#(x-3)(x-4)#
How?
(You don't need to write all of this, but it is the reason we get the answer we get.)
#(ax+b)(cx+d)=(ab)x^2+(ad)x+(bc)x+(bd)=(ab)x^2+(ad+bc)x+(bd)#
This expression has : #1x^2+(-7)x+(12)#, so we want #ac=1#.
With whole numbers that means we have #a=1# and #b=1#. That makes this easier (or less difficult).
#(x+b)(x+d)=x^2+(d+b)x+bd#
Now we need #bd=12#. With whole numbers let's look at the possibilities:
#1*12=12#
#2*6=12#
#3*4=12#
#4*# no, stop. We already see #4# on the right, so we can stop
Thinking a bit more, we want #bd=+12#, so we need either both #b# and #d# are positive or both are negative.
Now look at the middle term #-7x#.
The #-7# needs to be #d+b#. We already knew that either both are positive or both are negative, but if both are positive, they would add up to a positive, so we need both to be negative and they have to add up to #-7#. Looking at the list of possibiities we see #3*4#. If we make both negative, they multiply to give us #+12# and the add up to #-7# That's what we wanted, so let's make sure it works:
#(x-3)(x-4)=x^2-4x-3x+12=x^2-7x+12# and that's what we wanted.
Notes:
Explaining it takes a lot longer than doing it.
The more practice you get, the faster and easier it gets to factor.
For more complicated problems, there are other methods you can learn.
