If a quadratic expression in the standard form contains an a coefficient which is not 1 (eq. 1), it may be difficult to factor via the usual process, as you are not going to know which bracket needs to contain the 2x and which contains the x (or for more complicated decompositions, which combination of factors you need to get ax2). Instead, it is better to factor by decomposition.

Eq. 1: \(y=2x^2+10x+12\)

The first step of factoring by decomposition is to multiply the a and c coefficients together (in this case 2 and 12) to give you another number (24). You then need to find a pair of factors of this number which add together to give the b value. You can do this by listing all of the factors, along with their sums (Table 1).

Table 1: Searching for a pair of numbers where the product of the numbers is equal to the product of the a and c values of the quadratic expression in the standard form, while the sum of the values is equal to the b value of the quadratic expression in the standard form. Finding factors of 24, in both the positive and negative helps to exhaustively identify the pair of numbers, which in this case is 4 and 6.

Once you have a pair of numbers which the sum of is the b coefficient in the expression (in this case 4 and 6), you can decompose the bx term into two terms in your equation (eq. 2). This equation means exactly the same as the original equation, however rather than combining like terms, you have split the combined like terms into separate like terms.

Eq. 2: \(y=2x^2+4x+6x+12\)

Once you have decomposed the bx expression, you can place the first two and last two terms into separate brackets (eq. 3).

Eq. 3: \(y=(2x^2+4x)+(6x+12)\)

Next you need to find the greatest common factor for both brackets. These values are 2x and 6. Next, divide the contents of both of these brackets by the greatest common factors of each (eq. 4) to give equivalent expressions.

Eq. 4: \(y=2x(x+2)+6(x+2)\)

If your equation can be factored by decomposition you will notice something. That is, that the contents of both brackets are identical! If this is the case, you can rearrange the equation into a two brackets, with the greatest common factors in one bracket and the identical contents of the brackets in the other bracket (eq. 5).

Eq. 5: \(y=(2x+6)(x+2)\)

This completes factoring by decomposition, to give you a quadratic equation in factored form which .