Why This Matters for Grade 11#
Factorisation is the single most-used skill in Grade 11:
- Quadratic equations: You factorise to solve $x^2 - 5x + 6 = 0$
- Quadratic inequalities: You factorise to find critical values
- Trig identities: $\cos^2\theta - \sin^2\theta = (\cos\theta - \sin\theta)(\cos\theta + \sin\theta)$
- Functions: Finding $x$-intercepts of the parabola means factorising $ax^2 + bx + c = 0$
- Surds: Rationalising $\frac{1}{\sqrt{3} + 1}$ uses the DOTS pattern
If you can’t factorise fluently, Grade 11 will feel impossibly hard.
The Factoring Order (Always Follow This)#
Every time you factorise, work through this checklist in order:
Step 1: Common Factor (ALWAYS check first)#
Take out the highest common factor of ALL terms.
$6x^2 + 9x = 3x(2x + 3)$
$2x^3 - 8x = 2x(x^2 - 4)$ ← then keep going (DOTS inside!)
Step 2: Difference of Two Squares (DOTS)#
$$ a^2 - b^2 = (a - b)(a + b) $$$x^2 - 16 = (x - 4)(x + 4)$
$9a^2 - 25b^2 = (3a - 5b)(3a + 5b)$
$x^4 - 1 = (x^2 - 1)(x^2 + 1) = (x - 1)(x + 1)(x^2 + 1)$
Key: There is NO “sum of two squares” factorisation. $a^2 + b^2$ cannot be factorised over the reals.
Step 3: Trinomials ($ax^2 + bx + c$)#
When $a = 1$: Find two numbers that multiply to $c$ and add to $b$.
$x^2 + 7x + 12 = (x + 3)(x + 4)$ because $3 \times 4 = 12$ and $3 + 4 = 7$
$x^2 - 5x + 6 = (x - 2)(x - 3)$ because $(-2)(-3) = 6$ and $(-2) + (-3) = -5$
When $a \neq 1$: Use the cross-method or the “ac-method”:
$2x^2 + 5x + 3$: Find two numbers that multiply to $2 \times 3 = 6$ and add to $5$ → that’s $2$ and $3$.
$= 2x^2 + 2x + 3x + 3 = 2x(x + 1) + 3(x + 1) = (2x + 3)(x + 1)$
Step 4: Grouping (4 terms)#
Group into two pairs and factor each pair separately.
$x^3 + x^2 + 2x + 2 = x^2(x + 1) + 2(x + 1) = (x^2 + 2)(x + 1)$
Key: After grouping, the brackets MUST be the same. If they’re not, try a different grouping.
Step 5: Sum & Difference of Cubes (Grade 12, but useful to know)#
$a^3 + b^3 = (a + b)(a^2 - ab + b^2)$
$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$
Quick-Check Table#
| Expression looks like… | Method |
|---|---|
| All terms share a factor | Common factor first |
| $a^2 - b^2$ (two perfect squares, minus) | DOTS |
| $x^2 + bx + c$ | Trinomial ($a = 1$) |
| $ax^2 + bx + c$ with $a \neq 1$ | Trinomial (cross/ac-method) |
| 4 terms | Grouping |
| $a^3 \pm b^3$ | Sum/difference of cubes |
🚨 Common Mistakes#
- Forgetting to take out the common factor FIRST: $2x^2 - 8 = 2(x^2 - 4) = 2(x-2)(x+2)$. If you skip the common factor, you’ll miss the DOTS.
- Sign errors in trinomials: For $x^2 - 5x + 6$, both numbers must be negative: $(x-2)(x-3)$, not $(x+2)(x-3)$.
- DOTS needs a MINUS: $x^2 + 9$ cannot be factorised. Only $x^2 - 9 = (x-3)(x+3)$.
- Not factorising completely: $x^4 - 16 = (x^2 - 4)(x^2 + 4) = (x-2)(x+2)(x^2+4)$. Don’t stop at the first step.
- Cancelling terms instead of factors: $\frac{x^2 + 2x}{x} = \frac{x(x+2)}{x} = x + 2$. You CANNOT cancel the $x$ from $\frac{x^2 + 2x}{x}$ without factoring first.