The factorisation process varies slightly depending on the properties of a, the coefficient of x2.
a = 1
If you multiply two linear factors, (x+m)(x+n) then the product is x2 + (m+n)x + mn so going back the other way, to factor x2 + kx + c, you need to find numbers m and n so that m + n = k and mn = c.
Example To factor x2 + 8x + 12 you need two numbers m and n such that
\[ m+n=8 \ \text{ and }\ mn=12 \quad \]
Now the factors of 12 are 3 and 4, 6 and 2 or 12 and 1. Only one pair: 6 and 2, also satisfy the requirement that their sum is 8, and \[ x^2+8x+12 = (x+6)(x+2) \ \]
a has only one pair of factors
A similar approach works if a has only one pair of factors.
Example To factor 5x2 + 2x - 3. Here a = 5 x 1, so you need two numbers m and n such that \[ 5x^2+2x-3 = (5x+m)(x+n) \quad \] which means \[ m+5n=2 \ \text{ and }\ mn=-3 \quad \]
Now the factors of -3 are 3 and -1 or -3 and 1. Only one pair: -3 and 1, also satisfy the requirement that m + 5n = 2, and \[ 5x^2+2x-3 = (5x-3)(x+1)\ \]
Guided Examples
a has more than one pair of factors
This is best explained by an example.
Example To factor 8x2 + 10x - 3. Here a can be factored as 4 x 2 or 8 x 1.
First, form the expression \[ \frac{(8x + m)(8x + n)}{8} \quad \] Call this a preliminary factorisation.
Next, find the relationships between m and n. The constant term in the preliminary factorisation must match the constant term in the unfactorised expression. That is, mn/8 = -3, or mn = -24. Similarly for the coefficient of x, that is, (8mx + 8nx)/8 = 10x, or m + n = 10.
Now solve for m and n. The pair 12, -2 satisfies.
Substitute this into the expression: \[ \frac{(8x + 12)(8x - 2 )}{8} \quad \] Extract factors in the numerator to cancel the denominator. \[ \frac{4(2x + 3)2(4x - 1 )}{8} \quad \] to get \[ 8x^2 + 10x - 3 = (2x + 3)(4x - 1) \quad \]
Factorisations are easiest when the factors of the coefficients are integers. If this is not the case there are other methods.