You already know how to factorise a quadratic when the x^2 coefficient is 1. But what about 2x^2 + 7x + 3, where the leading coefficient a is not 1? The same idea works, with one extra step. A quadratic in the form

is factorised by the AC method: find two numbers that multiply to a \cdot c and add to b, use them to split the middle term, then factor by grouping.

The two numbers

For 2x^2 + 7x + 3 we have a = 2, b = 7 and c = 3, so a \cdot c = 2 \times 3 = 6. We need two numbers that multiply to 6 and add to 7 — that is 1 and 6:

(Notice that when a = 1 this is exactly the rule you already use — multiply to c, add to b — because a \cdot c = c.)

Splitting and grouping

Use the pair to rewrite the middle term 7x as 1x + 6x, giving four terms. Then group them into two pairs and pull a common factor out of each — the same bracket pops out of both, and that's the factorisation. Step through it below.

The working, line by line:

You can always check the result by multiplying it back out: (2x + 1)(x + 3) = 2x^2 + 6x + x + 3 = 2x^2 + 7x + 3. ✓

See it explained

Khan Academy works through factoring a harder quadratic by grouping: