Reading every factor of a number off its prime factorisation

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What to assume they know

Breaking a number down to primes, and why the result is unique — producing a prime factorisation, and the fact that the primes you get do not depend on where you started splitting
Rearranging a product without changing its value
Reading a product of several primes as a single number, and back again
Class 6 Prime Time: a factor divides exactly

What they should be able to do

Decide whether a stated product of primes divides a stated number, by looking at its factorisation instead of by dividing
Say what the other half of the pair must be when a subpart is pulled out
Explain why a subpart of the factorisation must be a factor
Explain why nothing outside the factorisation can be a factor — and identify where the previous topic's result is doing that work
List every factor of a number by taking its prime factors none at a time, one at a time, two at a time, and so on
Say why 1 has to be put into that list separately
Assess a claim of the pattern-spotting kind by looking for a single example that breaks it
State, with an example, why a larger number need not have a longer factorisation or more factors

Where it usually goes wrong

"Every subpart of the factorisation is a factor — so 3 × 3 × 3 is a factor of 840." It is not a subpart. 840's factorisation holds a single 3, so you cannot cut three of them out of it. The word subpart is doing precise work: you may take a prime only as many times as it is actually there. Make the running-out visible.
"You could still find a factor that is not made of the number's own primes." You could not, and this is exactly where the previous topic pays off: if the factorisation were not forced, a factor built from other primes would be conceivable.
"With four primes there are four ways to choose two of them." 225 = 3 × 3 × 5 × 5 has four prime factors and only three distinct products from a pair, because 3 × 3, 3 × 5 and 5 × 5 exhaust the possibilities and the two 3s are interchangeable. Repeats collapse. Any enumeration that treats the copies as different produces duplicates and an inflated count.
"1 is not really a factor." It is, it belongs in the list, and it is the one entry that is not a positive-length subpart — it is what you get by taking no primes at all. The book adds it explicitly for that reason.
"A larger number must have more factors." 121 against 96 kills this in one line. So does the observation that primes of any size have exactly two.
"A claim that works for the examples I tried is true." It is a conjecture until it is argued for every case. The book prints Anshu's claim precisely because it is plausible and false.

Questions to check understanding

Given a number's prime factorisation, decide whether a stated product is a factor of it, and justify the decision
Given a factor, state the co-factor without dividing
List a number's factors in full, working from its prime factorisation
State how many factors a given number has
Judge a stated claim about numbers and, if it is false, supply one instance that breaks it
Explain why two numbers with the same number of digits can have very different numbers of factors
Items of this shape are the whole of the "Figure it Out" block on p.51 of Part II and recur in the closing exercise block (Part II, p.63, questions 2 and 3 — both sit on p.63; neither runs on to p.64)

Examples worth working on the board

840 (Part II, §3.1, pp.50–51). Its factorisation as the book prints it: 2 × 2 × 2 × 3 × 5 × 7. The regrouping the book performs: 840 = (2 × 2 × 7) × (2 × 3 × 5), read as 840 = 28 × 30. Then three further questions on the same number, each answerable by inspection alone — is 2 × 7 a factor? is 2 × 2 × 2? is 3 × 3 × 3? Two of the three are settled by what the factorisation contains; the third by what it does not.
225 (Part II, §3.1, p.51). Checked against the printed page. Its division ladder is printed in the right margin: 5 | 225, 5 | 45, 3 | 9, 3 | 3, ending at 1 — so 225 = 5 × 5 × 3 × 3. The enumeration the book then runs, in its own order: the primes themselves, 3 and 5; two at a time, 3 × 3, 5 × 5 and 3 × 5; three at a time, 3 × 3 × 5 and 3 × 5 × 5; four at a time, 3 × 3 × 5 × 5. Then 1 is added to the list. The book prints the finished list of nine and asks the reader to check that nothing is missing.
The five numbers to enumerate (Part II, §3.1, "Figure it Out", p.51). 90, 105, 132, 360 and 840. Their factorisations, as inputs: 90 = 2 × 3 × 3 × 5; 105 = 3 × 5 × 7; 132 = 2 × 2 × 3 × 11; 360 = 2 × 2 × 2 × 3 × 3 × 5; 840 = 2 × 2 × 2 × 3 × 5 × 7. The book supplies two of the counts itself — it says 360 has 24 factors and 840 has 32 — and supplies no others. Those two printed counts are worth using as a check on an added enumeration: a movement that lands on a different number for 360 or 840 has a bug.
Anshu's claim (Part II, §3.1, p.51). The claim, paraphrased: the bigger the number, the longer its prime factorisation. The book's two numbers: 96 = 2 × 2 × 2 × 2 × 2 × 3 and 121 = 11 × 11.
The side box on p.52 (Part II, §3.1). Checked against the printed page. A pale blue panel with a small owl-in-a-mortarboard character at its left. Its content is the naming of conjecture and the naming of the killing move as a counterexample. It is set as an aside, not as a step of the argument.

Figures to have open

A row of prime factors that can be bracketed and re-bracketed, and from which subparts can be lifted out. Standard schematic; this is the single most important visual in the topic.
The 225 division ladder, redrawn. Standard schematic — do not lift the book's margin art.
A selection grid for 225: rows for "how many primes taken", cells for each distinct product. Standard schematic; the book prints the same information as running text, not as a table.
No numbered figure exists to cite: this chapter prints no "Fig. 3.x" label anywhere on pp.47–66 (all twenty pages checked).

Where this sits in the book

NCERT Ganita Prakash, Class 7, Part II, printed Chapter 3 "Finding Common Ground", §3.1 "The Greatest of All", "Factors of a Number Using Prime Factorisation", pp.50–51 — the 840 regrouping, the three test questions, and the 225 enumeration
Same section, "Figure it Out", Part II, p.51 — the five numbers to enumerate, Anshu's claim, and the counterexample — none of it runs on to the next page, which opens with the side box
Same section, the side box naming conjecture and counterexample, Part II, p.52
Same chapter, SUMMARY, Part II, p.65, bullet 3
Backward pointer: Breaking a number down to primes, and why the result is unique, whose result this topic depends on
Forward pointer: HCF: take the fewest occurrences of each prime, which applies the same cut to two factorisations at once