Study Companion: Programme F.2 — Introduction to Algebra
Algebra is arithmetic done before the numbers arrive. Every move in this programme — collecting terms, removing brackets, using indices and logarithms, dividing polynomials, adding fractions, factorizing — is a rule you already obey with numbers, now applied to a letter that stands for a number not yet chosen. So none of the rules is new. What this file shows is why each one is forced: a rectangle that can be counted two ways, a row of factors that can be counted, a ruler that adds lengths, a numeral whose base was never fixed.
The rules, procedures and drill are in the companion reference sheet, which follows the book's topic order. This file tells the story in seven units instead, because the programme's many topics are really seven 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: Letters Are Numbers Not Yet Chosen
The problem
Try this on a friend. Think of a number. Add 15. Double the result. Add the number you first thought of. Divide by 3. Take away the number you first thought of. Whatever they picked, you announce the answer: 10.
Start with 7: 7\to22\to44\to51\to17\to10. Start with 12: 12\to27\to54\to66\to22\to10. How can you know the answer to a calculation when you do not know the number it started from?
First attempt
Try more numbers. 100 gives 100\to115\to230\to330\to110\to10. Every test agrees — but a thousand agreeing tests do not prove that the next number will behave, and they say nothing about why the 15 turned into a 10.
The picture
Put the unknown number in a closed box and never open it. Each instruction then acts on boxes and loose tiles, and the whole calculation can be carried out without knowing what is inside:
Now the trick is visible. Doubling made two boxes, adding the number made three, and dividing by three was chosen to leave exactly one box — which the last step removes. The loose tiles went 15\to30\to10, and that is the answer, whatever the box held.
Call the box a and the table becomes a line of algebra:
A letter is a number not yet chosen. Because every step obeys the ordinary rules of arithmetic, one calculation with the letter covers every number at once.
Check it with numbers
Run the box table with a=7: the boxes hold 7 each, so the rows read 7, 7+15=22, 14+30=44, 21+30=51, 7+10=17, 10. The loose-tile column is identical for every starting number; only the box contents change, and they cancel.
The rule it gives
Vocabulary. With letters standing for numbers: the sum a+b, the difference a-b, the product a\times b, written a.b or simply ab (the multiplication sign is suppressed), the quotient a\div b=a/b=\frac ab provided b\ne0, and the power a^b. Using letters and numerals together this way is algebra.
Constants and variables. A letter that stands for one number, possibly unknown, like a in the trick, is a constant. A letter that may stand for any one of a collection of numbers is a variable. By convention letters from the start of the alphabet (a,b,c,\ldots) are constants and letters from the end (\ldots,x,y,z) are variables, but the role is set by the problem, not the letter: in V=IR with a fixed resistor, R is a constant; in a temperature sweep of the same resistor, it is a variable.
The rules of algebra are the laws of arithmetic from F.1, stated for every number at once:
Subtraction and division have neither property: x-y\ne y-x unless x=y, and x-(y-z)\ne(x-y)-z unless z=0. The rules of precedence (brackets, powers, \times\div, +-) carry over unchanged.
Terms and coefficients. In 8x-3xy the terms are the x term and the xy term, and the coefficients are the numbers multiplying them — with their signs: 8 and -3.
Like terms contain exactly the same letters. They are the same kind of box, so they combine by adding their coefficients: 3pq+qp=4pq, because qp is pq by commutativity. Letters in a term are written in alphabetical order. Unlike terms do not combine: 4pq-pr is as simplified as "four boxes minus one bag".
Similar terms share some letters but not all, like ab and ac. The shared part is a common factor, and it can be factored out: ab+ac=a(b+c). This is factorization, and numerical factors come out the same way: 9st-3sv-6sw=3s(3t-v-2w).
Worked example
Check. At p=2, q=3, r=5: the original is 18+50-30+6-60=-16 and the result is 24-10-30=-16. A substituted value can prove two expressions different, but never the same — they might agree at that point by accident — so the check catches errors without replacing the argument.
Where this shows up
The heat needed to warm a body is Q=mc\,\delta T. One line covers every material, mass and temperature rise, because each letter stands for a column of a table nobody has to write down. For 0.5 kg of water (c=4186 J kg^{-1} K^{-1}) warmed by 30 K, Q=0.5\times4186\times30=62\,790 J, about 62.8 kJ; change the material and only the value of c changes.
Narration spine. Run the trick with a closed box: the box doubles, gains a third copy, is divided back to one and is taken away, while the loose tiles go 15\to30\to10. Run the same trick with 7, then replace the box with a and write the steps as algebra.
Unit 02: Brackets Are Rectangles
The problem
A workshop floor is 5.0 m deep and has two bays, 3.2 m and 1.8 m wide. The concrete order can be worked out as one slab, 5.0\times(3.2+1.8), or as two, 5.0\times3.2+5.0\times1.8. Everyone expects the same answer. Why must it be the same — and does the same thing happen with division?
First attempt
"Multiplying spreads over a bracket" — so presumably dividing does too. Test it: is \frac{12}{3+1} equal to \frac{12}3+\frac{12}1? The left side is 3; the right side is 16. So operations do not simply spread over brackets. Something specific to multiplication is doing the work, and we need to see what.
The picture
A product is an area. Draw a rectangle a high and b+c wide; its area is a(b+c). Now draw one vertical line at distance b:
Nothing was added or removed, so the two ways of counting the area must agree:
That is the whole reason multiplication distributes: area does not care how it is counted. Subtraction takes a piece away instead of adding one, a(b-c)=ab-ac, and a negative multiplier reverses every term it reaches: -(x-3)=-x+3. The minus sign multiplies the whole bracket.
Now the division puzzle. \frac{x+y}z does split, \frac{x+y}z=\frac xz+\frac yz, because it is \frac1z\times(x+y) — a multiplication, so the rectangle picture applies. But \frac x{y+z} is not a multiplication over the bracket at all: the bracket is the divisor. Division distributes from the right only:
With two brackets, cut the rectangle both ways. A rectangle a+b high and c+d wide splits into four cells, each the product of one piece from each bracket:
There are four cells, and a missed term is a missed cell.
Check it with numbers
The floor: 5.0\times(3.2+1.8)=5.0\times5.0=25 m^2, and 5.0\times3.2+5.0\times1.8=16+9=25 m^2. The two cells of the picture are the two bays.
The rule it gives
Expanding (removing) brackets. Multiply — or divide — every term inside the bracket by the term outside; if the term outside is negative, every term inside changes sign:
A bracket divided by a term is expanded the same way, since dividing by 8x is multiplying by \frac1{8x}:
Nested brackets. Remove the innermost brackets first, then work outwards — each bracket is one rectangle drawn inside another.
Worked example
Check. At x=5: the original is 2\{15-[10-1]\}=2\times6=12, and 4x-8=12.
A second, with a negative multiplier inside:
And two brackets, by the four cells:
Where this shows up
The floor again, finished: the slab is 0.15 m thick, so the order is 25\times0.15=3.75 m^3 of concrete whether the two bays are poured together or separately. The same four-cell grid, with more cells, multiplies out the denominators of two connected systems in control engineering; a missed cell there is not a small error but a wrong polynomial, with the wrong roots.
Narration spine. Draw the floor with its two bays. Try splitting \frac{12}{3+1} and strike it out: 3 against 16. Then cut the floor at the bay line, count 16+9=25 m^2, and relabel the sides a, b, c so that the two cells build a(b+c)=ab+ac.
Unit 03: Powers Count Factors, So Fractional Powers Are Forced
The problem
The torsional stiffness of a round steel shaft grows as the fourth power of its diameter. Swap a 20 mm shaft for a 25 mm one and the stiffness is multiplied by \left(\frac{25}{20}\right)^4. Fine — but engineers also write d^{1/2}, d^{-1} and d^0 without blinking. What can "half a multiplication" possibly mean?
First attempt
Take the definition at its word: a^n means n factors of a multiplied together. That works for n=1,2,3,\ldots — and then stops. a^0 would be "no factors at all", a^{-2} "minus two factors", and a^{1/2} "half a factor". The definition has nothing to say, so it looks as if we are free to make something up.
The picture
Write the factors out in a row. Then the rules of indices are just counting:
| Law | Write the factors out | Result | |
|---|---|---|---|
| a^3\times a^2 | (a\,a\,a)(a\,a)=a\,a\,a\,a\,a | a^5 | count: 3+2 |
| a^5\div a^2 | \frac{a\,a\,a\,a\,a}{a\,a}=a\,a\,a | a^3 | cancel: 5-2 |
| (a^2)^3 | (a\,a)(a\,a)(a\,a)=a\,a\,a\,a\,a\,a | a^6 | groups: 2\times3 |
These three rules were proved for counting numbers. Now make one demand: rule 1, a^m\times a^n=a^{m+n}, must keep working for the new indices. Then nothing is left to choose:
- a^m\times a^0=a^{m+0}=a^m, so a^0 must be 1.
- a^m\times a^{-m}=a^0=1, so a^{-m} must be \frac1{a^m}.
- a^{1/2}\times a^{1/2}=a^1=a, so a^{1/2} must be a number whose square is a: \sqrt a. In the same way \left(a^{1/m}\right)^m=a, so a^{1/m}=\sqrt[m]a.
Every new value is forced by the counting rule. Nobody chose them; they were the only values that would not break it.
The same argument can be seen as a staircase. Draw a bar of height 2^n at each whole number n. Each step to the right multiplies by 2, so each step to the left must divide by 2 — and walking left past n=1 gives 2^0=1, then 2^{-1}=\frac12 and 2^{-2}=\frac14:
Halfway between two steps, the steps must still be equal: two half-steps, each multiplying by the same number k, have to make one whole step, so k\times k=2 and 2^{1/2}=\sqrt2=1.414\ldots. The tempting guess "half of 2", which is 1, fails at once: the first half-step would multiply by 1 and the second by 2.
Check it with numbers
8^{2/3} can be read two ways, and both must agree: \left(\sqrt[3]8\right)^2=2^2=4, and \sqrt[3]{8^2}=\sqrt[3]{64}=4.
\frac{x^7y^{-2}}{x^3y^{-5}}=x^{7-3}y^{-2-(-5)}=x^4y^3. At x=2, y=3: the left side is \frac{128/9}{8/243}=\frac{128\times243}{9\times8}=432, and 2^4\times3^3=16\times27=432.
The rule it gives
In 5a^3, a is the base, 3 the index (or exponent, or power) and 5 the coefficient. The rules of indices, in the book's numbering:
and from 3 and 6 together, fractional indices:
For a>0, \sqrt a and a^{1/2} mean the positive root. Rules 4 and 5 need a\ne0.
Worked example
A larger one — raise each bracket to its power term by term, then collect each letter:
Where this shows up
The shaft: \left(\frac{25}{20}\right)^4=1.25^4=2.44, so a 5 mm increase in diameter makes the shaft 2.44 times as stiff in twist. Fractional powers are just as routine: the natural frequency of a mass on a spring is proportional to k^{1/2}, so a spring four times as stiff raises the frequency by 4^{1/2}=2.
Narration spine. Write 2^3\cdot2^2 as factors: the indices count them. Build the staircase 2,4,8 and walk a probe left, halving at each step, until 2^0=1 and 2^{-2}=\frac14 appear. Strike out the guess 2^{1/2}=1, then move the probe to n=\frac12 and read 1.414. Finish on 8^{2/3}=4 computed both ways.
Unit 04: Logarithms Turn Multiplying Into Adding
The problem
A radio receiver takes a 2.0 mW signal. An amplifier multiplies its power by 100; the cable after it passes only about a quarter. Engineers never multiply these numbers: they write +20 dB and -6 dB and add, getting +14 dB. How can adding two numbers do a multiplication?
First attempt
It cannot, if the numbers being added are the gains themselves: 100+0.25 is not 100\times0.25. So whatever is being added must be something else — some stand-in for each gain that turns products into sums.
The picture
Unit 03 has already built the machine. For powers of one base,
multiplication on the left, addition of the indices on the right. The catch is that this only works for numbers already written as powers of a. So ask the question that makes it general: for a given number x, which power of a is it? Give that power a name — the logarithm of x to base a — and every number has a stand-in that adds when the numbers multiply.
A slide rule is this idea built in wood. Each ruler is marked not in equal steps but with the number x at distance \log x from the left end:
Sliding the lower ruler along adds its lengths to the upper one's, so under the upper 2, the lower 3 lands at distance \log2+\log3 — which the upper scale labels 6. Adding lengths multiplies numbers.
The proof. Let x=a^p and y=a^q, so \log_ax=p and \log_ay=q. Then xy=a^{p+q}, and reading off the index,
Check it with numbers
\log_2(8\times4)=\log_232=5, and \log_28+\log_24=3+2=5.
The receiver: a decibel figure is 10\log_{10} of a power ratio, so +20 dB is a ratio of 10^{2}=100, and -6 dB is 10^{-0.6}=0.251 — close to a quarter, not exactly (a quarter is -6.02 dB). The sum +14 dB is 10^{1.4}=25.12, and indeed 100\times0.2512=25.12. The output is 2.0\times25.12\approx50.2 mW.
The rule it gives
Definition. If a=b^c, with a>0, b>0 and b\ne1, then c is the logarithm of a to the base b: c=\log_ba. So \log_525=2 because 5^2=25, and \log_x81=4 means x^4=81, x=3. The restrictions are forced: a positive base raised to any power is positive, so only positive numbers have logarithms, and base 1 is excluded because every power of 1 is 1.
Rules of logarithms, each an index rule read backwards:
| Rule | Because | |
|---|---|---|
| (a) | \log_axy=\log_ax+\log_ay | rule 1 |
| (b) | \log_a(x\div y)=\log_ax-\log_ay | rule 2 |
| (c) | \log_a(x^n)=n\log_ax | rule 3 |
| (d) | \log_a1=0 | a^0=1 |
| (e) | \log_aa=1 | a^1=a |
| (f) | \log_aa^x=x | the definition |
| (g) | a^{\log_ax}=x | the definition |
| (h) | \log_ab=\frac1{\log_ba} | change of base with x=b |
Common and natural logarithms. Base 10 gives common logarithms, written without the base: \log5.321=0.726. Base e=2.71828\ldots gives natural logarithms, written \ln: \ln13.45=2.599. For any base greater than 1: the log of 1 is 0; the log of a number greater than 1 is positive; the log of a number between 0 and 1 is negative (\log0.278=-0.556); the log of 0 or of a negative number is not a real number.
Change of base. Let a=b^c and x=a^d, so c=\log_ba and d=\log_ax. Then x=(b^c)^d=b^{cd}, so cd=\log_bx:
In practice, with b=10: \log_ax=\frac{\log x}{\log a}. So \log_23.66=\frac{0.5635}{0.3010}=1.872. On the slide rule, changing the base just rescales every length by the same factor, which is why one set of scales serves every base.
Logarithmic equations. Combine each side into a single logarithm with rules (a)–(c), then equate what is inside. Every logarithm in the original equation must have a positive argument, so check the answer against each: \log_a(x-3) exists only for x>3.
Formulas in log form. Taking logs turns a formula made of products and powers into a sum:
Read backwards, \log K=\log P-\log T+1.3\log V becomes K=\frac{PV^{1.3}}T.
Worked example
Solve 2\log_ax-\log_a(x-1)=\log_a(x+3).
Check validity. At x=\frac32 the arguments are x=\frac32, x-1=\frac12 and x+3=\frac92, all positive, so the answer stands.
Where this shows up
The receiver chain is the everyday case: gains and losses in dB add along a signal path, and the 2.0 mW input leaves at about 50.2 mW. Log form is the other: a power law y=kx^n becomes \log y=\log k+n\log x, a straight line on log–log paper whose slope is the index n — which is how an index is read off experimental data.
Narration spine. Strike out the idea of adding the gains themselves. Build a slide rule from two log scales and slide 1 under 2 so that the lower 3 lands on 6; brace \log2 and \log3 end to end to reach \log6, and check 0.301+0.477=0.778. Build \log_a(xy)=\log_ax+\log_ay from the braces and end on the receiver: +20 dB and -6 dB add to +14 dB, about 25 times.
Unit 05: Polynomial Arithmetic Is Long Arithmetic in Base x
The problem
Divide 4x^3+13x+33 by 2x+3 and the working looks exactly like the long division you learned at school, down to the layout on the page. Is that a teaching convenience, or is it the same calculation?
First attempt
Call it an analogy: "divide the leading terms, multiply back, subtract, bring down — just like numbers". That describes the method but explains nothing, and the analogy seems to break at once, because x^3\div x has nothing to do with 4\div2.
The picture
F.1 Unit 01 showed that a numeral is a polynomial in its base: 4163=4(10^3)+1(10^2)+6(10)+3. Read that backwards: a polynomial is a numeral whose base was never fixed. Put x=10 and
Two things show up. The missing x^2 term is the digit 0: a placeholder that keeps the columns lined up. And the "digits" 13 and 33 are too big to be digits. That is the only difference between the two calculations: polynomial arithmetic is long arithmetic with the carrying switched off. A column may hold any coefficient, and nothing ever carries into the next column. Everything else — the columns, the placeholders, the four repeated steps — is the same.
The layout makes the correspondence exact. Here is multiplication in columns, with the powers of x as place values:
At x=10 this is 134\times25=3350, and 2000+1100+230+20=3350. The column totals 11 and 23 are where ordinary arithmetic would carry; polynomial arithmetic leaves them standing.
Check it with numbers
The division, laid out like long division, with 0x^2 inserted:
The quotient is 2x^2-3x+11, the three multipliers read down the left.
Put x=10 everywhere: 4163\div23=181, and the quotient is 2(100)-3(10)+11=181. The same calculation, digit for digit.
The rule it gives
Multiplication. Multiply the second expression by each term of the first and add, setting like powers in the same column. Where a power is missing, insert it with a zero coefficient.
Division. Set out as long division: divide the leading term of what remains by the leading term of the divisor, multiply the divisor by that result, subtract, bring down the next term, and repeat. Insert any missing power with a zero coefficient. The answer is the quotient. In this programme every division comes out exactly; what is left over when it does not — the remainder — is the subject of F.3.
The divisor may itself have several terms; the steps do not change:
A sum or difference of cubes divides exactly by the matching binomial:
Worked example
(12x^3-2x^2-3x+28)\div(3x+4). Divide 12x^3 by 3x: 4x^2; subtract 4x^2(3x+4)=12x^3+16x^2 to leave -18x^2-3x. Divide by 3x: -6x; subtract -18x^2-24x to leave 21x+28. Divide: +7; subtract 21x+28 to leave 0.
Check. At x=10: 11\,798\div34=347=400-60+7.
Where this shows up
A hollow steel cube of outer side q cm with a cubic cavity of side 2 cm contains q^3-8 cm^3 of metal. The division above says this is (q-2)(q^2+2q+4); for q=10 cm that is 8\times124=992 cm^3, and directly 1000-8=992 cm^3. The same machinery runs inside every computer algebra system and big-number library: long integers are multiplied as polynomials in a large base, and the carries are resolved only at the end.
Narration spine. Give each power of x a column and write the dividend 4,0,13,33, lighting up the zero placeholder. Run the division one quotient term at a time: divide, multiply, subtract. Then turn the column headings into 1000,100,10,1: the dividend becomes 4163, the division 4163\div23=181, and the quotient at x=10 is 181 too.
Unit 06: Algebraic Fractions Are Fractions
The problem
Two resistors, 6\ \Omega and 3\ \Omega, are connected side by side, in parallel. The rule is \frac1R=\frac1{R_1}+\frac1{R_2}. What single resistor does the pair behave like — and why is the standard answer \frac{R_1R_2}{R_1+R_2}?
First attempt
Add them: 9\ \Omega. But two side-by-side paths let more current through than either alone, so the pair must resist less than either branch — less than 3\ \Omega. A meter across the pair reads 2\ \Omega. Adding the resistances answers a different circuit.
The picture
What adds in parallel is \frac1R, the current per volt. So the question is F.1's fraction question, \frac16+\frac13, and the answer is F.1's picture: re-cut both bars into a common number of parts.
With letters nothing changes: cut one bar into R_1 parts and the other into R_2, and both fit a bar cut into R_1R_2 parts:
An algebraic fraction obeys the fraction rules because it is a fraction: its letters are numbers not yet chosen.
Check it with numbers
R_1=6, R_2=3: \frac{6\times3}{6+3}=\frac{18}9=2\ \Omega, matching the bar picture's \frac12 flipped.
The book's example with polynomial denominators:
At x=1: \frac22+\frac43=\frac73, and \frac{14}6=\frac73.
The rule it gives
Addition and subtraction need a common denominator, the LCM of the denominators:
With three terms the LCM of a, b and d^2 is abd^2: \frac ab-\frac c{d^2}+\frac da=\frac{a^2d^2-abc+bd^3}{abd^2}.
Multiplication: \frac ab\times\frac cd=\frac{ac}{bd}. Division: the reciprocal of \frac cd is \frac dc, and to divide by a fraction multiply by its reciprocal: \frac ab\div\frac cd=\frac{ad}{bc}.
Precedence. \div and \times are done left to right. In \frac{2a}{3b}\div\frac{a^2b}6\times\frac{ab}2 the division comes first:
Multiplying first gives \frac8{a^2b^3}, which is a different expression: at a=b=1 one is 2 and the other 8.
Simplifying by factorizing. Factorize top and bottom and cancel common factors:
Cancelling changes the domain. \frac{x^2-4}{x^2-x-2}=\frac{(x-2)(x+2)}{(x-2)(x+1)}=\frac{x+2}{x+1} — except at x=2, where the left side is \frac00, undefined, while the right side is \frac43. The two agree everywhere else. Record the excluded values from the original denominators before cancelling: here x\ne2 and x\ne-1.
Worked example
Where this shows up
The parallel pair: 6\ \Omega with 3\ \Omega gives 2\ \Omega. Let R_2 shrink towards 0 and \frac{R_1R_2}{R_1+R_2} goes to 0 — a short circuit, all current through the zero-resistance path. The limiting cases of a formula are where the hardware does something drastic, and they are worth checking every time. In control engineering the same cancellation appears when a factor of a transfer function's numerator matches one of its denominator: on paper they cancel, but the physical behaviour behind that factor has not gone away — it has only stopped showing in the input–output relation.
Narration spine. Show the two resistors and the wrong sum 9\ \Omega struck through. Draw \frac16 and \frac13 as shaded bars, re-cut both into sixths, stack them to \frac36=\frac12 and flip to 2\ \Omega. Then replace 6 and 3 by R_1 and R_2 and rebuild the same picture as algebra.
Unit 07: Factorizing Is Expanding Run Backwards
The problem
A steel washer has outer radius 30 mm and a hole of radius 20 mm. Its face area is \pi(30^2-20^2) mm^2. Separately, work out 53\times47 in your head. Both become easy once a^2-b^2 is seen as a product — but why is a^2-b^2=(a-b)(a+b)?
First attempt
Expand the right side: (a-b)(a+b)=a^2+ab-ab-b^2=a^2-b^2. That confirms the identity, but only after someone handed us the factors. It gives no way to find factors we were not given, and no reason why the two middle terms should cancel.
The picture
Take a square of side a and cut a square of side b from one corner. What is left has area a^2-b^2. Cut it into two rectangles:
Strip I is a wide and a-b high. Strip II is a-b wide and b high. Stand strip II on its end — now b wide and a-b high — and put it beside strip I:
One rectangle, a+b wide and a-b high, and no area gained or lost:
The cancelling middle terms are the move itself: strip II's area b(a-b) was removed from the bottom and added at the side.
Quadratics work the same way, with algebra tiles: one x\times x square, some x\times1 strips and some unit squares. To factorize x^2+5x+6, arrange the one square, five strips and six units into a single rectangle:
The strips must split between the two sides of the square — here 2 below and 3 beside — and then the units exactly fill the corner, 2\times3=6. The rectangle is x+2 by x+3. In general the strips split as a+b and the corner needs ab units:
So factorizing x^2+5x+6 is not guessing; it is two conditions — sum 5, product 6 — read off the picture.
Check it with numbers
Put a=8, b=3 into the washer picture: 64-9=55, and the rectangle is 11\times5=55.
53\times47=(50+3)(50-3)=50^2-3^2=2500-9=2491.
The washer: 30^2-20^2=(30-20)(30+20)=10\times50=500, so the area is 500\pi\approx1570.8 mm^2.
The rule it gives
Factorization is expansion run backwards, and the useful products of two simple factors are the pictures of this unit:
the last being the difference of two squares. Read right to left they factorize: x^2+10x+25=(x+5)^2, 4a^2-12a+9=(2a-3)^2, 25x^2-16y^2=(5x-4y)(5x+4y). A bracket can play the part of a or b:
Common factors. Take out the highest common factor: the HCF of the coefficients, then the lowest power of each letter present in every term. 35x^2y^2-10xy^3=5xy^2(7x-2y).
Common factors by grouping. With four terms, pair them, take a common factor from each pair, then the common bracket:
If the pairs share no bracket, rearrange the terms first.
Quadratics as the product of two factors. With a and b natural numbers, the three sign patterns are
A positive constant term means both signs are the same, set by the sign of the x term; a negative constant term means the signs differ. List the pairs with the right product and pick the one with the right sum or difference: x^2-2x-24 needs ab=24, a-b=-2, so (x+4)(x-6).
Leading coefficient other than 1. The tiles still work, but there are several squares; the arithmetic shortcut is to split the middle term. For px^2+qx+r, find two numbers with product pr and sum q, split qx into those two terms, and group.
Not every quadratic factorizes this way: x^2-2x-1 has no pair of whole numbers with product -1 and sum -2. What to do then is the subject of F.3.
Worked example
Factorize 12x^2-25x+12. The product is 12\times12=144 and the sum -25: the pair is -16 and -9.
Grouping after rearranging:
Where this shows up
The washer's face area, \pi(R^2-r^2)=\pi(R-r)(R+r), is the ring's width times its mean circumference \pi(R+r): for R=30 mm and r=20 mm, 10\times50\pi\approx1570.8 mm^2. More broadly, factorizing turns a sum into a product, and a product is zero only when a factor is — which is how the roots of an equation, and with them equilibrium points and natural frequencies, are found in F.3 and F.6.
Narration spine. Cut the small square from the big one, split the L into two strips, stand the lower strip on end and slide it into place beside the upper one: a^2-b^2 becomes (a+b)(a-b). Check with a=8, b=3, and finish on 53\times47=50^2-3^2=2491.