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)#