Study Companion: Programme F.3 — Expressions and Equations

An expression is a recipe: put numbers in, follow the operations, and a number comes out. Almost everything in this programme is that recipe used in a new direction. An equation asks which input gives a stated output; transposing a formula runs the recipe backwards; nesting is the cheapest order in which to follow it; the remainder and factor theorems read a division off a single evaluation; and factorizing a polynomial finds the inputs that make the output zero. The one idea underneath is that evaluating is running a process, and each skill here is a way of running it forwards, backwards or more efficiently.

The rules, procedures and drill are in the companion reference sheet, which follows the book's topic order. This file tells the story in six units instead, because the programme's many topics are really six 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: An Equals Sign Can Mean Five Different Things

The problem

Four lines, all with an equals sign:

x^2=4,\qquad 2(5-x)=10-2x,\qquad A=\pi r^2,\qquad \texttt{B3 = B2*2}.

The first is true for only two values of x. The second is true for every x. The third tells you the area of any circle. The last is a spreadsheet cell being given a value. Should all four be read the same way — and if not, how can you tell which is which?

First attempt

"An equals sign means the two sides are the same number." For x^2=4 that is false at almost every x: at x=3 the sides are 9 and 4. For the spreadsheet cell it is not even a claim — B3 had no value until the line gave it one. One reading does not cover all four.

The picture

Treat each side as a function of x and draw both. What the equals sign means is what the two graphs do:

The graphs meet at a few points: the equation is true only there. This is a conditional equation — usually just called an equation — and solving it means finding the meeting points. x^2=4 holds at x=2 and x=-2.
The graphs coincide everywhere: the two sides are one expression written two ways. This is an identity, written with \equiv: 2(5-x)\equiv10-2x.
There is only one side to draw, because the equation names something rather than comparing two things. A formula states a fact connecting well-defined quantities, A=\pi r^2. A defining equation fixes what a piece of notation means: a^2 means a\times a. An assigning equation gives a variable a value, written := — as in p:=4, or the spreadsheet cell.

An identity holds wherever both sides are defined, and no further. \frac{r^3-s^3}{r-s}\equiv r^2+rs+s^2 is an identity, but the left side has no value at r=s, so the identity carries the condition r\ne s.

Check it with numbers

Test both sides at three values of x:

x	x^2	4	2(5-x)	10-2x
0	0	4	10	10
1	1	4	8	8
2	4	4	6	6

The conditional equation's sides agree only in the last row; the identity's agree in every row. A table can never prove an identity — it checks finitely many values — but one disagreeing row proves that a claimed identity is false.

The rule it gives

Type	Meaning	Symbol	Example
Conditional equation	true only for certain values	=	x^2-2x=0 (only x=0,2)
Identity	true for all values where both sides are defined	\equiv	2(5-x)\equiv10-2x
Defining equation	fixes the meaning of notation	a "defined as" sign	4n means 4\times n
Assigning equation	gives a variable a value	:=	v:=23.4
Formula	states a fact linking well-defined quantities	=	A=\pi r^2, V=IR

In practice = is written for all five, and the context says which is meant.

Subject, dependent and independent variables. In r=2s^3+3t, the single letter on the left is the subject of the equation and the dependent variable; the letters inside the expression, s and t, are the independent variables. The formula says how the dependent variable is found from the independent ones.

Evaluating an expression. Substitute numbers for the letters and follow the usual order of operations. State the precision — a number of decimal places (dp) or significant figures (sf) — and the value of \pi used (3.142, 3.14 or the calculator's 3.14159\ldots), and round once, at the end.

Worked example

The volume of a spherical cap of height h on a sphere is V=\frac{\pi h}6(3R^2+h^2), where R is the radius of the cap's base. For h=2.85 and R=6.24, with \pi=3.142:

V=\frac{3.142\times2.85}{6}\,(3\times38.938+8.123)=1.49245\times124.937=186.46.

Where this shows up

V=IR is a formula: it links three measurable quantities for an ideal resistor, and any of them may be made the subject. The modelling assumption hides in the word ideal — a real resistor's R drifts as it warms — and that limits where the formula applies, not how it may be rearranged. A circuit simulator's netlist line R1 := 470 is an assignment: it is not claiming anything, it is setting a value. A resistor of 470\ \Omega carrying 0.02 A has V=470\times0.02=9.4 V across it by the formula, whatever assignment set the 470.

Narration spine. Draw y=x^2 and y=4 and slide a probe along x with both sides read out: they agree only at the two meeting points. Replace them with y=2(5-x) and y=10-2x: the graphs lie on top of each other, and the readouts agree at every position. End on the five types, sorted by what their graphs do.

Unit 02: Transposition Is Undoing, in Reverse Order

The problem

A pendulum clock should tick once a second, so its pendulum should take T=2 s for a full swing. The formula is

T=2\pi\sqrt{\frac lg},\qquad g=9.81\ \text{m s}^{-2}.

It gives T from l. The clockmaker needs the other direction: what length l gives T=2 s?

First attempt

Guess and check. l=1 m gives T=2\pi\sqrt{1/9.81}=2.006 s — close, but slow. l=0.9 m gives 1.903 s — now fast. Each guess costs a calculation, the answer is only ever bracketed, and a new T means starting again. We want a formula that gives l directly.

The picture

Read the formula as a process — a chain of boxes that l passes through to become T:

Every box can be undone by its opposite: dividing by multiplying, a square root by squaring, adding by subtracting. To get from T back to l, run the chain backwards: take the boxes in reverse order, and replace each by its opposite. It is the order in which you take off shoes and socks — the last thing put on comes off first.

T=2\pi\sqrt{\frac lg}\;\Rightarrow\;\frac T{2\pi}=\sqrt{\frac lg}\;\Rightarrow\;\frac{T^2}{4\pi^2}=\frac lg\;\Rightarrow\;l=\frac{gT^2}{4\pi^2}.

Each step does the same operation to both sides, so the equation stays true — an equation is a balance.

Check it with numbers

Run the backward chain with T=2: 2\div2\pi=0.31831; squared, 0.101321; times 9.81, l=0.994 m. Now run the forward chain with l=0.994: 0.994\div9.81=0.101325; square root, 0.31832; times 2\pi, T=2.000 s. The backward chain undid the forward one.

The rule it gives

Evaluation as a process. A formula is a process acting on its input. Writing f for the process and x for the input, the output is f(x), read "f acting on x". For f(x)=3x-4, the process "multiply by 3, then subtract 4" turns the input 5 into f(5)=11. The letter f names the process, so results can be tabulated: f(4)=8, f(5)=11, f(6)=14.

Transposition (changing the subject). To make another letter the subject:

Locate the target letter and the operations applied to it, from the inside out.
Remove them in reverse order, each by its opposite operation: addition ↔ subtraction, multiplication ↔ division, powers ↔ roots. Taking the reciprocal of both sides is allowed too.
Do every operation to both sides.
Write the new subject on the left.

When the target appears twice, no chain of single undos can isolate it. Clear any fractions, multiply out, gather every term containing the target on one side, and factor it out:

n=\frac{IR}{E-Ir}\;\Rightarrow\;nE-nIr=IR\;\Rightarrow\;nE=I(R+nr)\;\Rightarrow\;I=\frac{nE}{R+nr}.

Evaluating an independent variable. When the known values include the subject, and one independent variable is missing, transpose (or undo step by step with numbers) to find it.

Dividing both sides by an expression assumes that expression is not zero, and a square root taken while transposing means the positive root unless the physics allows a negative one.

Worked example

A battery of n=6 cells, each of e.m.f. E=2.01 V and internal resistance r, drives a current I=0.98 A through a load R=12\ \Omega, where I=\frac{nE}{R+nr}. Find r.

0.98=\frac{12.06}{12+6r}\;\Rightarrow\;12+6r=\frac{12.06}{0.98}=12.306\;\Rightarrow\;r=\frac{0.306}{6}=0.051\ \Omega.

Check in the original formula: \frac{12.06}{12+6\times0.051}=\frac{12.06}{12.306}=0.980 A.

Where this shows up

The clock: a seconds pendulum is 0.994 m long, which is why a longcase clock needs a case well over a metre tall. The battery: measuring I for a known load and solving for r is how the internal resistance of a sealed cell is found on a bench. The same formula, transposed three ways, answers three different questions — how much current, how many cells, how bad each cell is.

Narration spine. Draw the three boxes \div g, \sqrt{\ }, \times2\pi and send l=0.994 through them to get T=2.000. Then send T=2 back through the opposite boxes in reverse order — \div2\pi, square, \times g — and recover 0.994. Collect the reversed boxes into l=\frac{gT^2}{4\pi^2}.

Unit 03: Nesting Evaluates a Polynomial Without Writing Anything Down

The problem

A temperature sensor's calibration curve is a polynomial, and a small microcontroller has to evaluate it thousands of times a second. By hand, the same question: find f(4) for

f(x)=5x^3+2x^2-3x+6.

First attempt

Term by term: 5\times4\times4\times4=320, 2\times4\times4=32, 3\times4=12, then 320+32-12+6=346. It works, but it takes 3+2+1=6 multiplications, and three partial results have to be written down and added — each a chance for an error. For a polynomial of degree n the multiplications grow like \frac{n(n+1)}2.

The picture

Factor out x from everything except the constant, then do it again inside, and again:

5x^3+2x^2-3x+6=[(5x+2)x-3]x+6.

Now evaluation is a single accumulator that runs left to right. Start with the leading coefficient; then repeatedly multiply by x and add the next coefficient:

\begin{array}{r|cccc} \text{coefficients} & 5 & 2 & -3 & 6 \ \text{step at } x=4 & 5 & 5(4)+2 & 22(4)-3 & 85(4)+6 \ \text{running value} & 5 & 22 & 85 & 346 \end{array}

One multiplication per coefficient after the first — three instead of six — and nothing to record, because each result feeds straight into the next step.

Check it with numbers

The accumulator gives 5\to22\to85\to346, the same as the term-by-term total 320+32-12+6=346. At x=2: 5\to12\to21\to48, and 40+8-6+6=48.

The rule it gives

A polynomial in x is a sum of terms in powers of x, normally written in descending powers. Its degree is the highest power present: degree 1 is linear, 2 quadratic, 3 cubic, 4 quartic (fourth-order).

Nesting. Write the polynomial in descending powers; include every missing power with a zero coefficient; then bracket from the inside out, using (\ ), [\ ], \{\ \} for successive levels. Evaluate from the innermost bracket outwards.

The zero coefficient matters because the accumulator multiplies by x once for every power: skipping a power drops a factor of x from everything before it.

3x^4+2x^2-4x+5=\{[(3x+0)x+2]x-4\}x+5.

Worked example

f(x)=3x^4+2x^2-4x+5 at x=2: 3\to3(2)+0=6\to6(2)+2=14\to14(2)-4=24\to24(2)+5=53. Term by term: 48+8-8+5=53.

Without the placeholder the accumulator would run 3\to3(2)+2=8\to8(2)-4=12\to12(2)+5=29 — wrong, because 3x^4 has been treated as 3x^3.

Where this shows up

The microcontroller: each link of the chain is one multiply-and-add, a single instruction on most signal-processing chips, and nothing needs storing between links. For a degree-10 calibration polynomial that is 10 multiplications instead of 55. The same chain is the engine of the next unit, where its last value turns out to be a remainder.

Narration spine. Show the four coefficients 5,2,-3,6 in a row and a counter of multiplications. Run the accumulator with x=4: 5\to22\to85\to346, three multiplications. Replay term by term beside it and watch the counter reach six for the same 346.

Unit 04: The Remainder Is One Substitution Away

The problem

Dividing f(x)=x^3+3x^2-13x-10 by x-3 with the long-division layout of F.2 takes three rounds of divide, multiply and subtract, and leaves a remainder. Is there a quicker way to know the remainder — without doing the division at all?

First attempt

Do the division every time. It gives the right answer — quotient x^2+6x+5, remainder 5 — but when the aim is only to test whether x-3 divides exactly, it is a lot of work to find one number. And trying x-1, x-2, x+1 in turn would mean a full division for each.

The picture

Whatever the division gives, it can be written as

f(x)=(x-3)\,g(x)+R,

where g(x) is the quotient and R is a plain number. This is true for every x — it is an identity. So put x=3 into it. The first term is (3-3)\,g(3)=0, whatever g is, and what is left is

f(3)=R.

On the graph of y=f(x), the remainder on dividing by x-3 is simply the height of the curve above x=3. As the divisor x-a changes, the remainder is the height at a: positive where the curve is above the axis, negative where it is below, and zero where the curve meets the axis.

Check it with numbers

Nested form: f(x)=[(x+3)x-13]x-10. At x=3: 1\to6\to5\to5. So R=5, matching the long division.

The nesting chain also carries the quotient: its values before the last, 1,6,5, are exactly the coefficients of g(x)=x^2+6x+5. This shortcut is synthetic division — an alternative to the long-division layout, not a replacement for it.

The rule it gives

Remainder theorem. If f(x) is divided by (x-a), the remainder is f(a):

f(x)=(x-a)\,g(x)+R,\qquad R=f(a).

Watch the sign: dividing by x+2 is dividing by x-(-2), so a=-2.

Factor theorem. If f(a)=0, the remainder is zero, so (x-a) is a factor of f(x); the other factor g(x) is found by long division (F.2). Conversely, if (x-a) is a factor then f(a)=0.

A sketch of the graph shows roughly where the curve meets the axis, so it suggests candidates for a. Only an exact evaluation, f(a)=0, confirms a factor.

Worked example

Is (x-3) a factor of f(x)=3x^3-11x^2+10x-12? Nested: [(3x-11)x+10]x-12. At x=3: 3\to-2\to4\to0. The remainder is 0, so yes, and the chain's earlier values give the other factor:

3x^3-11x^2+10x-12=(x-3)(3x^2-2x+4).

The remainder on dividing 2x^3+3x^2-x+4 by (x+2) is f(-2)=-16+12+2+4=2.

Where this shows up

A vibrating system's natural frequencies and decay rates are the roots of its characteristic polynomial. Checking whether a suspected value is a root costs one nested evaluation — one pass of the accumulator — rather than a long division, and the same pass hands over the smaller polynomial left to solve.

Narration spine. Plot y=x^3+3x^2-13x-10 and mark x=3: a brace shows the height 5. Write f(x)=(x-3)g(x)+R and put x=3: the (x-3) factor turns to zero and R=f(3)=5 remains. Then slide the divisor's a along the axis with a live readout of the remainder, which passes through zero exactly where the curve meets the axis.

Unit 05: The Quadratic Formula Is a Completed Square

The problem

A rectangular gasket must be 5 cm longer than twice its width and have an area of 3 cm^2. With width x cm, x(2x+5)=3, so

2x^2+5x-3=0.

What width is needed?

First attempt

Look for two whole numbers with sum 5 and product -3, as for x^2+px+q in F.2. The candidates (1,-3) and (-1,3) give sums -2 and 2: nothing works. It looks as if 2x^2+5x-3 has no simple factors — yet it does, (x+3)(2x-1). The sum-and-product search was built for a leading coefficient of 1, and it fails here. We need a method that works for every quadratic.

The picture

Divide by 2 so the x^2 term stands alone: x^2+\frac52x=\frac32. Draw the left side as area: a square of side x, and the strip \frac52x cut into two equal strips of \frac54 by x, one along each side of the square.

The shape is almost a square of side x+\frac54 — it lacks only the corner, a square of side \frac54. Add that corner, \left(\frac54\right)^2=\frac{25}{16}, to both sides:

\left(x+\frac54\right)^2=\frac32+\frac{25}{16}=\frac{49}{16}\quad\Rightarrow\quad x+\frac54=\pm\frac74.

The corner's side is half the coefficient of x, because the strip was split in two.

Check it with numbers

x=-\frac54+\frac74=\frac12 or x=-\frac54-\frac74=-3. A width cannot be negative, so the gasket is 0.5 cm wide and 2(0.5)+5=6 cm long: area 0.5\times6=3 cm^2.

In decimals: the strips are 1.25 wide, the corner is 1.5625, and 1.5+1.5625=3.0625=1.75^2.

The rule it gives

The same completion for the general quadratic ax^2+bx+c=0, a\ne0: divide by a, split the strip \frac bax in two, add the corner \left(\frac b{2a}\right)^2 to both sides:

x^2+\frac bax=-\frac ca\;\Rightarrow\;\left(x+\frac b{2a}\right)^2=\frac{b^2-4ac}{4a^2}\;\Rightarrow\;x+\frac b{2a}=\frac{\pm\sqrt{b^2-4ac}}{2a},

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.

This is the general quadratic equation's solution formula. When b^2-4ac<0 the square would need a negative area, and there are no real roots.

Roots to factors. If the roots are r_1 and r_2, then x^2+\frac bax+\frac ca=(x-r_1)(x-r_2), so multiply by a to factor the original:

ax^2+bx+c=a(x-r_1)(x-r_2).

Worked example

2x^2+5x-3: a=2, b=5, c=-3.

x=\frac{-5\pm\sqrt{25+24}}{4}=\frac{-5\pm7}{4}=\frac12\ \text{or}\ -3,

2x^2+5x-3=2\left(x-\frac12\right)(x+3)=(2x-1)(x+3).

When b^2-4ac is not a perfect square the roots are irrational, and the formula gives them to any accuracy: 3x^2-4x-2=0 gives x=\frac{4\pm\sqrt{40}}{6}=1.721 or -0.387 (3 dp).

Where this shows up

The gasket is the everyday case: a size fixed by an area gives a quadratic, and the formula returns one physical answer and one to reject. The same completed square locates the peak of a projectile's height or the minimum of a cost curve, because the vertex of y=ax^2+bx+c sits at x=-\frac b{2a} — the point where the completed square is zero.

Narration spine. Draw the square x^2 and the two strips \frac54x, and show the empty corner. Fill it with the \frac54\times\frac54 square and relabel the whole as \left(x+\frac54\right)^2=\frac{49}{16}. Read off x=\frac12, the gasket's width, then replace the numbers with a, b, c to reach the formula.

Unit 06: Factorizing Cubics and Quartics One Root at a Time

The problem

Where does the curve y=2x^3+x^2-13x+6 meet the x-axis? Equivalently, what are the linear factors of the cubic?

First attempt

Read the crossings off a sketch. The curve appears to meet the axis near -3, near \frac12 and near 2 — but a sketch cannot tell 0.5 from 0.48, and a factor must be exact. The sketch gives candidates; it does not give factors.

The picture

Test candidates with the factor theorem, in the book's order x=1,-1,2,-2,\ldots, each by one nested evaluation of [(2x+1)x-13]x+6:

trial x	1	-1	2
f(x)	-4	18	\mathbf{0}

f(2)=0: the curve meets the axis, so (x-2) is a factor.

Each trial is the height of the curve at that x. Once f(2)=0, divide by (x-2):

2x^3+x^2-13x+6=(x-2)(2x^2+5x-3).

The quotient is a parabola that passes through the curve's other two meeting points: dividing out (x-2) removed one meeting point and left the rest. The remaining quadratic is Unit 05's, so 2x^2+5x-3=(x+3)(2x-1).

Check it with numbers

2x^3+x^2-13x+6=(x-2)(x+3)(2x-1).

At x=1: the left side is -4 and the right side is (-1)(4)(1)=-4. At x=-1: 18 against (-3)(2)(-3)=18.

The rule it gives

Factorizing a cubic.

Write f(x) in nested form.
Try x=1,-1,2,-2,\ldots until f(k)=0; then (x-k) is a factor.
Divide f(x) by (x-k) (long division) to leave a quadratic.
Factorize the quadratic, by the formula if necessary, remembering the leading coefficient.

Fourth-order polynomials work the same way, provided at least one simple factor exists. Two routes:

Route 1: find one factor, divide to leave a cubic, and factorize the cubic as above.
Route 2: find two factors (x-p) and (x-q), multiply them into a quadratic, and divide once, so that f(x)=(x-p)(x-q)(ax^2+bx+c).

A root may be repeated. The curve then touches the axis there instead of crossing it, which is why a graph is said to meet the axis at a root.

Beyond the book: the only rational numbers worth trying are fractions whose numerator divides the constant term and whose denominator divides the leading coefficient — here \pm1,\pm2,\pm3,\pm6,\pm\frac12,\pm\frac32 — which is why \frac12 had to appear.

Worked example

Route 2 on f(x)=2x^4-3x^3-14x^2+33x-18=\{[(2x-3)x-14]x+33\}x-18.

f(1)=0, so (x-1) is a factor; f(-1)=-60, so (x+1) is not; f(2)=0, so (x-2) is a factor. Then (x-1)(x-2)=x^2-3x+2, and one long division gives

\frac{2x^4-3x^3-14x^2+33x-18}{x^2-3x+2}=2x^2+3x-9=(x+3)(2x-3),

f(x)=(x-1)(x-2)(x+3)(2x-3).

Route 1 on 2x^4-x^3-8x^2+x+6: f(1)=0 gives (x-1)(2x^3+x^2-7x-6); for the cubic g(x), g(1)=-10 and g(-1)=0 give (x+1)(2x^2-x-6); so f(x)=(x-1)(x+1)(x-2)(2x+3).

Where this shows up

A mechanical system with characteristic equation s^3+6s^2+11s+6=0 has trials s=-1,-2,-3 all giving zero, so (s+1)(s+2)(s+3)=0. Each root is a mode that decays like e^{st}: time constants of 1 s, 0.5 s and 0.333 s. Factorizing is how a polynomial handed over by a model becomes the decay rates the physical system will actually show.

Narration spine. Plot y=2x^3+x^2-13x+6. Trial x=1, -1, 2 as dots on the curve with their heights, -4, 18, 0. At the zero, divide out (x-2) and draw the quotient parabola 2x^2+5x-3 through the remaining two meeting points, -3 and \frac12. End on (x-2)(x+3)(2x-1).