IB Maths AA HL Topic 2 — Functions Paper 1 & 2 ~7 min read

Factorising Quadratics

Factorising means rewriting ax2 + bx + c as a product of brackets — usually two of them. Why bother? Because once it’s factorised, the roots fall out for free: each bracket gives you one. There are three common cases you need to be fluent with — monic (leading coefficient 1), non-monic (leading coefficient bigger), and the special difference of two squares. Let me show you the practical move for each.

📘 What you need to know

Why factorise at all?

Two reasons:

Find the roots
(xp)(xq) = 0  ⟹  x = p or q
each bracket gives an x-intercept of the graph
Simplify expressions
cancel common brackets in fractions
essential later for rational functions and limits

Monic case — x2 + bx + c

“Monic” just means the coefficient of x2 is 1. The trick is short:

Monic factorisation find p, q  with  p + q = b  and  pq = c
⟹  x2 + bx + c = (x + p)(x + q)
Sign rules:  if c > 0, both p and q have the same sign (matching b).  If c < 0, they have opposite signs.

Non-monic case — ax2 + bx + c

When a is not 1, the rule changes slightly: the product is ac instead of just c.

🧭 Recipe — non-monic by splitting the middle

  1. Find m, n with m + n = b  and  mn = ac.
  2. Split the middle term bx as mx + nx.
  3. Group the four terms in pairs and factor each pair.
  4. Take out the common bracket — that’s your factorisation.
If a, b, c share a common factor, pull it out first. The remaining quadratic is smaller and easier — you can often spot the factors by inspection from there.

Difference of two squares

Spot this pattern when the quadratic has no middle term and the constant is negative:

Difference of two squares a2x2c2 = (axc)(ax + c)

Square-root each part, then write the sum-and-difference brackets. For example, 9x2 − 16 = (3x − 4)(3x + 4). No middle term needed.

The GDC shortcut

On Paper 2, this is the fastest method. Solve the quadratic on your calculator to get the roots, then write the factors:

If your GDC gives roots x = p and x = q for ax2 + bx + c = 0, then it factorises as a(xp)(xq). Don’t forget the a in front when a ≠ 1.

For fractional roots, clear the denominators. If x = 32 is a root, the bracket is (2x − 3). If x = −15 is a root, the bracket is (5x + 1).

Worked examples

WE 1

Factorise a monic quadratic — both terms positive

Factorise x2 + 9x + 20.

Find p, q with p + q = 9 and pq = 20 try pairs of 20: 1×20, 2×10, 4×5 4 + 5 = 9 ✓ and 4 × 5 = 20 ✓ (x + 4)(x + 5) check: x² + 5x + 4x + 20 = x² + 9x + 20 ✓
WE 2

Factorise a monic quadratic with a negative constant

Factorise x2 − 5x − 14.

Find p, q with p + q = −5 and pq = −14 since pq is negative → opposite signs try 1 & 14, 2 & 7 −7 + 2 = −5 ✓ and −7 × 2 = −14 ✓ (x − 7)(x + 2) check: x² + 2x − 7x − 14 = x² − 5x − 14 ✓
WE 3

Factorise a non-monic quadratic by splitting the middle

Factorise 6x2 − 7x − 5.

Step 1: Find m, n with m + n = −7 and mn = ac = 6 × (−5) = −30 opposite signs since mn < 0 −10 + 3 = −7 ✓ and −10 × 3 = −30 ✓ Step 2: Split the middle: −7x = −10x + 3x 6x² − 10x + 3x − 5 Step 3: Group and factor each pair 2x(3x − 5) + 1(3x − 5) Step 4: Take out the common bracket (3x − 5) (3x − 5)(2x + 1) (3x − 5)(2x + 1) check: 6x² + 3x − 10x − 5 = 6x² − 7x − 5 ✓
WE 4

Factor out a common factor first

Factorise fully: 4x2 + 8x − 60.

Step 1: All terms divisible by 4 — pull it out 4(x² + 2x − 15) Step 2: Factorise the monic quadratic inside need p + q = 2 and pq = −15 5 + (−3) = 2 ✓ and 5 × (−3) = −15 ✓ 4(x + 5)(x − 3) always pull the common factor out first — turns a non-monic into a monic for free
WE 5

Difference of two squares

Factorise: (a) 25x2 − 49,  (b) 8x2 − 50.

(a) No middle term, minus sign — DOTS pattern √(25x²) = 5x and √49 = 7 (5x − 7)(5x + 7) (b) Pull out common factor 2 first 8x² − 50 = 2(4x² − 25) now DOTS: √(4x²) = 2x, √25 = 5 2(2x − 5)(2x + 5) always check for a common factor — without it, 8x² − 50 doesn’t fit DOTS cleanly
WE 6

Use GDC roots to factorise

Using your GDC, the equation 10x2 + 11x − 6 = 0 has roots x = 25 and x = −32. Hence factorise 10x2 + 11x − 6.

Step 1: Convert each root into a bracket — clear denominators x = 2/5 → 5x = 2 → bracket (5x − 2) x = −3/2 → 2x = −3 → bracket (2x + 3) Step 2: Multiply the brackets and check the leading coefficient matches (5x − 2)(2x + 3) = 10x² + 15x − 4x − 6 = 10x² + 11x − 6 ✓ 10x² + 11x − 6 = (5x − 2)(2x + 3) leading coefficients of the brackets multiply to give a — that’s why clearing denominators automatically gives the right answer

💡 Top tips

⚠ Common mistakes

Factorising is the fastest way to find roots — but only when the quadratic factors nicely with integers. When the numbers get ugly, you need a different tool. The next note covers completing the square, which always works regardless of whether the quadratic factorises cleanly, and gives you the vertex as a bonus.

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