Grouping: 2 by 2

Step ① Split into two pairs

Step ② Factor each pair — aim for a common bracket

Step ③ Pull out the shared bracket

Example (a):

$3ac - 9c - 4ab + 12b$

• $= 3c(a - 3) - 4b(a - 3)$
• $= (a - 3)(3c - 4b)$

Example (b):

$(1 - 2a)5b + (2a - 1)$

• $= (1-2a)5b - 1(1-2a)$
• $= (1 - 2a)(5b - 1)$

Watch out: If brackets differ only by sign, e.g. $(2a-1)$ vs $(1-2a)$, flip using $-(1-2a) = (2a-1)$ so both pairs share the same bracket.

Question. Factorise $50x^3y - 18xy^3$ completely.

Solution.

Step 1 — Pull out the GCF. Coefficients $50, 18$ have HCF $= 2$. Variable powers: $x$ appears as $x^3$ and $x^1$ → take $x$. $y$ appears as $y^1$ and $y^3$ → take $y$.

GCF $= 2xy$:

Step 2 — Recognise the leftover. $25x^2 - 9y^2 = (5x)^2 - (3y)^2$ — a difference of squares.

Step 3 — Apply $a^2 - b^2 = (a-b)(a+b)$:

Answer: $2xy(5x - 3y)(5x + 3y)$.

Lesson: if you'd jumped straight to DOS without pulling out $2xy$, the numbers $50$ and $18$ aren't perfect squares — you'd be stuck.

Question. Given $6x^2 - xy = 7y^2$ with $x, y > 0$, find $\dfrac{2x}{y}$.

Solution — factorise then read off a ratio.

Rearrange:

Treat as a quadratic in $x$. Try cross method with $a = 6, c = -7$:

Diagonal products: $6(1) + 1(-7) = 6 - 7 = -1$ ✓ (matches the $-1$ coefficient of $xy$).

So:

Thus $6x = 7y$ or $x = -y$. Since $x, y > 0$, take $6x = 7y$, i.e. $x = \dfrac{7y}{6}$.

Answer: $\dfrac{2x}{y} = \dfrac{7}{3}$.

Lesson: factorising can collapse an equation in two variables into a single linear relation — perfect for finding ratios.

Question. Factorise $p^4 - 16q^4$ completely.

Solution.

Step 1 — First DOS. Treat as $(p^2)^2 - (4q^2)^2$:

Step 2 — Check each factor.

• $p^2 - 4q^2 = p^2 - (2q)^2$ → another DOS!
• $p^2 + 4q^2$ → sum of squares, does not factorise over the reals.

Step 3 — Apply DOS again:

Final answer:

Lesson: every time you get a new pair of factors, ask "is one of these still a DOS?" Don't stop early.

Question. Factorise $4xy^2 - z - x + 4zy^2$ completely.

Solution — rearrange first so pairs become obvious.

Step 1. Rearrange to group the $y^2$-terms together and the bare terms together:

Step 2 — Factor each pair:

• Pair 1: $4xy^2 + 4zy^2 = 4y^2(x + z)$
• Pair 2: $-x - z = -(x + z)$

Both pairs share $(x + z)$:

Step 3 — Always factorise completely. $4y^2 - 1 = (2y)^2 - 1^2$ is a DOS:

Answer: $(2y - 1)(2y + 1)(x + z)$.

Lesson: rearranging the four terms before grouping is half the battle. After grouping, always re-check each factor.

Question. Factorise $c^2 - 9a^2 + 30ab - 25b^2$ completely.

Solution.

Step 1 — Spot the 3-term perfect square. $-9a^2 + 30ab - 25b^2$ has 3 terms but all with awkward signs. Factor out $-1$ first:

Step 2 — Factorise the trinomial. $9a^2 - 30ab + 25b^2 = (3a - 5b)^2$ (check: $2 \cdot 3a \cdot 5b = 30ab$ ✓).

Step 3 — Apply DOS with $a' = c$, $b' = (3a - 5b)$:

Answer: $(c - 3a + 5b)(c + 3a - 5b)$.

Lesson: the minus sign trap — when the 3-term group has all negative signs, factor out $-1$ first so the trinomial is in standard form. The signs inside the bracket then flip.

Question. Factorise $2x^3 - 7x^2 - 18x + 63$ completely.

Solution — group 2-by-2.

Step 1. Group:

Step 2 — Factor each pair:

• $2x^3 - 7x^2 = x^2(2x - 7)$
• $-18x + 63 = -9(2x - 7)$

Both share $(2x - 7)$:

Step 3 — Always check each factor. $x^2 - 9 = x^2 - 3^2$ — DOS:

Answer: $(2x - 7)(x - 3)(x + 3)$.

Lesson: a cubic with 4 terms often hides a 2-by-2 grouping. After grouping, the leftover quadratic is frequently a DOS in disguise.

Question. Simplify $\dfrac{9x^2 - 4}{15ax + 3bx - 10a - 2b}$.

Solution.

Numerator. DOS: $9x^2 - 4 = (3x)^2 - 2^2 = (3x - 2)(3x + 2)$.

Denominator. Four terms — try 2-by-2 grouping. Group $a$-terms and $b$-terms:

Combine:

Answer: $\dfrac{3x + 2}{5a + b}$.

Lesson: factorise top and bottom completely first. The shared factor $(3x - 2)$ only appears once you do.

Question. Without a calculator, evaluate $1234^2 - 1244 \times 1224$.

Solution — recognise the algebraic skeleton.

Notice that $1244 = 1234 + 10$ and $1224 = 1234 - 10$.

Let $a = 1234$ and $b = 10$. Then:

The product $(a+b)(a-b) = a^2 - b^2$, so:

Answer: $100$.

Lesson: when a numerical expression looks scary, look for the DOS shape $(a+b)(a-b)$ hiding in the product. Then $a^2 - (a^2 - b^2) = b^2$ collapses everything to one easy square.