If a quadratic with 33 terms; one with a coefficient of xx, for example, x^2, 2x^2, 3x^2x2,2x2,3x2 etc, a value of xx and a constant usually the factorised form includes 22 brackets.
When factorising that has 22 brackets we need 22 numbers that add up to make the second term and the same 22 numbers to multiply to get the second term.
I usually start by listing the factors of the third term which is 1212:
1212 and 11
As 1212 and 11 cannot add or subtract to make 77 this pair does not work.
66 and 22
As 66 and 22 cannot add or subtract to make 77 this pair does not work.
44 and 33
44 and 33 add to make 77 so we can use this.
With all of the signs being positive within x^2+7x+12x2+7x+12, both 44 and 33 have to be both positive.
-> (x+3)(x+4)→(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))(x+3)(x+4)
color(red)(x) xx color(red)(x)=color(lightgreen)(x^2)x×x=x2
color(red)(x) xx color(blue)(4)=color(red)(4x)x×4=4x
color(blue)(3) xx color(red)(x)=color(red)(3x)3×x=3x
color(blue)(3) xx color(blue)(4)=color(blue)(12)3×4=12
-> color(lightgreen)(x^2)color(red)(+4x)color(red)(+3x)color(blue)(+12)→x2+4x+3x+12
->-> color(lightgreen)(x^2)color(red)(+7x)color(blue)(+12)→→x2+7x+12
This is the same as what we started with, therefore, it is correctly factorised.
-> (x+4)(x+3)→(x+4)(x+3)
