Solving by factoring
What you'll learn
Use the zero-product rule to solve quadratics — and see the solutions as the points where the parabola touches the x-axis.
A quadratic equation sets a degree-2 polynomial to zero: x² + 2x − 3 = 0. Unlike the linear equations of module 1, you can't just undo your way to x — the x² and the x can't be merged. The way in is a fact so simple it looks like a trick.
The zero-product rule
If a product of numbers is zero, at least one factor is zero:
No other number does this — a product equal to 12 tells you almost nothing about its factors (1 × 12? 3 × 4? −2 × −6?), but a product equal to zero pins a factor down. This is why we factor: it converts one hard quadratic into two easy linear equations.
The method
Solve x² + 2x − 3 = 0.
| Step | Work |
|---|---|
| Everything on one side, zero on the other | x² + 2x − 3 = 0 ✓ |
| Factor (sum 2, product −3) | (x + 3)(x − 1) = 0 |
| Set each factor to zero | x + 3 = 0 or x − 1 = 0 |
| Solve both | x = −3 or x = 1 |
Two solutions — quadratics generally have two, one per factor.
The picture
Graph y = (x + 3)(x − 1) and the solutions become visible: they're the points where the parabola crosses the x-axis, because that's where y = 0. Solving a quadratic and finding a parabola's x-intercepts are the same act.
The non-negotiable zero
The rule needs an actual 0 on one side. Faced with x² + 2x = 3, moving the 3 over comes first — factoring x(x + 2) = 3 and trying "x = 3 or x + 2 = 3" produces nonsense, because a product of 3 doesn't constrain its factors. Zero on one side, always, before you factor.
Why this matters
This is the fastest way to solve any quadratic that factors — but not all of them do. The final two lessons build the tools for the rest: completing the square, and the formula it gives birth to.
Check your understanding
Question 1 of 2
Solve: x² − 5x + 6 = 0