Mathematical Foundations

Mathematics is usually taught as though the numbers were simply there, waiting, and the only question was what to do with them. That is backwards: every kind of number was invented under protest, because someone wrote down a problem that the numbers already available could not answer.

This lesson follows that chain of protests from one end to the other. It assumes nothing beyond arithmetic, and what it produces is the object the rest of the course lives on, the real line, along with an honest account of what that line still cannot do.

Counting, and the first thing it cannot do

Start with what a child starts with: $1,2,3,4,\mathrm{dots}$, the natural numbers, written $\mathbb{N}$. They answer one question, "how many", and they are closed under two operations. Add two naturals and you get a natural. Multiply two naturals and you get a natural. Closure is worth naming, because it is exactly what fails next.

Now ask a question of the same shape as one they can answer. The equation $x+3=7$ has the answer $4$, and nothing is wrong. The equation $x+7=3$ has no answer at all in $\mathbb{N}$. It is not that the answer is hard to find; there is no natural number that works, so within this system the question is malformed.

There are two honest responses. One is to declare the question illegal and keep the numbers small and safe, which is roughly what Greek mathematics did. The other is to invent whatever object the equation demands and then check that the new object does not wreck the arithmetic that already worked. Every step in this lesson is that second response, taken again and again.

Zero and the negatives

The demand of $x+7=3$ is for a number that undoes addition. Grant it: for every natural $n$ introduce an object $-n$ with the defining property $n+(-n)=0$, and introduce $0$ itself as the number that changes nothing when added. The result is the integers, $\mathbb{Z}$, running $\mathrm{dots},-2,-1,0,1,2,\mathrm{dots}$ and closed under addition, subtraction and multiplication.

The first systematic account of how these behave is Brahmagupta's, in the Brahmasphutasiddhanta of 628 CE, which states rules for zero and for what he calls fortunes and debts: a debt subtracted from zero is a fortune, and the product of two debts is a fortune. That second rule is the one every student finds arbitrary, and it is not arbitrary at all. It is forced, and the next lesson shows exactly what forces it.

Resistance lasted a very long time. Cardano, solving cubics in the Ars Magna of 1545, still called negative roots fictae, fictitious, and European mathematicians were arguing about whether negative quantities were legitimate into the eighteenth century. Two facts sit behind this. A negative number answers no question of the form "how many", so it has no direct counterpart in a heap of stones. And a debt, a temperature below freezing, or a displacement to the left are all perfectly concrete, so the objection was never about the world. It was about what a number was allowed to be.

Fractions, and a system that looks complete

Multiplication now has the same problem addition had. The equation $3x=12$ is fine. The equation $3x=7$ has no solution among the integers, and the fix is the same trick: for every non-zero integer $n$ introduce an object $1/n$ with the defining property $n\times (1/n)=1$. What comes out is the rational numbers, $\mathbb{Q}$, every number expressible as $p/q$ with $p$ and $q$ integers and $q$ not zero.

The rationals are closed under all four operations, division by zero excepted, and they are also dense: between any two distinct rationals there is another, since their average is rational and sits strictly between them. Take $1/3$ and $2/5$; their average is $11/30$, which lies between the two, and the same move repeats forever. There is no such thing as the next rational after a given one.

Density makes the system feel finished. Pick any point on a ruler and you can name a rational as close to it as you please, so it is natural to assume the rationals fill the line completely. That assumption is false, and the discovery that it is false is the most important event in this lesson.

Every rational has a decimal expansion that either terminates or eventually repeats, and every terminating or repeating decimal is a rational. The forward direction is long division: dividing by $q$ can only produce $q$ possible remainders, so a remainder must recur, and once it does the digits cycle. The reverse direction is a trick worth having.

Example. Write $0.272727\mathrm{dots}$ as a fraction.

Call the number $x$. The repeating block is two digits long, so multiply by $100$ to shift the expansion by exactly one block: $100x=27.272727\mathrm{dots}$. Subtracting the original, the infinite tails cancel exactly, since they are identical: $100x-x=27$, so $99x=27$ and $x=27/99$. Dividing top and bottom by $9$ gives $x=3/11$. Checking by division, $3$ divided by $11$ is $0.2727\mathrm{dots}$, as claimed.

Now you. Write $0.135135\mathrm{dots}$ as a fraction in lowest terms.

The diagonal that is not a fraction

Draw a square of side $1$ and ask for the length $d$ of its diagonal. Pythagoras' theorem gives $d^2=1^2+1^2=2$, so $d$ is a number whose square is $2$. The Pythagoreans, working in the fifth century BCE, could prove that no fraction has this property. Aristotle refers to the argument as a familiar one, and a version of it survives in Book X of Euclid's Elements.

The full proof is the business of the last lesson of this course, where the technique it uses is the point. What matters here is the consequence, and it is severe. The diagonal of a perfectly ordinary square has a length, you can draw it, and that length is not any ratio of whole numbers. The rationals, dense as they are, have a hole in them exactly where the diagonal lands.

The tradition that the discovery was a scandal, and that Hippasus was drowned at sea for divulging it, is a late story and probably not history. The mathematical damage, however, was real. A school whose slogan was that all is number had assumed that any two lengths share a common unit small enough to measure both, and the diagonal of a square shows they need not.

You can watch the failure numerically. Fractions approximate $\sqrt{2}$ beautifully, and never hit it.

Example. How close does $99/70$ come to having a square of $2$?

Square it: ${99}^2=9801$ and ${70}^2=4900$, so $(99/70{)}^2=9801/4900$. Now $2\times 4900=9800$, so the square is $9801/4900=2+1/4900$, missing $2$ by exactly one part in $4900$, about $0.0002$. Close, and not equal, and the argument of the last lesson says the miss can never be zero however clever the fraction.

Now you. By how much does the square of $577/408$ miss $2$?

The real line

Patch the holes and you get the real numbers, $\mathbb{R}$: every point on a continuous line has a number, and every number a point. Informally, a real number is a decimal expansion, allowed to run forever without repeating. The rationals are the expansions that terminate or repeat; the irrationals are all the rest, and $\sqrt{2}=1.41421356\mathrm{dots}$ is one of them.

Making that respectable took until the 1870s, when Dedekind and Cantor gave constructions of the reals in terms of the rationals alone, so that the line stopped being a picture and became a definition. The property their constructions deliver is completeness: there are no gaps left, and any quantity you can squeeze arbitrarily tightly between rationals is itself a real number. Completeness is what makes limits work, so it is the foundation the whole of calculus is built on, and it is why the course you are reading stops here and calculus starts here.

Two facts about the irrationals are worth carrying. First, they come in two kinds. $\sqrt{2}$ is algebraic, meaning it solves a polynomial equation with whole-number coefficients, here $x^2-2=0$. Numbers that solve no such equation at all are transcendental, and both of the famous constants are: Hermite proved it for $e$ in 1873 and Lindemann for $\pi$ in 1882. Lindemann's result settled squaring the circle, an open problem of two thousand years, in the negative.

Second, the irrationals are not rare. Cantor showed in 1874 that the rationals can be listed in an infinite sequence while the reals cannot, so the two infinities differ in size, and in a precise sense almost every real number is irrational. The numbers we can name are the exceptions.

Decimals, precision and error

In practice, every real number that is not a whole number gets replaced by a decimal that stops, so it is worth being able to say how much that costs. The absolute error is the difference between the approximation and the true value; the relative error is that difference divided by the true value, and it is the one that usually matters, because being wrong by a metre is trivial for a road and fatal for a doorframe.

Example. $\pi =3.14159265\mathrm{dots}$ and the schoolroom approximation is $22/7=3.14285714\mathrm{dots}$. What are the absolute and relative errors?

Subtracting, the absolute error is $3.14285714-3.14159265=0.00126449$, so $22/7$ is a little too large. Dividing by $\pi$ gives a relative error of $0.00126449/3.14159265=0.000402$, about four parts in ten thousand, or $0.04\%$. On a circle the size of a dinner plate, say $26$ cm across, the circumference is about $817$ mm, so that is an error of roughly a third of a millimetre.

Now you. The better approximation $355/113=3.14159292\mathrm{dots}$ was known to Zu Chongzhi in the fifth century. What is its absolute error, and roughly its relative error?

The equation the line cannot solve

The chain of this lesson has a shape: name an equation with no solution, invent the solution, check that nothing already working breaks. Run it once more and it does not terminate.

The equation is $x^2=-1$. On the real line it has no solution, and for a reason stronger than mere absence: any real number, positive or negative, has a square that is zero or positive, so no arrangement of real numbers will ever produce $-1$. The same demand can be made anyway, and it was, by Bombelli in 1572, who found that solving cubic equations sometimes required carrying square roots of negatives through the middle of a calculation even when the final answers were ordinary whole numbers. Grant the demand and you get the complex numbers, and everything that worked before still works.

This course does not follow that road, and the reason is worth saying plainly rather than hiding. The material ahead, functions, graphs, growth, decay and the geometry of angles, is about quantities you can measure and plot: a length, a temperature, a balance, a height. Those live on the line. The complex numbers are the natural home of a different set of questions, and taking them on here would double the machinery without touching the goal.

So the working set is $\mathbb{R}$, with $\mathbb{N}$, $\mathbb{Z}$ and $\mathbb{Q}$ sitting inside it, and one honest gap left at $x^2=-1$.

There is something conspicuously unfinished about all of this. Solving $x+7=3$ and $3x=7$ meant subtracting from both sides and dividing both sides, and no reason was ever given for why those moves are allowed. They were used because they are familiar. The rules that licence them are few, they can be written down completely, and once they are, results such as Brahmagupta's rule about two debts stop being conventions to memorise and become consequences. That is the next lesson.

Solving $x+7=3$ meant subtracting seven from both sides, and nobody said why that was allowed. The move is so familiar that the question sounds pedantic, but the whole of algebra is built out of a handful of such moves, and if they are taken on faith then every result downstream rests on faith too.

This lesson writes the licence down. It assumes only the number systems of the previous lesson, the naturals inside the integers inside the rationals inside the reals, and it produces a list short enough to memorise from which everything else follows, including the rule about two negatives that every student is told to accept.

The moves nobody justified

School algebra is taught as a set of permissions handed out one at a time. You may add the same thing to both sides. You may multiply out a bracket. You may cancel a common factor. You may not divide by zero. Each arrives as a separate instruction, and the impression left is of a large body of rules held together by nothing.

The truth is the reverse. There are about eight facts about addition and multiplication, all of them things you would guess, and every legitimate manipulation is a consequence of them. Nothing else is needed and nothing else is allowed. When a manipulation goes wrong, and one will go wrong later in this lesson in a way that produces the proof that $1=2$, the fault is always that some step used a law outside the conditions the law actually carries.

The name for a system carrying these laws is a field, and $\mathbb{Q}$ and $\mathbb{R}$ are both fields. The integers are not, because they lack multiplicative inverses: $3x=7$ has no integer solution, which is exactly the failure that forced the rationals into existence.

The list

Addition is commutative, $a+b=b+a$, and associative, $(a+b)+c=a+(b+c)$. It has an identity, the number $0$ with $a+0=a$, and every $a$ has an additive inverse $-a$ with $a+(-a)=0$. Multiplication has the same four properties: $ab=ba$, $(ab)c=a(bc)$, the identity $1$ with $a\times 1=a$, and for every $a$ except zero a multiplicative inverse $a^{-1}$ with $a\times a^{-1}=1$.

That is four laws twice over, and they say nothing about how the two operations interact. One law does: distributivity,

This is the only bridge between addition and multiplication in the whole list, and it is the busiest law in mathematics. Expanding brackets is distributivity read left to right; factoring is the same law read right to left; long multiplication of ordinary numbers is distributivity applied to place value.

The laws about equations follow from these plus one fact about equality: if $a=b$, then $a$ and $b$ are the same number, so anything true of one is true of the other. Adding $c$ to both sides is legal because $a+c$ and $b+c$ are the same number when $a$ and $b$ are. That is the licence that was missing in the previous lesson, and it costs one sentence.

Subtraction and division are shorthand

Notice what the list does not contain. There is no law of subtraction and no law of division, and this is not an oversight. Subtraction is defined out of what is there: $a-b$ means $a+(-b)$, the addition of the additive inverse. Division likewise means multiplication by the multiplicative inverse, $a÷b=a\times b^{-1}$.

This is worth more than tidiness, because it explains a fact every student meets as a nuisance. Addition is commutative and associative; subtraction is neither. $7-3$ is not $3-7$, and $(10-4)-3=3$ while $10-(4-3)=9$. The rules did not fail. Subtraction was never on the list, and the operation it abbreviates is commutative only in the sense that $a+(-b)=(-b)+a$, which is not the statement that swapping the visible numbers is safe.

The same holds for division, which is why $12÷4÷2$ is ambiguous without a convention, while $12\times 4^{-1}\times 2^{-1}$ is not ambiguous at all. Reading a subtraction as the addition of a negative, and carrying the sign with the number rather than leaving it stranded between two terms, removes most sign errors permanently.

What the laws force

The rule that a negative times a negative gives a positive is presented to most people as a convention, and Brahmagupta's fortunes and debts do read like a convention. They are not. Given the laws above, the rule is the only possibility, and here is the argument.

First, anything times zero is zero. Since $0+0=0$, multiply both sides by $a$ and distribute: $a\times 0=a(0+0)=a\times 0+a\times 0$. Now add $-(a\times 0)$ to both sides, which is legal, and the left becomes $0$ while the right becomes $a\times 0$. So $a\times 0=0$, derived rather than assumed.

Now take $(-1)\times (-1)$. Start from $(-1)+1=0$ and multiply through by $-1$:

Distribute the left side: $(-1)(-1)+(-1)(1)=0$, which is $(-1)(-1)+(-1)=0$. Add $1$ to both sides and $(-1)(-1)=1$. If a negative times a negative were anything else, distributivity would fail, and with it every bracket anyone has ever expanded. The rule is not a choice made for convenience; it is the price of keeping the arithmetic consistent.

Example. Use distributivity to compute $97\times 103$ mentally.

Write it as $(100-3)(100+3)$ and expand: $100\times 100+100\times 3-3\times 100-3\times 3$. The two middle terms cancel exactly, leaving $10000-9=9991$. This is the difference of two squares, $(a-b)(a+b)=a^2-b^2$, and it is nothing but distributivity applied twice.

Now you. Compute $58\times 62$ the same way.

Why zero has no inverse

The one prohibition in school algebra is the ban on dividing by zero, and it is usually taught as a rule with a vague warning attached. It is a theorem. The laws say every number except zero has a multiplicative inverse, and the exception is forced, because granting zero an inverse contradicts a result already proved.

Suppose $0^{-1}$ existed, meaning some number $c$ with $0\times c=1$. But it was proved two sections ago that $a\times 0=0$ for every $a$, so $0\times c=0$, and therefore $0=1$. Once zero and one are the same number, every number is zero, since $a=a\times 1=a\times 0=0$, and arithmetic collapses into a single point. Division by zero is not undefined out of squeamishness. Defining it destroys the system.

This is also where the famous fake proof that $1=2$ hides. Start with $a=b$, multiply by $a$ to get $a^2=ab$, subtract $b^2$ to get $a^2-b^2=ab-b^2$, factor both sides into $(a-b)(a+b)=b(a-b)$, then cancel the common factor $(a-b)$ to get $a+b=b$. With $a=b=1$ that reads $2=1$. Every step is legal except the cancellation, which is multiplication by $(a-b{)}^{-1}$, and $a-b$ is zero. Cancelling a common factor is only ever multiplication by an inverse, so it always carries the condition that the factor is not zero, and forgetting that condition is the commonest way to lose a solution to an equation.

Distributivity in both directions

Expanding is mechanical: every term in the first bracket multiplies every term in the second. Factoring is the same law used backwards, and it is harder because it is a search rather than a procedure. It is worth the effort because a factored expression displays its zeros: a product is zero exactly when one of its factors is, which is the fact the whole of equation solving rests on.

For a quadratic $x^2+bx+c$ the search has a shape. If it factors as $(x+p)(x+q)$, then expanding gives $x^2+(p+q)x+pq$, so $p$ and $q$ are two numbers that multiply to $c$ and add to $b$. Since there are finitely many integer pairs multiplying to $c$, the search terminates.

Example. Expand $(2x+3)(x-5)$, then factor $x^2-x-12$.

Expanding: $2x\times x+2x\times (-5)+3\times x+3\times (-5)=2x^2-10x+3x-15=2x^2-7x-15$. Factoring: two numbers multiplying to $-12$ and adding to $-1$ are $-4$ and $3$, so $x^2-x-12=(x-4)(x+3)$. Check by expanding: $x^2+3x-4x-12=x^2-x-12$.

Now you. Expand $(3x-4)(2x+5)$, then factor the result back.

Order of operations, and what is actually a convention

Some of what looks like a law is only notation. The agreement that multiplication binds tighter than addition, so $2+3\times 4$ is $14$ and not $20$, is a convention about reading, not a fact about numbers. It could have gone the other way, and expressions would be written with more brackets. Associativity, by contrast, is a fact: $(2+3)+4$ and $2+(3+4)$ are equal whatever anyone agrees.

Distinguishing the two matters when a manipulation feels forbidden. Rearranging $a+bc$ into $bc+a$ is licensed by commutativity of addition. Rewriting it as $(a+b)c$ is licensed by nothing and is false, as $2+3\times 4=14$ against $(2+3)\times 4=20$ shows. When in doubt, put in the brackets the convention allowed you to omit, and check which law the intended step is appealing to.

Example. Simplify $5(2x-3)-2(4x-7)$.

Distribute each bracket, taking the second sign with the $2$: $10x-15-8x+14$. The $-2\times -7=+14$ is the negative-times-negative rule proved above. Collecting like terms, which is distributivity backwards on $10x-8x=(10-8)x$, gives $2x-1$.

Now you. Simplify $3(4x+5)-4(2x-3)$.

Where the laws stop

Two honest limits are worth stating. First, the laws describe addition and multiplication only. Inequalities carry their own rules, and one of them famously breaks the pattern: multiplying both sides of $a<b$ by a negative number reverses it, since $2<3$ but $-2>-3$. Nothing in the field laws covers order, and an ordered field is a stronger structure than a field.

Second, commutativity is not a property of everything worth multiplying. It fails for rotations in space, for the matrices met in linear algebra, and for the operators of quantum mechanics, where the failure is the physics. That such systems exist is the reason the laws are stated explicitly rather than assumed: writing them down is what makes it possible to notice which ones a new system keeps and which it abandons.

Within $\mathbb{R}$, though, the list holds without exception, and it licences every rearrangement the rest of this course performs. Rearranging is only half of what algebra does. The other half begins by taking two expressions and asserting they are equal, which turns a manipulation into a question with an answer, and answering that question is the next lesson.

An expression such as $2x^2-7x-15$ is not true or false; it is just a recipe waiting for a number. Set it equal to something and it becomes a claim about $x$, and the claim is true for some values and false for others. Finding exactly which ones is what solving means.

This lesson takes that from the easiest case to the first genuinely hard one. It assumes the laws of the previous lesson, since every step in a solution is one of them applied to both sides, and it ends with a formula for every quadratic that is derived here rather than quoted.

From expression to statement

An equation asserts that two expressions name the same number. Its solutions are the values of the unknown that make the assertion true, and to solve it is to replace it with a chain of equations having the same solutions until the last one reads $x=\text{something}$.

The chain works because of the equality fact from the previous lesson: doing the same thing to two names for one number leaves two names for one number. Adding, subtracting, and multiplying by anything non-zero all preserve the solution set exactly, so nothing is gained or lost.

Two operations do not preserve it, and both are traps worth naming now. Multiplying by an expression that might be zero can invent solutions, since $x=3$ has one solution but $x(x-5)=3(x-5)$ has two. Squaring both sides does the same, because $\sqrt{x+6}=x$ becomes $x+6=x^2$, whose solutions are $3$ and $-2$, and $-2$ fails in the original since the square root sign means the non-negative root. The fix is not to avoid these moves but to check every candidate in the original equation, which costs seconds.

Linear equations by inverse operations

A linear equation has the unknown to the first power only, and it always yields to the same strategy: undo the operations applied to $x$, in reverse order. Collect all the $x$ terms on one side, all the numbers on the other, and divide by whatever multiplies $x$.

Example. Solve $5(x-3)+2=3x+7$.

Distribute: $5x-15+2=3x+7$, so $5x-13=3x+7$. Subtract $3x$ from both sides: $2x-13=7$. Add $13$: $2x=20$. Divide by $2$: $x=10$. Check in the original, which is not optional: the left is $5\times 7+2=37$ and the right is $30+7=37$.

Now you. Solve $4(2x+1)-3=2(x-4)+11$ and check it.

Not every linear equation has exactly one solution. Reduce $2(x+3)=2x+6$ and the unknown vanishes, leaving $6=6$, true for every $x$: the equation was an identity in disguise. Reduce $2(x+3)=2x+5$ and you get $6=5$, false for every $x$, so there is no solution. A vanishing unknown is information, not a mistake, and which of the two you have is decided by whether what remains is true.

Quadratics, and why factoring is not enough

A quadratic equation is $a{x}^2+bx+c=0$ with $a$ not zero. When the left side factors over the integers the work is already done, because a product is zero exactly when one factor is: from $(x-4)(x+3)=0$ read off $x=4$ and $x=-3$ immediately.

The method is fast and it is unreliable, because most quadratics do not factor over the integers. The equation $x^2-6x+2.5=0$ has perfectly good solutions and no integer pair multiplies to $2.5$ while adding to $-6$. Searching for a factorisation that does not exist can take a long time before you conclude anything, so a method that always works is worth having.

The one that always works comes from noticing which quadratics are easy. Anything of the form $(x+p{)}^2=k$ is solved in one line: take square roots, remembering both signs, so $x+p=\pm \sqrt{k}$ and $x=-p\pm \sqrt{k}$. So the task is to turn any quadratic into that shape, and the technique for doing so is called completing the square.

Completing the square

The identity to exploit is $(x+p{)}^2=x^2+2px+p^2$. Given $x^2+bx$, matching the middle terms gives $2p=b$, so $p=b/2$, and the perfect square that starts with those two terms is $(x+b/2{)}^2=x^2+bx+b^2/4$. It has an extra $b^2/4$ on the end, so subtract it back off: $x^2+bx=(x+b/2{)}^2-b^2/4$. That single line is the whole technique.

Example. Solve $2x^2-12x+5=0$ exactly, then to four decimal places.

Divide through by $2$ so the leading coefficient is $1$: $x^2-6x+2.5=0$. Half of $-6$ is $-3$, so $x^2-6x=(x-3{)}^2-9$, and the equation becomes $(x-3{)}^2-9+2.5=0$, that is $(x-3{)}^2=6.5$. Take roots: $x=3\pm \sqrt{6.5}$. Since $\sqrt{6.5}=2.5495$, the solutions are $x=5.5495$ and $x=0.4505$. Check the first: $2(5.5495{)}^2-12(5.5495)+5=61.594-66.594+5=0$ to the precision shown.

Now you. Solve $3x^2+12x-7=0$ by completing the square, exactly and to four decimals.

The formula, derived

Run that procedure once on the general equation and you never have to run it again. Start from $a{x}^2+bx+c=0$ and divide by $a$, which is legal because $a$ is not zero:

Complete the square on the first two terms, with half the middle coefficient being $b/2a$:

Move the constants across and put them over the common denominator $4a^2$:

Take square roots of both sides. The right-hand denominator is a perfect square, so its root is $2a$, and

That is the quadratic formula, and it is not a fact to be memorised on authority: it is the completed square, done once with letters instead of numbers. Anyone who forgets it can rederive it in five lines. Check it against the worked example above, where $a=2$, $b=-12$, $c=5$: the formula gives $x=(12\pm \sqrt{144-40})/4=(12\pm \sqrt{104})/4$, and $\sqrt{104}=10.198$, so $x=5.5495$ or $0.4505$ as before.

The discriminant tells you before you start

The quantity under the root, $b^2-4ac$, is the discriminant, and it decides the character of the answer before any arithmetic. If it is positive there are two distinct real solutions. If it is zero the $\pm$ collapses and there is one, a repeated root at $x=-b/2a$. If it is negative there are no real solutions at all, because no real number squares to a negative, which is precisely the gap left open in the first lesson at $x^2=-1$.

So $x^2+4x+7=0$ can be dismissed in one step: $16-28=-12$, negative, no real solutions. This is not a failure of technique. The equation asks for a number whose square plus four times itself is $-7$, and on the real line there is none. The complex numbers supply two, and this course leaves them alone.

A repeated root is the boundary case between the two, and it is what a projectile problem gives when the object just grazes the height asked about. Discriminant zero is the algebraic signature of tangency, a fact that becomes visible as a picture in the next lesson but one.

A quadratic that came from somewhere

Formulas earn their keep on real problems. Throw a stone upward at $15$ m s⁻¹ from a cliff $40$ m above the sea. Ignoring air resistance, its height in metres after $t$ seconds is $h=40+15t-4.9t^2$, where $4.9$ is half the gravitational acceleration of $9.8$ m s⁻².

Example. When does the stone hit the water?

Set $h=0$: $-4.9t^2+15t+40=0$, or $4.9t^2-15t-40=0$ after multiplying by $-1$. The discriminant is $225+4\times 4.9\times 40=225+784=1009$, whose root is $31.765$. So $t=(15\pm 31.765)/9.8$, giving $t=4.772$ s and $t=-1.711$ s. Both are solutions of the equation; only one is a solution of the problem, since the stone was not in flight before it was thrown. Discarding a mathematically valid root on physical grounds is a normal part of modelling, and it should be done explicitly rather than silently.

Now you. A ball is thrown upward at $20$ m s⁻¹ from a $25$ m balcony, so $h=25+20t-4.9t^2$. When does it land?

The same quadratic answers other questions about the flight. The stone is highest when the two roots of $h=H$ coincide, which is when the discriminant vanishes, and by symmetry that is halfway between the roots of any horizontal cut: $t=15/9.8=1.531$ s, at which $h=51.48$ m.

What the formula reveals, and cannot do

Look at the formula rather than through it. The solutions are expressed in terms of $a$, $b$ and $c$, so as those coefficients vary the answers vary with them, and this is the first time in the course that a quantity has been described as depending on other quantities in a way no single equation captures. That dependence is the subject of the next lesson.

Two honest limits close this one. The formula is exact but not always the best way to compute: when $b^2$ is enormously larger than $4ac$, subtracting two nearly equal numbers in the numerator loses precision, and numerical libraries compute one root by the formula and the other from the fact that the roots multiply to $c/a$. And it does not generalise as far as one would hope. There is a formula for the cubic, published by Cardano in 1545, and one for the quartic, but Abel proved in 1824 that no formula in radicals exists for the general fifth-degree equation. Solving by formula stops at degree four, permanently.

The quadratic formula says that the solutions of $a{x}^2+bx+c=0$ depend on the three coefficients, and that dependence is not itself an equation. It is a rule: hand it three numbers and it hands back an answer, every time, without ambiguity.

Mathematics has a name for such a rule and a machinery for handling it, and building that machinery is what this lesson does. It assumes only the algebra of the previous two lessons, and it produces the object that the rest of the course studies almost exclusively.

What a function is, exactly

A function is a rule that assigns to each input exactly one output. Three things must be given before the rule is fully specified: the set of allowed inputs, called the domain; the set the outputs are drawn from, the codomain; and the rule itself. Writing $f(x)=3x+1$ gives only the third, and the other two are usually left implicit as "whatever real numbers make sense".

The demand for exactly one output is the whole of the definition and it is stricter than it looks. The rule "$y$ is a number whose square is $x$" is not a function, because $x=9$ yields both $3$ and $-3$. This is why $\sqrt{9}$ is defined to mean $3$ and not $-3$: the square root symbol has to name one number, and the choice of the non-negative one is a convention adopted to rescue functionhood. It is also why the quadratic formula carries an explicit $\pm$ rather than hiding two answers inside one symbol.

The notation $f(x)$ is worth reading carefully, because it is not multiplication. It means the output of $f$ at the input $x$, and $f$ alone is the name of the function while $f(x)$ is the name of a number. Sloppiness here is harmless in easy cases and disastrous when functions start being fed to other functions, which happens later in this lesson.

Domain: where the rule refuses to run

Two things stop a formula in its tracks over the reals, both met already. Division by zero is barred, so $f(x)=1/(x-2)$ has domain every real except $2$. Even roots of negatives do not exist on the real line, so $g(x)=\sqrt{x-4}$ needs $x-4\ge 0$, giving domain $x\ge 4$. Everything else is allowed, and a polynomial has every real number in its domain.

Domain can also be imposed from outside rather than discovered from the formula. If $A(r)=\pi r^2$ is the area of a circle of radius $r$, then $A(-3)=9\pi$ is arithmetically fine and physically meaningless, so the domain is $r>0$ because of what the function is for. Modelling almost always narrows a domain this way, and forgetting to say so is how a model gets used outside the range where it was ever true.

The range is what actually comes out: the set of values $f(x)$ takes as $x$ runs over the domain. Range is generally harder to find than domain, because it requires knowing the behaviour of the whole rule rather than checking for prohibited operations. For $f(x)=x^2$ with domain all reals, the range is $y\ge 0$, since squares are never negative and every non-negative number is a square.

Example. Give the domain and range of $g(x)=\sqrt{x-4}+1$.

The root demands $x\ge 4$, so that is the domain. As $x$ runs from $4$ upward, $x-4$ runs from $0$ upward, its square root runs from $0$ upward taking every non-negative value, and adding $1$ shifts all of it, so the range is $y\ge 1$. A check: $g(13)=\sqrt{9}+1=4$, and $g(20)=\sqrt{16}+1=5$.

Now you. Give the domain and range of $h(x)=3-\sqrt{x+5}$.

A function need not be a formula

The formula is the commonest presentation but not the definition, and holding the two apart pays off. A table of values is a function if no input appears twice with different outputs. The mapping from a date to the closing price of an index is a function with no formula at all. A rule given in words, such as "round to the nearest integer", is a function.

Piecewise definitions matter in practice because so much of the world is written in bands. United Kingdom income tax in the 2024/25 year charges nothing on the first $12\{,\}570$ pounds, $20$ per cent on income between there and $50\{,\}270$, and $40$ per cent from there to $125\{,\}140$. That is one function of income, defined by different formulas on different stretches.

On an income of $30\{,\}000$ pounds the tax is $0.20\times (30\{,\}000-12\{,\}570)=3\{,\}486$ pounds. On $60\{,\}000$ it is $0.20\times (50\{,\}270-12\{,\}570)+0.40\times (60\{,\}000-50\{,\}270)=7\{,\}540+3\{,\}892=11\{,\}432$ pounds. Note what the calculation shows: the higher rate applies only to the part of the income above the threshold, so crossing a band boundary never reduces take-home pay. That single fact, which is a property of the function, is the one most often got wrong in public argument about tax.

Composition, and why order matters

Feeding the output of one function into another gives a new function, written $(f\circ g)(x)=f(g(x))$ and read "$f$ after $g$". The inner function runs first, which is the reverse of the reading order, and it is a common source of error.

Take $f(x)=3x+1$ and $g(x)=x^2$. Then $f(g(2))=f(4)=13$, while $g(f(2))=g(7)=49$. Composition is not commutative, and unlike the commutativity of multiplication this failure is normal rather than exotic: doubling then adding three is a different rule from adding three then doubling, as anyone who has applied a discount before or after tax knows.

Composition also inherits domain restrictions from both parts. In $f(g(x))$ the input must be in the domain of $g$, and the value $g(x)$ must then be in the domain of $f$. With $g(x)=x-4$ and $f(u)=\sqrt{u}$, the composition needs $x\ge 4$ even though each function separately is happy with any real. Checking both conditions, not just the visible one, is the discipline.

Example. With $f(x)=2x-5$ and $g(x)=x^2+1$, find $f(g(3))$, $g(f(3))$, and a formula for $f(g(x))$.

$g(3)=10$, so $f(g(3))=20-5=15$. In the other order, $f(3)=1$, so $g(f(3))=1+1=2$. In general $f(g(x))=2(x^2+1)-5=2x^2-3$, and a check at $x=3$ gives $18-3=15$, agreeing.

Now you. With the same two functions, find a formula for $g(f(x))$ and evaluate it at $x=4$.

Running a function backwards

An inverse of $f$ is a function $f^{-1}$ that undoes it: $f^{-1}(f(x))=x$ for every $x$ in the domain, and $f(f^{-1}(y))=y$ for every $y$ in the range. The superscript is unfortunate notation, since $f^{-1}$ does not mean $1/f$, and only context distinguishes them.

Not every function has one, and the reason is the same strictness that defined functions in the first place. If two different inputs give the same output, the inverse would have to send that output back to both, and it would not be a function. So $f$ is invertible exactly when it is injective, meaning distinct inputs always give distinct outputs. The function $f(x)=x^2$ on all the reals is not injective, since $f(3)=f(-3)=9$, and it therefore has no inverse.

The standard repair is to shrink the domain until injectivity holds. Restrict $f(x)=x^2$ to $x\ge 0$ and it becomes injective, with inverse $\sqrt{y}$. The restriction is a genuine choice: $x\le 0$ would have worked equally well and would give $-\sqrt{y}$ as the inverse. Conventions of this kind, chosen for convenience and then treated as inevitable, are behind the odd-looking domain restrictions on the inverse trigonometric functions two lessons from the end of this course.

Finding an inverse algebraically is a routine. Write $y=f(x)$, solve for $x$ in terms of $y$ using the laws of lesson two, and rename the variables at the end.

Inverses in practice

Example. Find the inverse of $f(x)=(2x-1)/(x+3)$, and state where each function is defined.

Set $y=(2x-1)/(x+3)$ and multiply through by $x+3$, which is legal since $x\ne -3$: $y(x+3)=2x-1$, so $xy+3y=2x-1$. Gather the $x$ terms on one side: $xy-2x=-1-3y$, so $x(y-2)=-(3y+1)$ and $x=(3y+1)/(2-y)$. Hence $f^{-1}(x)=(3x+1)/(2-x)$. The domain of $f$ excludes $-3$ and the domain of $f^{-1}$ excludes $2$, and these correspond: $2$ is the value $f$ approaches but never takes, so it is missing from the range of $f$. Check numerically: $f(5)=9/8=1.125$, and $f^{-1}(1.125)=(3.375+1)/(0.875)=4.375/0.875=5$.

Now you. Find the inverse of $g(x)=(x+5)/(x-2)$ and check it at $x=7$.

The most familiar inverse pair in daily use is temperature conversion. Fahrenheit from Celsius is $F(c)=9c/5+32$, a function built from a stretch and a shift, and undoing those in reverse order gives $C(f)=5(f-32)/9$. Check: $F(37)=66.6+32=98.6$, and $C(98.6)=5\times 66.6/9=37$. The two scales agree at exactly one temperature, where $c=9c/5+32$, which solves to $c=-40$, a linear equation of the kind the previous lesson handled.

What is still missing

A function has now been defined, restricted, composed and inverted, all algebraically. What has not happened is seeing one. Asking whether $f(x)=x^3-3x+1$ is injective, or what its range is, or how many solutions $f(x)=0$ has, is hard work by algebra alone and nearly immediate once the function is drawn.

Turning a rule into a picture requires a device for holding an input and its output at once, and that device is the coordinate plane. It arrives in the next lesson, and with it the observation that injectivity, range, and the number of solutions of an equation are all visible at a glance.

A function is a rule, and a rule is invisible. Asking whether it ever repeats an output, or how many times it hits zero, or where it is largest, means grinding through algebra for answers that a picture would give away instantly.

The device that supplies the picture is the coordinate plane, and this lesson builds it and puts it to work on the two functions the course has met so far, the linear and the quadratic. It assumes only the algebra of the earlier lessons and the definition of a function.

Two numbers for one point

The idea appears in Descartes' La Géométrie of 1637, published as an appendix to the Discourse on Method, and independently in unpublished work of Fermat. Fix two perpendicular number lines crossing at a point called the origin. Any point in the plane is then named by two numbers, how far along and how far up, written $(x,y)$.

The consequence is larger than the construction. Geometry becomes algebra and algebra becomes geometry: a curve is the set of points whose coordinates satisfy an equation, and an equation is a curve. Two thousand years of Greek geometry, done with compass and straightedge on figures, becomes computation with letters, and problems that needed a clever construction each time fall to routine.

For a function this pairing has a specific use. Plot the point $(x,f(x))$ for every $x$ in the domain, and the resulting set of points is the graph of $f$. Everything the rule does is now laid out at once, which is why the rest of this course reasons about functions with one eye on their graphs.

Distance, and the first theorem to fall out

Take two points $(x_1,y_1)$ and $(x_2,y_2)$. The horizontal gap between them is $x_2-x_1$ and the vertical gap is $y_2-y_1$, and those two gaps are the legs of a right triangle whose hypotenuse joins the points. Pythagoras' theorem gives the length immediately:

The squares mean the order of subtraction does not matter, which is as it should be, since distance has no direction. The midpoint is even simpler, the average of the coordinates separately, $\mathrm{big}((x_1+x_2)/2,(y_1+y_2)/2\mathrm{big})$.

Example. Find the distance between $(2,-3)$ and $(7,9)$, and their midpoint.

The gaps are $7-2=5$ and $9-(-3)=12$, so $d=\sqrt{25+144}=\sqrt{169}=13$. The midpoint is $(4.5,3)$.

Now you. Find the distance between $(-4,5)$ and $(8,-11)$.

Circles come free from the same formula. The set of points at distance $r$ from $(a,b)$ is $(x-a{)}^2+(y-b{)}^2=r^2$, which is a definition turned directly into an equation. Note that a circle is not the graph of a function: the input $x=3$ has two points above it on a circle centred at the origin, and a function must have exactly one output. This is the vertical line test: a set of points is the graph of a function precisely when no vertical line meets it twice.

The straight line

The defining feature of a line is that the ratio of vertical change to horizontal change is the same everywhere on it. That constant is the slope or gradient $m=(y_2-y_1)/(x_2-x_1)$, and it is the rate at which the output changes per unit of input.

Given a slope $m$ and any point $(x_1,y_1)$ on the line, every other point $(x,y)$ satisfies $(y-y_1)/(x-x_1)=m$, which rearranges to the point-slope form $y-y_1=m(x-x_1)$. Expanding gives $y=mx+c$, where $c$ is the value at $x=0$, the intercept. Vertical lines are the exception: their slope is undefined, since the horizontal change is zero, and they are written $x=k$ and are not graphs of functions.

Two lines are parallel when their slopes are equal. They are perpendicular when $m_1{m}_2=-1$, which can be seen by rotating a slope triangle a quarter turn: a rise of $a$ over a run of $b$ becomes a rise of $-b$ over a run of $a$. So the perpendicular to a line of slope $3$ has slope $-1/3$.

Example. Find the line through $(2,7)$ and $(6,19)$.

The slope is $(19-7)/(6-2)=12/4=3$. Using the first point, $y-7=3(x-2)$, so $y=3x+1$. Check the second point: $3\times 6+1=19$.

Now you. Find the line through $(-3,11)$ and $(5,-5)$, and check both points.

A line through real data

Amos Dolbear noticed in 1897 that snowy tree crickets chirp faster when it is warmer, and that the relationship is close to linear over the range where the insects are active. His rule, still quoted, is that the temperature in degrees Fahrenheit is $50$ plus a quarter of the amount by which the chirp count per minute exceeds $40$.

As a function that is $T(N)=50+(N-40)/4$, a line of slope $1/4$ through the point $(40,50)$. At $120$ chirps per minute it predicts $70$ °F, at $160$ it predicts $80$ °F, and at $80$ it predicts $60$ °F. The slope carries the physical content: four extra chirps per minute for each additional degree, which is a statement about how insect metabolism responds to temperature.

The limits of the model are visible in the same equation, which is the point of writing it down. Set $N=0$ and it predicts $40$ °F, whereas silence in fact means only that the crickets have stopped, and below about $50$ °F they do. Extrapolating a fitted line outside the range of the data that produced it is the commonest abuse of a linear model, and the graph makes the abuse obvious: the line continues cheerfully into a region where no data ever existed.

What a graph shows at a glance

Draw a function and its algebraic properties become visual. The domain is the set of $x$ values under which the curve exists; the range is the set of $y$ values it reaches. Solutions of $f(x)=0$ are the points where the curve crosses the horizontal axis, so the number of solutions can be counted rather than derived. Solutions of $f(x)=g(x)$ are the crossings of two curves, which is why the intersection of $y=3x+1$ and $y=-2x+5$ at $x=0.8$, $y=3.4$ is both a picture and the solution of a pair of simultaneous equations.

Injectivity, the condition for having an inverse from the previous lesson, becomes the horizontal line test: $f$ is injective exactly when no horizontal line meets its graph twice. The parabola $y=x^2$ fails it at every positive height, which is the visual form of the fact that $3$ and $-3$ share a square.

The graph of an inverse is a reflection of the graph of the function in the line $y=x$, because reflecting swaps the roles of the coordinates, and swapping input and output is exactly what the inverse does. This is a good check on any inverse computed algebraically.

The parabola, read off its vertex form

The graph of $y=a{x}^2+bx+c$ is a parabola, opening upward when $a>0$ and downward when $a<0$. Its most useful feature is its turning point, and completing the square from the third lesson delivers it without further work.

Take $y=2x^2-12x+5$. Factor $2$ out of the first two terms: $y=2(x^2-6x)+5$. Complete the square inside: $x^2-6x=(x-3{)}^2-9$, so $y=2\mathrm{big}[(x-3{)}^2-9\mathrm{big}]+5=2(x-3{)}^2-13$.

Now everything is visible. The squared term is never negative, so $y$ is smallest when it is zero, at $x=3$, where $y=-13$. The vertex is $(3,-13)$, the curve is symmetric about the vertical line $x=3$, and since the minimum is below zero the parabola crosses the axis twice, at the two roots computed in the earlier lesson, $3\pm \sqrt{6.5}$. The discriminant and the vertex are saying the same thing in two languages: the sign of the discriminant is the question of whether the vertex sits on the far side of the axis from the opening.

Example. Put $y=x^2-6x+5$ into vertex form and describe the graph.

$x^2-6x=(x-3{)}^2-9$, so $y=(x-3{)}^2-4$. The vertex is $(3,-4)$, opening upward, symmetric about $x=3$. Since $y=0$ gives $(x-3{)}^2=4$, the crossings are at $x=1$ and $x=5$, and the vertex sits midway between them as symmetry requires.

Now you. Put $y=3x^2+12x-7$ into vertex form and give the vertex.

Shifting, stretching, reflecting

Vertex form is an instance of something general. Once the graph of $y=f(x)$ is known, four modifications give a family of related graphs, and knowing them removes most of the work of plotting.

Replacing $f(x)$ by $f(x)+k$ moves the graph up by $k$, which is obvious since every output gains $k$. Replacing it by $f(x-h)$ moves it right by $h$, which surprises people, and the reason is that the new function at $x=h$ does what the old one did at $x=0$: the input must be larger by $h$ to produce the same behaviour. Multiplying the output, $af(x)$, stretches vertically by a factor $a$, and a negative $a$ also flips the graph upside down. Multiplying the input, $f(bx)$, squeezes horizontally by a factor $b$, again the opposite of what the symbol suggests, for the same reason as the shift.

So $y=2(x-3{)}^2-13$ is the basic parabola $y=x^2$ moved three right, stretched by two, and dropped thirteen. Every quadratic is the same curve in a different position, which is a genuinely surprising fact: parabolas differ in scale and place but not in shape. Later lessons apply the same four moves to exponentials and to waves, where a horizontal shift becomes a phase and a horizontal squeeze becomes a frequency.

Where the picture misleads

A graph is evidence, not proof, and the honest limits are worth carrying. A plot shows a window, and behaviour outside it is invisible: $y=x^3-3x+1$ looks like a rising line if plotted from $-100$ to $100$, and only a window a few units wide reveals its two turning points. Features can also be too small to see, and a curve that appears to touch the axis may cross it just below the resolution of the drawing.

Nor does a graph settle exact values. It shows that $y=x^2-2$ crosses the axis somewhere near $1.41$; it cannot tell you that the crossing is irrational, which took the argument sketched in the first lesson. Pictures are for finding out what is true. Algebra, and eventually proof, is for establishing it.

Both curves in this lesson are members of one family: the linear function is degree one and the quadratic degree two, and nothing in the definition of a polynomial stops at two. What degree three and beyond look like, and how their roots and their factors are the same information twice, is the next lesson.

The line and the parabola are the first two members of a family, and there is nothing in either that suggests the family should stop at degree two. Cubics and quartics are written the same way, arise from the same kinds of problem, and are handled by one set of tools.

This lesson builds those tools. It assumes the algebra of the earlier lessons and the graphs of the previous one, and it produces the central fact about polynomials: a root and a factor are the same thing seen twice, so finding one is finding the other.

Degree, and what a polynomial is not

A polynomial in $x$ is a finite sum of terms $a_k{x}^k$ with whole-number exponents and constant coefficients, such as $3x^4-x^3+7x-2$. The largest exponent with a non-zero coefficient is the degree, and the coefficient sitting on it is the leading coefficient. Degree zero is a constant, one is linear, two quadratic, three cubic, four quartic.

The restriction to whole-number exponents is what makes the family well behaved, and it excludes a great deal. Neither $1/x$, which is $x^{-1}$, nor $\sqrt{x}$, which is $x^{1/2}$, is a polynomial, and neither is $2^x$, where the variable is in the exponent rather than the base. Each of those three has its own lesson later in this course, and each behaves quite differently.

Adding two polynomials gives a polynomial whose degree is at most the larger of the two, and can be less when leading terms cancel. Multiplying gives one whose degree is exactly the sum, since the leading terms multiply and nothing else can reach that power. So degree behaves under multiplication the way exponents do, which is the first hint of the index laws that a later lesson makes general.

At large $\mathrm{lvert}x\mathrm{rvert}$ the leading term swamps everything else, and this settles the shape of the far ends of the graph. For $p(x)=x^3-4x^2+2x+5$ at $x=100$, the leading term is $1\{,\}000\{,\}000$ while the rest contributes $-40\{,\}000+200+5$, four per cent of the total, and the fraction shrinks as $x$ grows. So an odd-degree polynomial with positive leading coefficient falls without limit to the left and rises without limit to the right, and an even-degree one goes the same way at both ends.

Division with a remainder

Whole numbers can be divided with a remainder smaller than the divisor: $17=5\times 3+2$. Polynomials do the same, with the degree of the remainder taking the place of the size: dividing $p(x)$ by a non-zero $d(x)$ gives a unique quotient $q(x)$ and remainder $r(x)$ with

where the degree of $r$ is less than the degree of $d$. Dividing by a linear $x-a$ therefore leaves a remainder of degree zero, a plain number.

The procedure is long division, arranged like the arithmetic version. Divide $x^3-4x^2+2x+5$ by $x-3$. The leading term $x^3$ divided by $x$ gives $x^2$; multiply $x-3$ by $x^2$ to get $x^3-3x^2$ and subtract, leaving $-x^2+2x+5$. Now $-x^2$ divided by $x$ gives $-x$; multiply and subtract, leaving $-x+5$. Finally $-x$ divided by $x$ gives $-1$; multiply and subtract, leaving $2$. So the quotient is $x^2-x-1$ and the remainder is $2$.

Multiplying back is the check, and it should always be done: $(x-3)(x^2-x-1)=x^3-x^2-x-3x^2+3x+3=x^3-4x^2+2x+3$, and adding the remainder $2$ recovers the original.

The remainder theorem, and why it is obvious in hindsight

There is a shortcut for the remainder that avoids the division entirely. Write the division statement for a linear divisor: $p(x)=(x-a)q(x)+r$, with $r$ a constant. This holds for every $x$, so it holds at $x=a$, where the first term dies:

The remainder on dividing by $x-a$ is simply $p(a)$. Checking against the division just done, $p(3)=27-36+6+5=2$, which is the remainder found the long way.

The factor theorem is the case $r=0$, and it is the reason the whole apparatus matters. If $p(a)=0$ then the remainder is zero, so $x-a$ divides $p$ exactly; and if $x-a$ divides $p$ exactly then $p(a)=0$. A number $a$ with $p(a)=0$ is a root, and the theorem says roots and linear factors are two descriptions of one fact. Finding a root hands you a factor, and dividing by that factor leaves a polynomial of one lower degree, which is a smaller problem of the same kind.

Example. Show that $2$ is a root of $p(x)=2x^3+x^2-13x+6$, and factor it completely.

Evaluate: $p(2)=16+4-26+6=0$, so $x-2$ is a factor. Dividing gives $2x^2+5x-3$, and that quadratic factors as $(2x-1)(x+3)$, since $-3\times 2=-6$ and the middle term works out. So $p(x)=(x-2)(2x-1)(x+3)$ and the roots are $2$, $1/2$ and $-3$. Check the middle one: $p(0.5)=0.25+0.25-6.5+6=0$.

Now you. Show that $1$ is a root of $q(x)=x^3-7x+6$ and factor it completely.

Where to look for a root

The factor theorem is useless without a root to start from, and guessing is only tolerable if the guesses are few. The rational root theorem makes them few. If a polynomial with integer coefficients has a rational root $p/q$ in lowest terms, then $p$ divides the constant term and $q$ divides the leading coefficient.

For $2x^3+x^2-13x+6$ the constant is $6$ and the leading coefficient is $2$, so the numerator is $\pm 1,\pm 2,\pm 3,\pm 6$ and the denominator is $1$ or $2$. Once repeats such as $2/2$ are struck out that is twelve candidates rather than infinitely many, and the roots found above, $2$, $1/2$ and $-3$, are all on the list as they must be.

The theorem is a genuine restriction and also a warning about how special rational roots are. Most polynomials with integer coefficients have none: $x^3-2$ has only $\sqrt[3]{2}$, which is irrational, and the candidates $\pm 1,\pm 2$ all fail on inspection. Textbook exercises are chosen to have rational roots, and this creates a false impression that finding roots is a matter of persistence. In the wild it is not, and numerical methods do the work.

Multiplicity, and what the graph does

A factor can repeat. In $p(x)=(x-2{)}^2(x+1)$ the root $2$ has multiplicity two, and the graph shows the difference plainly: at a root of odd multiplicity the curve crosses the axis, and at a root of even multiplicity it touches and turns back. The reason is a sign argument. Near $x=2$ the factor $(x-2{)}^2$ is positive on both sides, so the sign of the product is controlled by $(x+1)$ alone and does not change; with a single factor $(x-2)$ the sign flips as $x$ passes through.

Counting roots with their multiplicities makes the bookkeeping come out right. A polynomial of degree $n$ has at most $n$ roots, because each root contributes a linear factor and the degrees of the factors must add to $n$. It may have fewer real roots: $x^2+1$ has none, the gap left open in the first lesson, and $x^3-x$ has three while $x^3+x$ has one.

The fundamental theorem of algebra, proved by Gauss in his doctoral thesis of 1799 with several later and more rigorous proofs, says that over the complex numbers the count is always exact: every polynomial of degree $n\ge 1$ has exactly $n$ roots counted with multiplicity. This is the payoff of the extension the first lesson declined to make. Admitting the solutions of $x^2=-1$ does not merely patch that one equation, it makes every polynomial equation solvable at once, and no further extension is ever needed. On the real line the statement is weaker and messier, and the honest version is that the roots come in a bounded number and some of them may be missing.

Example. Factor $x^4-5x^2+4$ completely and describe its graph.

The expression is a quadratic in $x^2$, so treat $u=x^2$: $u^2-5u+4=(u-1)(u-4)$. Substituting back, $(x^2-1)(x^2-4)=(x-1)(x+1)(x-2)(x+2)$. Four distinct roots, all of multiplicity one, so the curve crosses the axis at $-2,-1,1,2$. The degree is even with a positive leading coefficient, so both ends rise, and the curve must therefore dip below the axis between $-2$ and $-1$, rise above between $-1$ and $1$, and dip again between $1$ and $2$.

Now you. Factor $x^3-3x^2-4x+12$ completely, given that $3$ is a root.

Building a polynomial from what it must do

The correspondence runs both ways, which is how polynomials get used as models. If a cubic must vanish at $-1$, $2$ and $5$, it is $a(x+1)(x-2)(x-5)$ for some constant $a$, and one further condition fixes $a$.

Example. Find the cubic with roots $-1$, $2$, $5$ passing through $(0,20)$.

Write $p(x)=a(x+1)(x-2)(x-5)$. At $x=0$ this is $a\times 1\times (-2)\times (-5)=10a$, and that must equal $20$, so $a=2$. Expanding, $p(x)=2(x+1)(x^2-7x+10)=2(x^3-6x^2+3x+10)=2x^3-12x^2+6x+20$. Check at $x=2$: $16-48+12+20=0$.

Now you. Find the quadratic with roots $3$ and $-4$ passing through $(1,-12)$.

The limits of exact solution

Everything above is exact, and it depends on someone handing over a root. When no rational root exists, degree three and four can still be solved in radicals by the formulas of Cardano and Ferrari, published in the Ars Magna of 1545, though the cubic formula is unpleasant enough that almost nobody uses it.

Beyond that the road ends. Ruffini argued in 1799 and Abel proved in 1824 that the general fifth-degree equation has no solution by radicals, and Galois in 1832 explained exactly which equations do and which do not, in work that founded modern algebra. This is not a gap awaiting a cleverer formula. It is a proof that no such formula can exist, and it is one of the places where mathematics stops offering better methods and starts explaining why a method is impossible.

What is left is numerical work, which is enough for every practical purpose: bisection, which uses the fact that a sign change traps a root, and faster iterative schemes built on the derivative. Those belong to calculus.

Polynomials share one further limitation, and it is the one that opens the next lesson. No polynomial doubles at fixed intervals. A quantity that grows by a fixed factor for each unit of time, as money at compound interest does and as a bacterial culture does, cannot be described by any of these functions, however high the degree. Describing it needs the variable in the exponent, which is where the family ends and a new one begins.

Money in an account earning five per cent does not grow by a fixed amount each year; it grows by a fixed factor. No polynomial does that, however high its degree, so the family of functions built in the previous lessons cannot describe interest, populations, or radioactive decay.

This lesson builds the family that can. It starts from repeated multiplication, which is arithmetic, and ends with a function defined for every real exponent, a specific irrational number that falls out of compounding, and a comparison that settles which kind of growth wins in the long run.

Counting the factors

Write $b^n$ for $b$ multiplied by itself $n$ times. With that reading the index laws are not rules to memorise but observations about counting. Multiplying $b^m$ by $b^n$ writes down $m$ factors and then $n$ more, so

Raising a power to a power writes down $n$ copies of a block of $m$ factors, giving $(b^m{)}^n=b^{mn}$. Distributing across a product, $(ab{)}^n=a^n{b}^n$, because multiplication is commutative and the factors can be sorted. That is the whole of the theory for whole-number exponents, and every one of the three laws is proved by counting.

Division follows from the first law: $b^m/b^n=b^{m-n}$ whenever $m$ exceeds $n$, since cancelling $n$ factors from $m$ leaves $m-n$. So $2^5\times 2^3/2^6=2^2=4$, which is worth checking directly: $32\times 8/64=4$.

The extensions are forced, not chosen

Nothing in the counting picture makes sense of $b^0$, or $b^{-3}$, or $b^{1/2}$: multiplying something by itself zero times, or minus three times, is not an operation. Yet all three have standard values, and the reason is a principle used repeatedly in the first lesson. Extend the notation in whatever way keeps the existing laws true, then check that nothing already working breaks.

Apply the first law with $n=0$: $b^m\times b^0=b^{m+0}=b^m$. Dividing by $b^m$, which is legal when $b$ is not zero, gives $b^0=1$. That is not a convention adopted for tidiness, it is the only value compatible with the law. Now apply the law with $n=-m$: $b^m\times b^{-m}=b^0=1$, so $b^{-m}$ must be $1/b^m$. A negative exponent is a reciprocal, forced.