Factorising quadratic equations

TODO — replace with Arosh’s own write-up. The Maths below is correct; the words are a starting point.

Factorising a quadratic means writing it as two brackets multiplied together. It matters because a product is zero exactly when one of its factors is zero — so once you have the brackets, the solutions fall straight out.

The easy case: $x^2 + bx + c$

When the coefficient of $x^2$ is 1, you are looking for two numbers that multiply to $c$ and add to $b$.

Take $x^2 + 7x + 12$. Which pairs multiply to 12? $1 \times 12$, $2 \times 6$, $3 \times 4$. Of those, $3 + 4 = 7$. So

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

The reason is just the expansion in reverse:

$$(x + p)(x + q) = x^2 + (p + q)x + pq,$$

so $p + q$ has to be $b$ and $pq$ has to be $c$.

Watch the signs. If $c$ is negative the two numbers have opposite signs; if $c$ is positive they share the sign of $b$. For $x^2 - 5x + 6$ you want two negatives: $-2$ and $-3$, giving $(x-2)(x-3)$.

The harder case: $ax^2 + bx + c$

When $a \neq 1$, guessing gets slow. The AC method — also called splitting the middle term — removes the guessing.

Factorise $6x^2 + 11x - 35$.

1. Multiply $a$ by $c$: $6 \times (-35) = -210$.
2. Find two numbers that multiply to $-210$ and add to $b = 11$. They are $21$ and $-10$.
3. Split the middle term using them: $$6x^2 + 21x - 10x - 35.$$
4. Factorise the two halves separately: $$3x(2x + 7) - 5(2x + 7).$$
5. The bracket $(2x+7)$ is now common to both, so take it out: $$(3x - 5)(2x + 7).$$

Expanding to check: $6x^2 + 21x - 10x - 35 = 6x^2 + 11x - 35$. Correct.

Why splitting the middle term always works

It is not a coincidence that the two halves share a bracket. Suppose the quadratic does factorise, as

$$ax^2 + bx + c = (px + q)(rx + s).$$

Expanding the right-hand side gives

$$prx^2 + (ps + qr)x + qs,$$

so $a = pr$, $b = ps + qr$ and $c = qs$.

Now take the two parts of the middle coefficient and multiply them:

$$(ps)(qr) = pqrs = (pr)(qs) = ac.$$

So $ps$ and $qr$ are two numbers whose product is $ac$ and whose sum is $b$ — precisely what step 2 asks you to find. Finding them is not a trick that happens to work; it is finding the two halves of the middle term directly.

When it does not factorise

Not every quadratic splits into brackets with whole numbers. The test is the discriminant:

$$\Delta = b^2 - 4ac.$$

• If $\Delta$ is a perfect square, the quadratic factorises over the integers.
• If $\Delta > 0$ but is not a perfect square, there are two real roots, but they are irrational — use the formula rather than hunting for factors.
• If $\Delta < 0$ there are no real roots at all, and no real factorisation.

For $6x^2 + 11x - 35$: $\Delta = 121 + 840 = 961 = 31^2$. A perfect square, which is why the factors came out whole.

Completing the square

When factorising fails, completing the square always works, and it is where the quadratic formula comes from:

$$ax^2 + bx + c = a\left(x + \frac{b}{2a}\right)^2 + c - \frac{b^2}{4a}.$$

Setting that to zero and rearranging gives

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.$$

Worth deriving once yourself rather than only memorising — the $\Delta$ above is sitting right there under the square root, which is exactly why it decides how many roots there are.