Study Companion: Programme F.6 — Polynomial Equations

In F.3 a polynomial was something to evaluate: choose x, run the nesting chain, read off the value. This programme runs the same process backwards. The value is fixed at zero, and the question is which inputs produce it. Every method here is a way of answering that question with tools already built: the formula finishes any quadratic (over the complex numbers; Unit 01 says when its roots are real); the factor theorem turns one lucky trial into a linear factor; long division strips that factor off and drops the degree by one. The one idea underneath is that a root found is a degree removed, so a cubic needs one root found before the formula can finish it, and a fourth-order equation needs two.

The rules, procedures and drill are in the companion reference sheet, which follows the book's topic order. This file tells the story in four units instead, one for each of the book's topics: quadratics, cubics, fourth-order equations, and the equations the method cannot touch.

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: Solving Is Evaluation Run Backwards

The problem

A stone is thrown straight up at 12 m/s from the edge of a flat roof 4 m above the ground. Taking g as 10 m s^{-2}, its height after t seconds is

Evaluating is easy: after 2 s the stone is h=4+24-20=8 m up. The question runs the other way. When does it hit the ground — for which t is h=0?

First attempt

Try to factorize 5t^2-12t-4, as in F.2: look for two whole numbers with product 5\times(-4)=-20 and sum -12. The pairs are (1,-20), (2,-10), (4,-5) and their negatives, with sums \pm19, \pm8 and \pm1. None gives -12, so the quadratic has no factors with whole-number coefficients.

Try values instead: h(2)=8 and h(3)=4+36-45=-5. The stone lands somewhere between 2 and 3 seconds — bracketed, but not found, and every closer guess costs another calculation.

The picture

Draw the height against time. Evaluating goes up from a time on the axis to the curve and across to a height. Solving goes the other way: start at the height 0, run along the axis, and drop down from wherever the curve meets it.

The time axis is drawn to scale, one column to 0.1 s, with the height axis at t=0: the moment of the throw, where the curve passes through the roof height 4 m. The curve is symmetric about its peak — a stone takes as long to rise from any height to the top as to fall back to it. So the two points where it meets h=0 sit an equal distance either side of the peak time. F.3 completed the square to show exactly this:

The peak is 11.2 m, at t=1.2 s. The stone is at h=0 when 5(t-1.2)^2=11.2, that is (t-1.2)^2=2.24, so

That is all the quadratic formula says. Its first part, -\frac b{2a}, is the centre — the time of the peak. Its second part, \frac{\sqrt{b^2-4ac}}{2a}, is the half-gap — here, the time to fall from the peak to the ground.

Check it with numbers

Write 5t^2-12t-4=0, so a=5, b=-12, c=-4:

So t=1.2+1.4967=2.697 s or t=1.2-1.4967=-0.297 s — the same centre and half-gap the picture gave, and \frac{224}{100}=2.24 is the (t-1.2)^2 of the completed square. The stone lands at t=2.697 s, between the bracketing guesses 2 and 3.

The rule it gives

Quadratic equations (the formula method). For ax^2+bx+c=0 with a\ne0,

The formula was derived in F.3 Unit 05 by completing the square; here it is simply used. Substitute a, b and c with their signs — for b=-12, -b is +12 and b^2 is (-12)^2=144. Work out b^2-4ac first, take its square root to four decimal places, and round the two answers once, at the end, usually to 3 dp.

Beyond the book: the number b^2-4ac (the discriminant) decides how many real roots there are, because it is 4a^2 times the square of the half-gap. When will the stone be 12 m up? 4+12t-5t^2=12 gives 5t^2-12t+8=0, with b^2-4ac=144-160=-16. A negative number has no real square root, so there is no real time: the stone peaks at 11.2 m and never gets there. Asked for exactly 11.2 m, b^2-4ac=0 and the two times merge into one, t=1.2 s. So "every quadratic can be solved" is true only over the complex numbers; over the reals a quadratic has two, one or no roots. Every quadratic in this programme has b^2-4ac>0.

Beyond the book: centre and half-gap also give two quick checks on a pair of roots. Write the roots as m\pm d, with centre m=-\frac b{2a} and half-gap d=\frac{\sqrt{b^2-4ac}}{2a}. Adding them, the half-gaps cancel:

Multiplying them is a difference of two squares:

For the stone, a=5, b=-12, c=-4: the landing times add to -0.297+2.697=2.4=\frac{12}5, and multiply to m^2-d^2=1.2^2-2.24=1.44-2.24=-0.8=-\frac45. A pair of answers that fails either check has a slip in it.

Worked example

Solve 2x^2-3x-4=0. Here a=2, b=-3, c=-4:

The centre is \frac34=0.75 and the half-gap 1.6008; the roots sit symmetrically about 0.75.

Where this shows up

The stone lands 2.697 s after it is thrown — time it with a phone next time something is tossed off a balcony. The other root, t=-0.297 s, is not nonsense: it is when a stone on the same parabola would have had to leave the ground to pass the roof edge at 12 m/s. The equation knows nothing about the roof, so it offers both times; the physics — the stone was thrown at t=0 — rejects the negative one. Every projectile, fountain jet and braking-distance question that asks "when" or "where" instead of "how high" is a quadratic run backwards in exactly this way.

Narration spine. Draw h=4+12t-5t^2. Evaluate once, up from t=2 to h=8; then run backwards from h=0 to the two meeting points. Mark the peak at t=1.2 and the two equal half-gaps of 1.497. Build the formula as centre \pm half-gap and land on t=2.697 s. Beyond the book: raise the target height and watch the two meeting points close together at 11.2 m and vanish above it.

Unit 02: A Cubic Is One Root Away From a Quadratic

The problem

A card 12 cm by 8 cm has a square of side x cm cut from each corner, and the flaps are folded up into an open tray. Its volume is

Which cut gives a tray holding exactly 64 cm^3? Setting V=64, multiplying out and dividing by 4:

First attempt

Try whole numbers: x=1 gives V=1\times10\times6=60, and x=2 gives V=2\times8\times4=64. A 2 cm cut works — done?

Not quite. Slide the cut from 0 to 4 cm (beyond 4 the 8 cm side is used up): the volume rises to about 67.6 cm^3 near x=1.57 and falls away again. It passes 64 twice — once at x=2 on the way down, and once somewhere between 1 and 1.57 on the way up. No whole-number trial hits that second cut. And there is no cubic formula to fall back on: one exists, but it is far too long to use by hand.

The picture

Write f(x)=x^3-10x^2+24x-16 and draw it. Its roots are the points where the curve meets the axis, and there are up to three.

Each trial is one nested evaluation of [(x-10)x+24]x-16, and its value is the height of the curve at that x:

By the factor theorem, f(2)=0 means f(x)=(x-2)\times(\text{a quadratic}), and long division gives the quadratic. (Alternatively, the chain's values before the final 0, namely 1,-8,8, are its coefficients: this is synthetic division, from F.3 Unit 04, an alternative to the long-division layout, not a replacement for it.)

Now the key step. A product is zero only when one of its factors is zero. So f(x)=0 exactly when x-2=0 or x^2-8x+8=0 — the other roots of the cubic are precisely the roots of the quadratic, no more and no fewer. On the graph, at every x the cubic's height is the line y=x-2's height times the parabola y=x^2-8x+8's height, so away from x=2 the cubic meets the axis exactly where the parabola does. Finding one root turned a cubic we cannot solve into a quadratic we can.

Check it with numbers

The formula on x^2-8x+8=0: x=\frac{8\pm\sqrt{64-32}}{2}=\frac{8\pm5.6569}{2}, so x=1.172 or x=6.828.

The first is the second cut. Its tray is 12-2.343=9.657 cm by 8-2.343=5.657 cm by 1.172 cm deep: 9.657\times5.657\times1.172\approx64.0 cm^3. The root x=6.828 is a real root of the cubic, but a cut of 6.828 cm from an 8 cm side is impossible, so the tray rejects it.

The rule it gives

Cubic equations having at least one simple linear factor.

1. Write f(x) in nested form, with a zero coefficient for any missing power.
2. Apply the remainder theorem: evaluate f(1),f(-1),f(2),f(-2),\ldots in that order until some f(k)=0. Record each failure. Then (x-k) is a factor — note the sign: f(-2)=0 gives the factor (x+2).
3. Long-divide f(x) by (x-k) (the layout of F.2 Unit 05) to get f(x)=(x-k)(ax^2+bx+c).
4. So x=k or ax^2+bx+c=0; solve the quadratic by the formula.

The three solutions are x=k, x=x_1, x=x_2. The whole method depends on step 2 succeeding: the cubic must have at least one simple linear factor.

Worked example

Solve 3x^3+12x^2+13x+4=0. Nested: f(x)=[(3x+12)x+13]x+4.

f(1)=32, so (x-1) is not a factor. f(-1): 3\to9\to4\to0, so (x+1) is a factor. Long division:

The quotient is 3x^2+9x+4.

Where this shows up

Fold both trays from card and fill them with rice: the 2 cm tray (8\times4 cm base) and the 1.172 cm tray (9.657\times5.657 cm base) hold the same 64 cm^3. A packaging engineer choosing between them trades depth against footprint, 32 cm^2 against 54.6 cm^2 of shelf. The whole-number trial found the deep one; only dividing out its root and finishing with the formula found the shallow one — and the same algebra produced a third "tray", at 6.828 cm, that only the card itself could rule out.

Narration spine. Show the card with its four corner squares and slide the cut: the volume readout climbs past 64 and comes back. Plot the cubic and trial 1, -1, 2 by nesting, as dots at their heights. At f(2)=0, write (x-2)(x^2-8x+8) and draw the parabola through the cubic's other two meeting points. Finish with the formula: 1.172 is the second tray, 6.828 is struck out by the card.

Unit 03: Fourth-Order Equations Come Down One Factor at a Time

The problem

A pine shelf L=2 m long is built into a wall at one end, so it leaves the wall level, and rests on a bracket at the other end, at the same height. It is 300 mm wide and 20 mm thick, so its bending stiffness is EI=(10\times10^9\ \text{N/m}^2)\times\frac{0.3\times0.02^3}{12}\ \text{m}^4=2000 N m^2, and books load it at w=300 N/m (about 31 kg per metre). Beam theory gives its sag y, measured downwards at distance x from the wall, as

Halfway along, at x=1 m, it sags y=\frac{1\times2}{160}=0.0125 m =12.5 mm. The sag is zero at the wall and at the bracket, yet sight along the shelf and it does two different things there: it leaves the wall flat, and it arrives at the bracket tilted. Setting y=0 gives the fourth-order equation x^4-5x^3+6x^2=0, which can have up to four roots. What are the other two, and what in the equation tells a flat meeting from a tilted one?

The shelf's equation shows part of its answer at sight: x^2 is plainly a factor. Most fourth-order equations hide their factors, so learn the method on one that does. Where, and how, does the curve

meet the x-axis? A fourth-order equation has at most four real roots, and the formula finishes only the last two. So two linear factors have to be found first.

First attempt

Trial in nested form, \{[(3x+2)x-15]x+12\}x-2: at x=1 the chain is 3\to5\to-10\to2\to0, so (x-1) is a factor and

Now hunt for a factor of the cubic F(x). The natural move is to skip x=1 — it has "already been used" — and carry on down the list:

Nothing. It looks as if the cubic has no simple factor and the method has run out after one root.

The picture

It has not. Try x=1 again: F(1): 3\to8\to-2\to0. The factor (x-1) divides f twice, and

A repeated root looks different on the graph, and the reason is the square. Near x=1 the quadratic factor is close to 3+8-2=9, a positive number, while (x-1)^2 is never negative. So f(x) is positive just left of 1, zero at 1, and positive just right of it: the curve comes down to the axis, touches it, and goes back up. At a simple root the factor (x-k) changes sign as x passes k, so the curve crosses.

Across, one column is 0.1, with the y-axis at x=0. Upwards the heights are squeezed: the curve dips to -54 at x=-2, but rises only to 0.69 at x=0.5 between the two right-hand roots.

This is why a graph is said to meet the axis at a root rather than cross it. Seen at full scale, the touch at x=1 is invisible — the curve only rises to about 0.08 at x=0.9 — so a sketch cannot tell a touch from a near miss. Only F(1)=0 settles it.

Check it with numbers

Either side of the touch: f(0.9)=0.076 and f(1.1)=0.104, both positive. Either side of the root near 0.23: f(0.2)=-0.179 and f(0.3)=0.328, a change of sign.

The quadratic factor, by the formula: x=\frac{-8\pm\sqrt{64+24}}{6}=\frac{-8\pm9.3808}{6}, so x=0.230 or x=-2.897. The four roots are

The rule it gives

Fourth-order equations having at least two linear factors — the sequential route, which is the book's method:

1. Find a first factor (x-k) as for a cubic: nested form, trials 1,-1,2,-2,\ldots.
2. Divide, so that f(x)=(x-k)F(x) with F(x) a cubic.
3. Then f(x)=0 gives x=k or F(x)=0.
4. Re-nest F(x) and solve it exactly as a cubic, starting the trials again from x=1, to get F(x)=(x-m)(ax^2+bx+c).
5. The four solutions are x=k, x=m and the two roots of the quadratic by the formula.

A repeated root is reported twice — x=1,\ x=1 — because a fourth-order equation has four roots over \mathbb C, counted with multiplicity. The "two linear factors" may be the same factor twice.

Route 2, from F.3. If two different trials f(p)=0 and f(q)=0 turn up during the first hunt, multiply (x-p)(x-q) into a quadratic and divide once. It gives the same quadratic factor as the sequential route, because both are f(x) divided by the same two linear factors, and division does not care about the order.

Worked example

Solve 4x^4-19x^3+24x^2+x-10=0. Nested, \{[(4x-19)x+24]x+1\}x-10; f(1): 4\to-15\to9\to10\to0.

F(1)=8, F(-1)=-18, F(2): 4\to-7\to-5\to0. So F(x)=(x-2)(4x^2-7x-5), and

Route 2 agrees: f(1)=0 and f(2)=0, (x-1)(x-2)=x^2-3x+2, and one division of f(x) by x^2-3x+2 gives the same 4x^2-7x-5.

Where this shows up

Back to the shelf. Its sag equation x^4-5x^3+6x^2=0 has zero constant and x terms, so x^2 comes out at once and leaves x^2-5x+6=0. The formula finishes it: x=\frac{5\pm\sqrt{25-24}}{2}=\frac{5\pm1}{2}, so x=3 or x=2. The sag is therefore

with roots x=0, x=0, x=2 and x=3 m. The double root is the wall: x^2 never changes sign, so the sag is positive just either side of x=0 (0.41 mm at x=-0.1 m on the curve's continuation, 0.34 mm at x=0.1 m), and the curve touches the level line. That is the shelf leaving the wall flat. The simple root is the bracket: the factor (x-2) changes sign, the sag goes from +2.48 mm at x=1.9 m to -2.48 mm at x=2.1 m on the continuation, and the curve crosses. That is the shelf arriving tilted. How tilted: near x=2 the other factors \frac{x^2(x-3)}{160} are close to \frac{4\times(-1)}{160}=-\frac1{40}, so there y\approx-\frac{x-2}{40}: a straight line, the shelf rising onto the bracket at 1 in 40 (at x=1.9 it gives 2.5 mm against the exact 2.48 mm). Near the wall the same reasoning gives y\approx\frac{6x^2}{160}, a parabola that is flat at x=0. The root x=3 m lies beyond the end of a 2 m shelf, and is rejected. Try it with a long plastic ruler: hold one end flat under a heavy book, rest the other end on a support of the same height, and press down along the middle. It leaves the book flat and rises onto the support at a slope, however the load is spread.

Narration spine. Plot the fourth-order curve. Trial x=1: zero, so divide out (x-1) to leave the cubic F(x). Skip 1 and trial -1, 2, -2: all struck through. Go back to x=1: zero again. Zoom in on x=1 to see the curve touch the axis instead of crossing, then finish 3x^2+8x-2 with the formula: x=1,\ 1,\ 0.230,\ -2.897.

Unit 04: When No Trial Works

The problem

A solid wooden ball of radius R=10 cm, with density 750 kg/m^3, floats in water. How deep does it sit?

By Archimedes, the water it pushes aside weighs as much as the ball. Measuring the depth as h=xR, the submerged cap has volume \frac{\pi h^2(3R-h)}{3}, and it must equal 0.75 of the ball's \frac43\pi R^3. That gives x^2(3-x)=3, or

First attempt

Use the method: nested form [(x-3)x+0]x+3, with a zero for the missing x term, and trial.

Every trial fails, and the list could go on forever.

The picture

Plot the curve anyway: it meets the axis three times. The roots exist; they just are not at any trial value.

Across, one column is 0.1, with the y-axis at x=0; each row is one unit of height. The curve turns at f(0)=3, on the y-axis, and again at f(2)=-1.

The signs in the trial table show where the crossings are. f(-1)=-1 but f(0)=3, so the curve crosses between -1 and 0. f(1)=1 but f(2)=-1: a crossing between 1 and 2. f(2)=-1 but f(3)=3: another between 2 and 3. The trial method needs a crossing to land exactly on a simple number, and here none does.

Beyond the book, this can be proved rather than suspected. Suppose a fraction x=\frac pq in lowest terms (whole numbers, q>0, no common factor) were a root. Substituting it and multiplying through by q^3 gives

Read this two ways. Written as p^3=q\,(3p^2-3q^2), it says q divides p^3. But q shares no factor with p, so it shares none with p^3 (a prime that divides p^3 divides p), and a number that divides p^3 while sharing no factor with it must be 1. So q=1. Written as 3q^3=p\,(3pq-p^2), it says p divides 3q^3; the prime factors of p must all be found in 3q^3, and none of them is in q^3, so p divides 3. The only rational numbers that could be roots are therefore \pm1 and \pm3. All four are in the trial table, and all have failed. So x^3-3x^2+3 has no simple linear factor, and no amount of trial will find one.

The same two readings work for any polynomial a_nx^n+\cdots+a_1x+a_0 with whole-number coefficients. Multiplying f\left(\frac pq\right)=0 by q^n gives

Every term but the last contains p, so p divides a_0q^n; it shares no factor with q^n, so p divides a_0. Every term but the first contains q, so q divides a_np^n; it shares no factor with p^n, so q divides a_n. This is the rational root theorem: a rational root \frac pq in lowest terms has p dividing the constant term and q dividing the leading coefficient. It turns the book's open-ended list of trials into a finite list of candidates.

Check it with numbers

The ball's depth cannot be negative or more than its diameter, so 0\le x\le2 and the answer is the crossing between 1 and 2. Narrow it by looking for the sign change: f(1.3)=0.127 and f(1.4)=-0.136, then f(1.34)=0.019 and f(1.35)=-0.007. The root lies between 1.34 and 1.35, but that alone does not fix the third decimal place. To round to 1.347 the root must lie between 1.3465 and 1.3475, and it does: f(1.3465)=0.0021>0 and f(1.3475)=-0.0005<0. So x=1.347 to 3 dp: the ball floats h=13.5 cm deep. Check in the original: 1.347^2\times(3-1.347)=1.814\times1.653=3.00.

Beyond the book: the three roots, -0.879, 1.347 and 2.532, add up to 3.000, the negative of the x^2 coefficient. They must. Call the roots p, q, r. Since f(p)=0, the factor theorem gives f(x)=(x-p)\,g(x) with g a quadratic; at x=q, 0=f(q)=(q-p)\,g(q) with q\ne p, so g(q)=0 and (x-q) divides g; likewise (x-r). After three linear factors only a constant is left, and matching the x^3 terms makes it 1:

Matching the x^2 terms gives p+q+r=3. For a cubic ax^3+bx^2+cx+d with roots p,q,r (counted with multiplicity) the constant left is a, and the same matching gives p+q+r=-\frac ba: the cubic companion of the quadratic's x_1+x_2=-\frac ba from Unit 01. It is a quick check that no root has been missed or given the wrong sign.

The rule it gives

The limits of the method. Everything in this programme relies on finding one or more simple factors by the remainder theorem: at least one for a cubic, at least two for a fourth-order equation. Many cubic and fourth-order equations have no simple factors at all. They still have real roots wherever their graph meets the axis, but finding them needs other, numerical methods, dealt with later in the course. Beyond the book: a change of sign between two trials shows where to look, and narrowing the interval, as for the ball, is where those numerical methods begin.

Worked example

Show that x^3-3x+1=0 has no simple linear factor, and locate its roots. Nested: [(x+0)x-3]x+1. Trials: f(1)=-1, f(-1)=3, f(2)=3, f(-2)=-1. Beyond the book: by the rational root theorem the only possible rational roots are \pm1 (the constant term and the leading coefficient are both 1), and both fail, so there is no simple factor. Sign changes: f(-2)<0<f(-1), f(0)=1>0>f(1) and f(1)<0<f(2), so there are roots between -2 and -1, between 0 and 1, and between 1 and 2 (in fact -1.879, 0.347, 1.532).

Those roots are the ball's, each less by exactly 1, and that is no accident. Put x=u+1 into the ball's cubic: (u+1)^3-3(u+1)^2+3=u^3+3u^2+3u+1-3u^2-6u-3+3=u^3-3u+1. So x^3-3x+1 is the ball's curve slid one unit to the left, and every crossing slides with it.

Where this shows up

Change the tray of Unit 02 to hold 66 cm^3 instead of 64 and its equation becomes 2x^3-20x^2+48x-33=0: now f(1)=-3, f(2)=-1, and (beyond the book) all sixteen candidates the rational root theorem allows, \pm1,\pm3,\pm11,\pm33 and those numbers halved, fail. The trays still exist — cuts of about 1.30 and 1.85 cm — but no trial finds them. Notice that f(1) and f(2) have the same sign although both cuts lie between 1 and 2: f(1.5)=0.75, so the curve rises above the axis and comes back between the two samples. A change of sign proves a root is there; no change of sign does not prove there is none. Real design equations almost never come with whole-number roots, which is why engineers find roots numerically; the method of this programme works when the numbers have been arranged to allow it, and the sign-change picture is where the numerical methods begin. The ball is the test anyone can run: drop a wooden ball in a sink, and it settles at a depth no trial value predicts, a little over two-thirds of its diameter.

Narration spine. Show the ball in water with its depth h marked. Write x^3-3x^2+3=0 and trial 1, -1, 2, -2, 3, -3, each struck through. Plot the curve: three crossings, none at a trial value. Mark the sign changes that bracket them, then narrow the one between 1 and 2 to 1.347, and sink the ball to 13.5 cm.