Solving quadratic equations by factorising: the zero product rule

If a quadratic equation is written as two brackets multiplied together that equal $0$, you don't need the quadratic formula. The zero product rule solves it in two lines. The usual hurdle is believing that it really is that simple.

When do I use this?

You use the zero product rule when the equation is in factorised form, like $(x - 3)(x + 5) = 0$, or when you've factorised it first. If the equation is written as $ax^2 + bx + c = 0$ instead, you use the quadratic formula.

The zero product rule

The zero product rule says that if we have two things multiplied together that have to give zero, then at least one of the things is equal to zero. Let's take an example straight away. We have

$$(x - 3)(x + 5) = 0$$

Here we have two brackets multiplied together that should give $0$. Because they're multiplied together, one of these brackets must be equal to $0$ (because everything multiplied by $0$ is equal to $0$). So we can write

$$x - 3 = 0$$

or

$$x + 5 = 0$$

We can say that because we know that one of the brackets has to be $0$ if the whole calculation is to give $0$. So we can just work out these two simple equations and find the two solutions. When we do that, we get

$$x = 3$$

or

$$x = -5$$

Why there are two solutions

For the calculation with the two brackets there are two solutions, because it's actually a hidden quadratic equation. If we tried multiplying the brackets together, we'd see that we get an $x^2$:

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

So we can see that the example is a quadratic equation in factorised form, and that the solutions are the roots of the quadratic function. That's also why the zero product rule works so well together with the factorised form: the roots are sitting in the brackets, and the rule pulls them out.

Another example

Say the factorised quadratic has a number in front and a plus in one of the brackets:

$$3(x + 6)(x - 2) = 0$$

Here three things are multiplied together: $3$, $(x + 6)$ and $(x - 2)$. The $3$ is never $0$, so one of the brackets has to be. That gives us

$$x + 6 = 0 \quad \text{or} \quad x - 2 = 0$$

and so

$$x = -6 \quad \text{or} \quad x = 2$$

Notice that the solutions have the opposite sign of what's in the brackets, exactly as when we read roots off the factorised form.

Common mistakes

• "$x - 3 = 0$ gives $x = -3$." Solve the small equation: $x - 3 = 0$ means $x = 3$. The solution has the opposite sign of the number in the bracket.
• "There's only one solution." Two brackets give two small equations and two solutions. It's a quadratic equation in disguise, so two solutions is exactly what we expect.
• "The brackets have to equal each other." No. They're multiplied together to give $0$, so each bracket on its own is set equal to $0$.

Related

Where the brackets come from, and how the roots hide inside them, is in factorising quadratics. If the equation isn't in brackets, the quadratic formula does the job. The bigger picture, the parabola and its roots, is in Quadratic functions and the parabola.