Solving Quadratics by Factorising
A linear equation has one answer. A quadratic like x² + 5x + 6 = 0 usually has two, and you can't get at them by "doing the same to both sides". Instead you rewrite the left side as two brackets multiplied together, then use the one fact that makes it all work: if two things multiply to give zero, one of them must be zero.
GCSE (Foundation and Higher): factorise quadratic expressions of the form x² + bx + c; solve quadratic equations algebraically by factorising; Higher adds ax² + bx + c1 · The zero-product rule
Try to find two numbers that multiply to make 0. You can't, unless one of them is 0. 7 × 0 = 0, 0 × (−3) = 0, but 2 × 5 is never 0. This is the whole reason factorising solves equations: once you have (something) × (something else) = 0, you know one of the brackets equals 0, and each bracket is a simple linear equation.
So if (x − 2)(x + 5) = 0, then either x − 2 = 0 (so x = 2) or x + 5 = 0 (so x = −5). Two solutions, read straight off the brackets, with the signs flipped. The hard part is getting the brackets in the first place.
2 · Expanding two brackets: where b and c come from
Before factorising, see what expanding does. (x + 2)(x + 3) is a rectangle with sides x + 2 and x + 3; split it into four regions. The x² is the big square, the 6 is the little rectangle, and the two x-strips add up to 5x. So the number term is the two numbers multiplied; the x coefficient is the two numbers added. Factorising is finding the two numbers from their sum and product.
3 · Factorising x² + bx + c
Given x² + 5x + 6, you need two numbers that multiply to 6 and add to 5. List the factor pairs of 6 and check which pair adds to 5. Signs matter and the table below handles them: when c is negative the numbers have different signs; when c is positive they share the sign of b.
| Signs in x² + bx + c | The two numbers are… | Example |
|---|---|---|
| c positive, b positive | both positive | x² + 7x + 10 = (x + 2)(x + 5) |
| c positive, b negative | both negative | x² − 7x + 10 = (x − 2)(x − 5) |
| c negative | one of each; the bigger one takes b's sign | x² + 3x − 10 = (x + 5)(x − 2) |
| c = 0 | take x out as a common factor | x² + 4x = x(x + 4) |
| b = 0, c negative square | difference of two squares: ±√ | x² − 16 = (x + 4)(x − 4) |
Always expand to check. (x + 2)(x + 3): x² + 3x + 2x + 6 = x² + 5x + 6 ✓. It takes ten seconds and catches every sign slip. The expanding panel above is the check for the factorising panel below it.
4 · Solve it: factorise, then zero-product
Put the two ideas together. Make sure the equation equals 0 (rearrange if it doesn't), factorise the left side, set each bracket to 0, solve each, and substitute back. Step through any of these.
It must equal zero first. x² + 3x = 10 cannot be solved by factorising x² + 3x = x(x + 3) and writing x = 10 or x + 3 = 10. The zero-product rule only works for zero: 2 × 5 = 10 and 1 × 10 = 10 and plenty of others. Move everything to one side, get = 0, then factorise.
5 · What the solutions mean on a graph
The solutions of x² + bx + c = 0 are exactly where the graph of y = x² + bx + c crosses the x-axis, because that's where y = 0. Factorised, the brackets tell you the crossing points directly: (x − 2)(x + 5) crosses at 2 and −5. Drag the roots and watch the equation rebuild itself.
6 · When x² has a coefficient (Higher)
2x² + 7x + 3 doesn't start with a plain x², so the brackets start (2x …)(x …). One method that always works: multiply a × c (here 2 × 3 = 6), find two numbers that multiply to 6 and add to 7 (those are 1 and 6), split the middle term 7x into 1x + 6x, then factorise in pairs.
7 · Where quadratics come from
Areas
A rectangle is 3 cm longer than it is wide and has area 40 cm²: x(x + 3) = 40 → x² + 3x − 40 = 0 → (x + 8)(x − 5) = 0 → x = 5 (the −8 is rejected: a width can't be negative).
Projectiles
Height h = 20t − 5t². When does it land? 5t(4 − t) = 0 → t = 0 or 4. Factorising answers "when is it on the ground" in one line.
Number puzzles
"Two consecutive numbers multiply to 56": n(n + 1) = 56 → n² + n − 56 = 0 → (n + 8)(n − 7) = 0 → 7 and 8 (or −8 and −7).
Computing
Game physics solves quadratics every frame to find when a moving object hits a surface; graphics engines solve them to find where a ray meets a sphere. The two roots are "entering" and "leaving".
Business
Profit = revenue − cost often comes out quadratic in price. The roots are the break-even prices; the turning point between them is the price that maximises profit.
Further maths
When factorising fails (x² + x + 1), the quadratic formula and completing the square take over. Both still rely on the zero-product rule underneath.
Worked examples
Solve x² − 7x + 12 = 0
Solve x² + 2x = 15
Solve x² = 9x
Exam-style: a rectangle has length (x + 4) cm and width x cm. Its area is 45 cm². Find x.
Quick check
What you've learned
If a product is zero, one factor is zero: that single fact turns a quadratic into two linear equations. To use it, get the equation equal to zero, then factorise x² + bx + c by finding two numbers that multiply to c and add to b, watching the signs. Set each bracket to zero and solve; the solutions are the sign-flipped bracket numbers, and they are the x-intercepts of the graph. Expand to check the factorising, substitute to check the solutions, and in a context, reject any root that makes no sense.