Teacher: one page per learner, printed A4. Sections A, B and C
count toward Written Works; see assessment.md §1. Section A asks
only for a judgement — no factoring — because telling finished from unfinished is the
competency and can be assessed on its own. The answer key at the foot is screen-only.
A Complete or not? — every one of these is already correct
Given
Complete?
If not, finish it
2(x² − 4)
5x(x + 3)
3(x² − 9)
(x − 1)(x + 1)
x(2x − 18)
4(x² + 2x + 3)
B One move each
Factor
Answer
6x² + 15x
x² − 64
x² + 9x + 14
x² − 12x + 36
C Two moves each
Factor completely
Answer
2x² − 72
3x³ − 3x
2x² + 14x + 20
The heading tells you there are two. In an examination it will not.
D Does it factor at all?
x² − 36
x² + 36
x² + 5x + 9
x² + 5x + 4
E True or false?
A sum of two squares can be factored.
The common factor should come out first.
(x + 3)² is completely factored.
Every trinomial factors.
★ The challenge
x⁴ − 16
Factor it completely.
How many separate factoring moves did it take?
One of your brackets cannot be factored any further. Which one, and
why?
Answer key — screen only, not printed. A · no → 2(x − 2)(x + 2) · yes · no → 3(x − 3)(x + 3) · yes ·
no → 2x(x − 9) · yes B · 3x(2x + 5) · (x − 8)(x + 8) · (x + 2)(x + 7) · (x − 6)² C · 2(x − 6)(x + 6) · 3x(x − 1)(x + 1) · 2(x + 2)(x + 5) D · yes · no · no · yes, (x + 1)(x + 4) |
E · false · true · true · false ★ · 1. (x − 2)(x + 2)(x² + 4) · 2. 2 moves · 3. x² + 4 — it is a
sum of two squares, and no two numbers multiply to 4 and add to 0. A row 6 is the hard one. 4(x² + 2x + 3) looks unfinished, and it is not: no two
whole numbers multiply to 3 and add to 2. Learners must be allowed to conclude "it stops
here" — training them that every trinomial factors is worse than teaching nothing. D3 is the same trap in a different coat — x² + 5x + 9 does not factor either.