ac is negative, so that the quadratic has no real roots at all, the same Bx+C template is the one control engineers use for a lightly damped, oscillating mode. That application needs the Laplace transform, met well after this programme. The book's quadratic factors, with b^2-4ac=40 and 12, have real roots and do not oscillate.)

Narration spine. A bare constant over the quadratic factor is tried and struck: one unknown for two degrees of freedom. The book's test is applied: b^2-4ac=12 is not a perfect square, so the roots are real but irrational and the quadratic stays whole. With a linear numerator Bx+C the split \frac3{x-4}+\frac{x+4}{2x^2-6x+3} follows, checked at x=0. Then the capstone: the cubic x^3-4x^2+x+6 is nested, the trial f(1)=4 is struck and f(-1)=0 accepted, and dividing out (x+1) leaves (x-2)(x-3). The final split \frac1{x+1}+\frac2{x-2}+\frac5{x-3} is checked at x=0.