Solving Quadratics by Factorising

The zero superpower

Quadratic equations show up whenever a real problem multiplies two changing quantities — sizing a phone screen to a fixed area, finding the price at which a stall breaks even, or working out when a launched drone falls back to the ground. Solving one means finding the values that make it equal zero, and the whole method turns on one surprising fact about zero itself.

Here is a question that sounds useless: two numbers multiply to give 12 — what are they? You can't answer it. It could be 3 \times 4, or 2 \times 6, or 1.5 \times 8, or -0.5 \times -24, or 100 \times 0.12… there are infinitely many boring possibilities, and knowing the product tells you almost nothing about the factors.

Now change one word: two numbers multiply to give zero — what are they? Suddenly you know something rock-solid: at least one of them must be zero. There is no other way. Try to build 0 out of two non-zero numbers and you will fail every time: 0.001 \times 0.001 is tiny, but it is not zero. Zero is the only number with this power — a product hitting zero fingers a culprit.

If A \times B = 0, then A = 0 or B = 0 (possibly both).
Conversely, no product of non-zero numbers can equal 0.
This holds for any number of factors: if A \times B \times C = 0, at least one of A, B, C is 0.

This one fact — also called the null-factor law — is the entire engine of this page. It is why we bother to factorise quadratics at all: a factorised quadratic set equal to zero splits into two easy equations, and a hard-looking problem falls apart in two lines.

The method

To solve x^2 + bx + c = 0, first rewrite the left-hand side as a product of two brackets:

x^2 + bx + c = (x + p)(x + q) = 0

Now the whole expression is two things multiplied together, and that product equals zero. The zero-product property does all the work: one of the brackets must be zero. So:

(x + p)(x + q) = 0 \quad\Longrightarrow\quad x + p = 0 \ \text{ or } \ x + q = 0

Each of those is a simple equation giving one solution: x = -p or x = -q. A quadratic usually has two solutions, called its roots — and both are genuine answers. Neither is "the" answer; the equation is satisfied by each of them separately.

The whole recipe, then, is four moves:

Rearrange so everything is on one side and the other side is 0.
Factorise the quadratic side.
Split: set each factor equal to 0.
Solve each little equation — and check both answers in the original.

Worked example: x^2 + 7x + 12 = 0

We need two numbers that multiply to 12 and add to 7 — that is 3 and 4. So it factorises and then splits into two pieces:

(x + 3)(x + 4) = 0 \quad\Longrightarrow\quad x + 3 = 0 \ \text{ or } \ x + 4 = 0

Solving each piece gives the two roots x = -3 or x = -4. Notice the sign flip: the bracket (x + 3) gives the root -3, not +3 — you're asking what value of x makes this bracket zero? Watch each stage build.

Worked example: x^2 + 3x - 10 = 0

This time c is negative, so the two numbers must multiply to -10 (one positive, one negative) and add to +3. Running through the factor pairs of 10: 5 and -2 multiply to -10 and add to 3. Done:

x^2 + 3x - 10 = (x + 5)(x - 2) = 0

x + 5 = 0 \ \text{ or } \ x - 2 = 0 \quad\Longrightarrow\quad x = -5 \ \text{ or } \ x = 2

Now the professional habit: check both roots in the original equation — it takes seconds and catches nearly every slip.

x = -5: \quad (-5)^2 + 3(-5) - 10 = 25 - 15 - 10 = 0 \ \checkmark

x = 2: \quad 2^2 + 3(2) - 10 = 4 + 6 - 10 = 0 \ \checkmark

Both work. There's a lovely picture hiding here too: plot y = x^2 + 3x - 10 and the roots are exactly the places where the curve crosses the x-axis — where y (the whole quadratic) equals zero.

Rearrange first: x^2 = 5x

Sometimes the equation doesn't arrive with a zero on one side — and the zero-product property is useless until it does. Take x^2 = 5x. The tempting shortcut is to divide both sides by x and get x = 5. Quick, tidy… and wrong — or rather, half-right. The correct start is to get everything on one side:

x^2 - 5x = 0

There's no number pair to hunt for here — both terms share a factor of x, so factor it out:

x(x - 5) = 0 \quad\Longrightarrow\quad x = 0 \ \text{ or } \ x = 5

Two solutions. Check them: 0^2 = 5 \times 0 ✓ and 5^2 = 5 \times 5 ✓. The divide-by-x shortcut silently threw away x = 0, because dividing by x is only legal when x \neq 0 — and x = 0 was one of the answers! Factorising never loses a root; dividing by the unknown can.

The repeated root: (x - 3)^2 = 0

What if the two brackets turn out to be the same? Solving x^2 - 6x + 9 = 0, the numbers multiplying to 9 and adding to -6 are -3 and -3:

(x - 3)(x - 3) = (x - 3)^2 = 0 \quad\Longrightarrow\quad x = 3

Both factors demand the same thing, so there is only one solution, x = 3. We call it a repeated root (or a double root): the root appears twice in the factorisation but names one value. On a graph, this is the case where the parabola doesn't cut through the x-axis at two points — it swoops down and just touches the axis at x = 3, then lifts away again.

From story to solution

The real power move is building the quadratic yourself. Try this: a rectangular lawn is 3 metres longer than it is wide, and its area is 40 \text{ m}^2. How wide is it?

Call the width x. Then the length is x + 3, and area = width × length gives:

x(x + 3) = 40 \quad\Longrightarrow\quad x^2 + 3x - 40 = 0

(Everything to one side first — always.) Two numbers multiplying to -40 and adding to 3: that's 8 and -5.

(x + 8)(x - 5) = 0 \quad\Longrightarrow\quad x = -8 \ \text{ or } \ x = 5

The algebra hands over two roots, but the story gets a veto: a lawn cannot be -8 metres wide, so we reject that root and keep x = 5. The lawn is 5 m wide and 8 m long — and 5 \times 8 = 40 ✓. In word problems, always solve for both roots first, then ask which ones make sense.

A neat special case

When the quadratic is a perfect square subtracted, like x^2 - 9 = 0, there's no middle term to balance at all — the factors are a difference of two squares:

x^2 - 9 = (x - 3)(x + 3) = 0 \quad\Longrightarrow\quad x = 3 \ \text{ or } \ x = -3

A tidy symmetric pair, \pm 3 — the same null-factor law finishes the job. (And notice you could rewrite x^2 - 9 = 0 as x^2 = 9: "what squares to nine?" has two answers, and the factorisation is what makes sure you don't forget the negative one.)

These three traps catch even strong algebra students:

Never divide both sides by x. x^2 = 5x has two solutions, x = 0 and x = 5. Dividing by x assumes x \neq 0 and silently deletes the root x = 0. Factor out the x instead: x(x - 5) = 0.
The equation must equal ZERO before you factorise. Faced with x^2 + 3x - 10 = 2, you may not factorise the left side and write (x+5)(x-2) = 2, "so a bracket is 2". Two numbers can multiply to 2 in infinitely many ways (1 \times 2, 4 \times 0.5, …) — the superpower only works for zero. Subtract the 2 first: x^2 + 3x - 12 = 0, then factorise what you actually have.
Check both roots in the ORIGINAL equation. Not in your rearranged version — if the rearrangement had a slip, checking against it just confirms your own mistake. Substitute each root back into the equation as it was first given; thirty seconds of arithmetic buys certainty.

Kick a ball straight up at 20 m/s and (with school-friendly gravity of 10 \text{ m/s}^2) its height after t seconds is h = 20t - 5t^2. "When does it land?" is exactly the question "when is h = 0?" — a quadratic equation, already begging to be factorised:

20t - 5t^2 = 0 \quad\Longrightarrow\quad 5t(4 - t) = 0 \quad\Longrightarrow\quad t = 0 \ \text{ or } \ t = 4

Both roots mean something! t = 0 is the kick itself (the ball starts on the ground), and t = 4 is the landing, four seconds later. Every thrown ball, every fountain jet, every artillery table since the 1600s is a factorised quadratic asking "where does height equal zero?" — which is why generals hired mathematicians long before schools taught this.

The Greeks, brilliant as they were, refused to treat zero as a number — "how can nothing be something?" — and medieval Europe was so suspicious of the strange new digit arriving from India (via the Arabic world) that Florence banned it from official records in 1299. Zero only won its place slowly, through merchants who found calculation with it irresistibly fast.

The irony is that zero's "spooky" behaviour — it annihilates anything it multiplies — turned out to be algebra's single best tool. Precisely because A \times B = 0 forces a factor to be zero (true of no other number on the right-hand side), every polynomial equation ever solved by factorising leans on the number people once refused to write down. The nothing-number does the heavy lifting.

See it explained

Sal Khan works a quadratic equation from factorised form to its roots.