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)#
