Not lectured — kept here for when the algebra bites
This chapter is the ground the rest of the course stands on. We will not spend class time on it. But every time a problem later in the semester goes wrong, there is a good chance the mistake happened here, in a step someone learned years ago and has not looked at since. Come back to this page when that happens. It is short on purpose.
Arithmetic asks what is the answer? Algebra asks what is the rule? Once you write \(x\) instead of a particular number, you stop solving one problem and start solving all problems of that shape at once. That move — from the instance to the pattern — is the single most powerful thing humans have ever done with symbols, and everything in this course is a consequence of it.
We count with the natural numbers \(1,2,3,\dots\). Allowing subtraction to always work gives the integers; allowing division to always work (except by zero) gives the rational numbers, the ratios \(a/b\). The rationals are not enough — \(\sqrt2\) and \(\pi\) are not ratios of whole numbers — and filling in every remaining gap gives the real numbers \(\mathbb{R}\). Chapter 1 takes that filling-in seriously; for now, a real number is anything with a decimal expansion.
What makes algebra possible is not the numbers themselves but the rules they obey. There are only three that matter.
For all real numbers \(a,b,c\):
Almost every algebraic manipulation you will ever do is the distributive law, used forwards or backwards. Used forwards it is called expanding. Used backwards it is called factoring. Nothing else is going on.
Why is \(3\times 47\) easy to do in your head?
Because \(47=40+7\), and the distributive law lets you split the job: $$3(40+7)=3\cdot 40+3\cdot 7=120+21=141.$$ You already use the distributive law constantly. Algebra just gives it a name and lets you apply it to symbols you do not know the value of.
The rules that get forgotten most often:
$$\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd},\qquad \frac{a}{b}\cdot\frac{c}{d}=\frac{ac}{bd},\qquad \frac{a/b}{c/d}=\frac{a}{b}\cdot\frac{d}{c}.$$A recipe uses \(\tfrac34\) of a cup per batch, and there are \(5\) cups of flour in the bag. How many batches? Do it two ways: by reasoning, and by the rule.
By reasoning: each cup holds one and a third three-quarter-cups, since \(\tfrac34+\tfrac14\) fills a cup and the leftover quarter is a third of another three-quarters. Five cups therefore hold \(5\times\tfrac43=\tfrac{20}{3}=6\tfrac23\) batches.
By the rule: \(\dfrac{5}{3/4}=5\cdot\dfrac43=\dfrac{20}{3}\).
The rule "invert and multiply" is not arbitrary. Dividing by \(\tfrac34\) asks how many three-quarters fit, and three-quarters is small, so the answer must be bigger than 5. If a division ever makes a number grow, check whether you were dividing by something less than 1 — usually you were.
\(\dfrac{a}{b+c}\) is not \(\dfrac{a}{b}+\dfrac{a}{c}\). Test it with \(a=1,b=c=1\): the left side is \(\tfrac12\), the right side is \(2\). Denominators do not distribute. If you take one thing from this page, take that.
An exponent is a count of repeated multiplication, and every rule below is just bookkeeping on that count.
For \(a\neq 0\) and integers \(m,n\):
$$a^ma^n=a^{m+n},\qquad \frac{a^m}{a^n}=a^{m-n},\qquad (a^m)^n=a^{mn},$$ $$(ab)^n=a^nb^n,\qquad a^0=1,\qquad a^{-n}=\frac{1}{a^n}.$$Write \(2^{5}\cdot2^{-3}\) as a single ordinary number.
decimals, fractions, pi and sqrt( ) are all fine
Why should \(a^0=1\)? Because \(a^3/a^3=1\) and the subtraction rule says it is \(a^{0}\). The definition is not a decision made for fun; it is forced, if we want the rules to keep working. The same argument forces \(a^{-n}=1/a^n\). This is how mathematics grows: a rule is discovered in one setting, and then the setting is stretched to keep the rule true.
Rational exponents extend the same idea: \(a^{1/n}=\sqrt[n]{a}\), so that \((a^{1/n})^n=a^{1}=a\), and \(a^{m/n}=\sqrt[n]{a^m}\).
Evaluate \(8^{2/3}\).
decimals, fractions, pi and sqrt( ) are all fine
Simplify \(\displaystyle \frac{(2x^{3}y^{-2})^{2}}{4x^{-1}y^{3}}\).
Square the top first: \((2x^3y^{-2})^2 = 4x^{6}y^{-4}\). Now divide: $$\frac{4x^{6}y^{-4}}{4x^{-1}y^{3}} = x^{6-(-1)}y^{-4-3}=x^{7}y^{-7}=\frac{x^{7}}{y^{7}}.$$
Evaluate \(16^{3/4}\) and \(125^{-2/3}\), and simplify \((8x^{6}y^{3})^{2/3}\).
Read \(a^{m/n}\) as "take the \(n\)th root, then raise to the \(m\)th power" — always root first, because the numbers stay small.
\(16^{3/4}=(\sqrt[4]{16})^{3}=2^{3}=8\).
\(125^{-2/3}=\dfrac{1}{(\sqrt[3]{125})^{2}}=\dfrac{1}{25}\).
\((8x^{6}y^{3})^{2/3}=8^{2/3}x^{4}y^{2}=4x^{4}y^{2}\).
Fractional exponents are not a new idea bolted on; they are the only way to keep \(a^{m}a^{n}=a^{m+n}\) true once you allow roots. Chapter 3 takes the last step and asks what \(2^{\pi}\) could possibly mean.
\(\sqrt{a+b}\neq\sqrt a+\sqrt b\) (try \(a=b=9\): \(\sqrt{18}\approx 4.24\), not \(6\)), and \((a+b)^2\neq a^2+b^2\). Powers and roots do not distribute over sums. Only multiplication does.
A polynomial is what you can build from a variable using only addition and multiplication: $$p(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0.$$ That sounds restrictive. It is not. Think of \(x\) as a task you perform, and the polynomial as a schedule.
\(x^2\) is do it, then do it again — repetition, composition, one thing feeding into the next. \(3x\) is do it three times side by side — addition, parallel copies. A polynomial is any schedule of nested repetitions and parallel copies, with a fixed cost \(a_0\) you pay whatever happens. A delivery fee plus a per-mile rate plus a term that grows with the square of the distance because traffic gets worse: that is a polynomial, and so is almost every cost, area, volume, and trajectory you will meet.
Adding polynomials means combining like terms. Multiplying them means the distributive law applied until nothing is left in brackets. Three products are worth memorizing, because they will appear hundreds of times:
Expand \((x+5)^{2}\). What is the coefficient of \(x\)?
decimals, fractions, pi and sqrt( ) are all fine
The middle term \(2ab\) in the first one is the term everyone drops. The picture below is there so you never drop it again: the square of side \(a+b\) genuinely contains two rectangles of area \(ab\), and you can see them.
Compute \(47\times 53\) and \(98^{2}\) without writing anything down.
\(47\times53=(50-3)(50+3)=50^{2}-3^{2}=2500-9=2491\).
\(98^{2}=(100-2)^{2}=10000-400+4=9604\).
These are not party tricks. They are the special products read right to left, and the habit of looking for that structure — "is this a difference of squares in disguise?" — is exactly the habit that makes factoring feel like recognition rather than search.
Expand \((x+2)(x^{2}-3x+4)\) and \((2x-1)^{3}\).
Distribute each term of the first over the whole second: $$x(x^{2}-3x+4)+2(x^{2}-3x+4)=x^{3}-3x^{2}+4x+2x^{2}-6x+8=x^{3}-x^{2}-2x+8.$$
For the cube, square first and then multiply again: $$(2x-1)^{2}=4x^{2}-4x+1,\qquad (4x^{2}-4x+1)(2x-1)=8x^{3}-12x^{2}+6x-1.$$
The degree of a product is the sum of the degrees, every time. That is worth knowing before you start: if the answer to the first part had not been a cubic, something went wrong.
Factoring is running multiplication backwards: given the expanded form, recover the pieces. It matters because a product tells you something a sum never can — if \(AB=0\), then \(A=0\) or \(B=0\). That is the only reliable way to solve an equation.
The standard moves, in the order you should try them:
Step 3 looks like guesswork. It is not. It is a rectangle.
Two numbers multiply to \(20\) and add to \(9\). What are they?
2 numbers, separated by a comma, in any order
Factor \(x^2+7x+12\) and \(2x^2-5x-3\).
For the first: two numbers with product \(12\) and sum \(7\) are \(3\) and \(4\), so \(x^2+7x+12=(x+3)(x+4)\).
For the second the leading coefficient is not 1, so look for two numbers with product \(2\cdot(-3)=-6\) and sum \(-5\): these are \(-6\) and \(1\). Split the middle term and group: $$2x^2-6x+x-3=2x(x-3)+1(x-3)=(2x+1)(x-3).$$
Factor \(x^{3}+2x^{2}-9x-18\) completely.
Four terms with no common factor: group them in pairs and hope the leftovers match. $$x^{2}(x+2)-9(x+2)=(x^{2}-9)(x+2)=(x-3)(x+3)(x+2).$$
The middle step is where people stop too early. \(x^{2}-9\) is a difference of squares, so it factors again. Completely means keep going until nothing left can be broken.
Note what you have gained: the roots of this cubic are \(3,-3,-2\), read straight off. Finding roots is the reason factoring exists, and in Chapter 2 you will meet the far more common situation in which no factoring is possible at all — and a method for finding the roots anyway.
A rational expression is one polynomial over another. They behave exactly like fractions, with one extra hazard: the denominator may not be zero, and that restriction survives even after cancelling.
Simplify \(\displaystyle \frac{x^2-9}{x^2+x-6}\) and state what \(x\) cannot be.
Factor both: \(\dfrac{(x-3)(x+3)}{(x+3)(x-2)}\). Cancel the common factor \(x+3\) to get \(\dfrac{x-3}{x-2}\).
The original expression was undefined at \(x=-3\) and \(x=2\), and it still is. Cancelling changed the formula, not the domain — a distinction Chapter 1 will make into a habit.
To add rational expressions, find a common denominator by factoring first — never by blindly multiplying the two denominators together.
Combine \(\dfrac{3}{x^{2}-1}+\dfrac{2}{x+1}\) into a single fraction.
Factor first: \(x^{2}-1=(x-1)(x+1)\). The common denominator is therefore \((x-1)(x+1)\) — not \((x^{2}-1)(x+1)\), which would be doing twice the work and then cancelling it back. $$\frac{3}{(x-1)(x+1)}+\frac{2(x-1)}{(x-1)(x+1)}=\frac{3+2x-2}{(x-1)(x+1)}=\frac{2x+1}{x^{2}-1}.$$
Valid for \(x\neq\pm1\).
Solving means peeling operations off the unknown, one at a time, doing the same thing to both sides. Two families cover most of what we need.
\(ax+b=c\) has exactly one solution, \(x=(c-b)/a\), as long as \(a\neq 0\). Geometrically: a straight line crosses a horizontal level exactly once unless it is parallel to it.
\(ax^2+bx+c=0\) can be attacked three ways: factoring (fast when it works), completing the square (always works, and explains everything), and the quadratic formula (which is completing the square, done once in general so you never have to do it again).
If \(a\neq 0\), the solutions of \(ax^2+bx+c=0\) are $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$ The quantity \(D=b^2-4ac\) is the discriminant: two real solutions if \(D>0\), one if \(D=0\), none at all among the reals if \(D<0\).
Solve \(x^2-6x+4=0\) by completing the square, and check against the formula.
Move the constant: \(x^2-6x=-4\). Half of \(-6\) is \(-3\); its square is \(9\). Add \(9\) to both sides: $$x^2-6x+9=5 \quad\Longrightarrow\quad (x-3)^2=5 \quad\Longrightarrow\quad x=3\pm\sqrt5.$$ The formula gives \(x=\dfrac{6\pm\sqrt{36-16}}{2}=\dfrac{6\pm\sqrt{20}}{2}=3\pm\sqrt5\). Same answer, because it is the same method.
Solve \(\dfrac{2}{x-3}=\dfrac{x}{x-3}-1\).
Multiply through by \(x-3\): $$2=x-(x-3)=3.$$ That is false for every \(x\), so there is no solution at all. Worse things can happen: if the algebra had produced \(x=3\), we would have to throw it away, because \(x=3\) makes the original denominators zero and the equation meaningless.
Multiplying both sides by something containing the unknown is the one algebraic move that can manufacture solutions out of nothing. Whenever you make it, check your answers against the original equation.
A ball thrown upward has height \(h(t)=-4.9t^{2}+12t+1.5\) metres after \(t\) seconds. Does it ever reach 12 metres?
Set \(-4.9t^{2}+12t+1.5=12\), that is \(4.9t^{2}-12t+10.5=0\). The discriminant is $$D=144-4(4.9)(10.5)=144-205.8=-61.8<0,$$ so there are no real solutions: the ball never gets there. Its peak is lower than 12 m.
The discriminant answered a physical question without our ever solving anything. That is the useful thing about it — it tells you whether an answer exists before you go looking, and "no real solution" is often the most informative result you can get.
"Completing the square" is not a metaphor. Look again at the picture in section 3. \(x^2+6x\) is a square of side \(x\) with two \(3\times x\) rectangles attached to its sides — an L-shape. To make it a genuine square you must fill in the missing \(3\times 3\) corner, which is why you add \(9\). The algebra is a literal description of a shape being completed. This is how the method was first found, and it is why it works.
Inequalities follow the same rules as equations with one exception, and it is the exception that gets people: multiplying or dividing by a negative number reverses the inequality. \(-2x<6\) gives \(x>-3\).
Solving \(-3x>12\) gives \(x\) .
For a nonlinear inequality, do not divide — factor, find where each factor changes sign, and test one point in each resulting interval.
Solve \(x^2-x-6>0\).
Factor: \((x-3)(x+2)>0\). The factors vanish at \(x=-2\) and \(x=3\), splitting the line into three intervals. Test \(x=-3\): \((-6)(-1)=6>0\), yes. Test \(x=0\): \((-3)(2)=-6\), no. Test \(x=4\): \((1)(6)=6>0\), yes.
Solution: \(x<-2\) or \(x>3\), written \((-\infty,-2)\cup(3,\infty)\).
Solve \(\dfrac{x-1}{x+2}\ge 0\).
Do not multiply by \(x+2\) — its sign is unknown, so you would not know whether to flip. Instead find where the expression can change sign: the numerator vanishes at \(x=1\), the denominator at \(x=-2\).
Test one point in each region: at \(x=-3\), \(\frac{-4}{-1}=4>0\) ✓. At \(x=0\), \(\frac{-1}{2}<0\) ✗. At \(x=2\), \(\frac{1}{4}>0\) ✓.
So the solution is \((-\infty,-2)\cup[1,\infty)\). Note the bracket at 1, where the expression is zero and therefore allowed, and the round bracket at \(-2\), where it is undefined and never allowed no matter what the inequality says.
The absolute value \(|a|\) is the distance from \(a\) to \(0\). Read every absolute value as a distance and these stop being memorization: \(|x-c|<r\) says \(x\) is within \(r\) of \(c\), which is the interval \((c-r,c+r)\).
Solve \(|2x-1|=7\).
By cases: \(2x-1=7\) gives \(x=4\), and \(2x-1=-7\) gives \(x=-3\).
By distance: divide inside by 2 first, \(2|x-\tfrac12|=7\), so \(|x-\tfrac12|=3.5\) — the points exactly \(3.5\) away from \(\tfrac12\), namely \(4\) and \(-3\). Same answers, and the second reading tells you in advance that there will be exactly two of them, symmetric about \(\tfrac12\).
None of the notation above is old. The Greeks did algebra in sentences and pictures; the Babylonians solved quadratics on clay tablets a thousand years before that, with recipes and no symbols at all.
Around 820 CE, at the House of Wisdom in Baghdad, Muhammad ibn Musa al-Khwarizmi wrote a book whose title contains the phrase al-jabr — the restoring, the moving of a subtracted term to the other side. The word became algebra. His own name, run through Latin, became algorithm. His book contains no symbols whatsoever: every equation is a paragraph, and every solution is justified by a drawing of a square being completed.
The same scholarly world carried the Hindu decimal numerals — including the zero, made a full-fledged number by Brahmagupta in India in 628 CE — west into the Mediterranean, where Fibonacci learned them from merchants in North Africa and argued for them in Europe in 1202. It took another three centuries for them to win. The symbols \(+\), \(-\), \(=\) and the use of letters for unknowns arrive only in the 1500s and 1600s, with Viete and Descartes.
So the compact line \(x^2-6x+4=0\) is a piece of technology roughly four hundred years old, assembled from parts made in Baghdad, India, and Renaissance Europe. It feels like it was always there. It was not. You are reading the current draft.
If you can do these without help, you are ready for Lecture 1. If one of them stops you, the relevant section is named beside it.
Next: Chapter 1, where a function stops being a formula and becomes a way of moving information from one place to another.