Factorising Quadratics

The height of a kicked football, the area of a rectangular garden, the point where a business stops losing money and starts making it — a huge range of real problems boil down to a quadratic, and cracking one usually starts with factorising it into brackets.

Expanding brackets is easy — almost boring. Hand anyone (x + 3)(x + 4), and as long as they follow the recipe from multiplying polynomials, out comes x^2 + 7x + 12, every time, no thinking required.

Factorising is the same journey run backwards — and that makes it detective work. Someone hands you the finished product, x^2 + 7x + 12, and asks: which two brackets were multiplied to make this? The evidence has been scrambled together — the 7x is really two terms merged into one — and your job is to un-scramble it:

x^2 + 7x + 12 = (\,x + \boxed{?}\,)(\,x + \boxed{?}\,)

Happily, the culprits always leave the same two fingerprints. In this example, the numbers hiding in the brackets multiply to 12 and add to 7 — and once you know to look for that pair of clues, every quadratic of this kind cracks the same way.

This lesson covers the cleanest case: a quadratic x^2 + bx + c whose x^2 has coefficient 1. (If every term shares a number first — like 2x^2 + 14x + 24 — pull out the common factor before you start; here that leaves 2(x^2 + 7x + 12), and the bracket is exactly the case we're about to crack.)

The two-number trick

Why do the clues work? Expand a general pair of brackets (x + p)(x + q) and collect like terms:

(x + p)(x + q) = x^2 + px + qx + pq = x^2 + (p + q)\,x + pq

Look at where p and q end up. Their sum becomes the number in front of x, and their product becomes the constant on the end. Expanding hides the two numbers — but it hides them in a completely predictable place.

For any numbers p and q:

So to factorise x^2 + bx + c, run the identity backwards: find two numbers that multiply to c and add to b. That single sentence is the whole method — everything below is just practising the hunt in different disguises.

Worked example: the full hunt

Factorise x^2 + 7x + 12. We need two numbers that multiply to 12 and add to 7. Don't guess at random — be systematic. The product clue is the stronger one, because 12 has only a few factor pairs. List them all, then check each pair's sum:

\begin{array}{ccc} \textbf{pair} & \textbf{product} & \textbf{sum} \\ 1,\ 12 & 12 & 13 \\ 2,\ 6 & 12 & 8 \\ 3,\ 4 & 12 & \boxed{7} \end{array}

Only 3 and 4 pass both tests, so p = 3, q = 4:

x^2 + 7x + 12 = (x + 3)(x + 4)

Now check — always. Factorising has a built-in answer key: expand your brackets back out. Here (x + 3)(x + 4) = x^2 + 4x + 3x + 12 = x^2 + 7x + 12 — exactly what we started with, so the factorisation is right. Ten seconds of expanding catches nearly every slip, especially the sign slips coming up next. Make it a reflex.

(The order of the brackets doesn't matter, by the way: (x + 4)(x + 3) is the same answer, just as 4 \times 3 = 3 \times 4.)

See it as an area model

Here's what factorising really does, drawn as a picture. Think of each term of x^2 + 7x + 12 as a piece of area: the x^2 is a square of side x; the 7x splits into two strips (a 3x strip and a 4x strip — that split is the two-number trick, since 3 + 4 = 7); and the 12 is a 3 \times 4 block of ones.

Slide the four pieces together and they tile one perfect rectangle, (x + 3) wide and (x + 4) tall. A factorisation is literally a rectangle whose side lengths are the two brackets. And notice why the corner closes up: the gap left between the strips is exactly 3 \times 4 = 12 — the product clue. If the constant term were, say, 11, the block wouldn't fit and no rectangle (no factorisation with whole numbers) would exist. Step through it below.

The area picture above isn't a modern teaching gimmick — it's how quadratics were originally done. Around 820 CE in Baghdad, the mathematician Muhammad ibn Musa al-Khwarizmi wrote a book on solving equations whose method was entirely geometric: he drew an actual square of side x, glued rectangular strips onto it for the x-terms, and then worked out what block of units was needed to complete a perfect shape — squares and rectangles, no minus signs, no symbols, just areas.

His book's title contained the word al-jabr ("restoring" — moving a term across an equation to balance it), which became our word algebra. And his own name, Latinised, became algorithm. One author, two of the biggest words in mathematics — and the diagram you just stepped through is essentially page one of his book.

When c is negative: the signs hunt

Factorise x^2 + 2x - 15. Same trick: two numbers that multiply to -15 and add to +2. But a negative product is a big clue in itself: two numbers only multiply to a negative when they have opposite signs — one positive, one negative. So one bracket will hold a + and the other a -.

And when the signs are opposite, adding the numbers really means taking their difference. So hunt through the factor pairs of 15 (ignoring signs for a moment) and look for a pair whose difference is 2:

\begin{array}{ccc} \textbf{pair} & \textbf{product} & \textbf{difference} \\ 1,\ 15 & 15 & 14 \\ 3,\ 5 & 15 & \boxed{2} \end{array}

The pair is 3 and 5. Which one goes negative? The sum we need is +2 — positive — so the bigger number keeps the plus: +5 and -3 (check: 5 \times (-3) = -15, 5 + (-3) = 2). So:

x^2 + 2x - 15 = (x + 5)(x - 3)

Check by expanding: (x + 5)(x - 3) = x^2 - 3x + 5x - 15 = x^2 + 2x - 15. ✓ Had the middle term been -2x instead, the same pair would flip: (x - 5)(x + 3) — the bigger number takes the sign of b.

When b is negative but c is positive

Factorise x^2 - 8x + 15. Two numbers that multiply to +15 and add to -8. Read the clues in order. A positive product means the two numbers have the same sign — both positive or both negative. A negative sum then settles it: two positives can't add to a negative, so both numbers are negative.

Now the hunt is the familiar one — factor pairs of 15 summing to 8 — with minus signs stuck on at the end: 1 + 15 = 16 (no), 3 + 5 = 8 (yes). So the numbers are -3 and -5:

x^2 - 8x + 15 = (x - 3)(x - 5)

Check: (x - 3)(x - 5) = x^2 - 5x - 3x + 15 = x^2 - 8x + 15. ✓ Notice the two minus signs multiplied together to give the plus 15 — which is exactly why a positive c can hide either two pluses or two minuses, and only b tells you which.

Putting the three cases side by side, the constant term is your first witness and the middle term your second:

\begin{array}{ccl} \textbf{c} & \textbf{b} & \textbf{signs in the brackets} \\ + & + & (+)(+) \quad\text{e.g. } x^2 + 7x + 12 = (x+3)(x+4) \\ + & - & (-)(-) \quad\text{e.g. } x^2 - 8x + 15 = (x-3)(x-5) \\ - & \pm & (+)(-) \quad\text{e.g. } x^2 + 2x - 15 = (x+5)(x-3) \end{array}

Three traps catch almost everyone at first:

A quadratic isn't just a string of symbols — its graph is a parabola, a U-shaped curve. Ask "where does y = x^2 + 7x + 12 cross the x-axis?" and the expanded form keeps mum. But the factored form answers instantly: the curve touches the axis exactly where a bracket hits zero, because a product is zero only when one of its factors is. (x + 3)(x + 4) vanishes at x = -3 and x = -4 — read straight off the brackets, no plotting required.

That's why factorising earns its keep: it is the x-ray that shows a parabola's skeleton — its roots — through the skin of the expanded form. Engineers finding where a projectile lands, economists finding a break-even point, game programmers finding when two objects collide: all of them are asking "where is this quadratic zero?", and the factored form is usually the fastest honest answer.

See it explained