If a quadratic with 3 terms; one with a coefficient of x, for example, x^2, 2x^2, 3x^2 etc, a value of x and a constant usually the factorised form includes 2 brackets.
When factorising that has 2 brackets we need 2 numbers that add up to make the second term and the same 2 numbers to multiply to get the second term.
I usually start by listing the factors of the third term which is 12:
12 and 1
As 12 and 1 cannot add or subtract to make 7 this pair does not work.
6 and 2
As 6 and 2 cannot add or subtract to make 7 this pair does not work.
4 and 3
4 and 3 add to make 7 so we can use this.
With all of the signs being positive within x^2+7x+12, both 4 and 3 have to be both positive.
-> (x+3)(x+4)
Remember you can always expand each term to check:
(color(red)(x)color(blue)(+3))(color(red)(x)color(blue)(+4))
color(red)(x) xx color(red)(x)=color(lightgreen)(x^2)
color(red)(x) xx color(blue)(4)=color(red)(4x)
color(blue)(3) xx color(red)(x)=color(red)(3x)
color(blue)(3) xx color(blue)(4)=color(blue)(12)
-> color(lightgreen)(x^2)color(red)(+4x)color(red)(+3x)color(blue)(+12)
->-> color(lightgreen)(x^2)color(red)(+7x)color(blue)(+12)
This is the same as what we started with, therefore, it is correctly factorised.
-> (x+4)(x+3)
