How do you factor #80x^2+68x+12#?

Given:

#80x^2+68x+12#

The difference of squares identity can be written:

#a^2-b^2 = (a-b)(a+b)#

We use this with #a=(40x+17)# and #b=7#

First let us multiply the whole expression by #20#, complete the square, then divide by #20# at the end:

#20(80x^2+68x+12) = 1600x^2+1360x+240#

#color(white)(20(80x^2+68x+12)) = (40x)^2+2(40x)(17)+17^2-49#

#color(white)(20(80x^2+68x+12)) = (40x+17)^2-7^2#

#color(white)(20(80x^2+68x+12)) = ((40x+17)-7)((40x+17)+7)#

#color(white)(20(80x^2+68x+12)) = (40x+10)(40x+24)#

#color(white)(20(80x^2+68x+12)) = 10(4x+1)(8)(5x+3)#

#color(white)(20(80x^2+68x+12)) = 20*4(4x+1)(5x+3)#

Then dividing both ends by #20# we get:

#80x^2+68x+12 = 4(4x+1)(5x+3)#

#color(white)()#
Why did I choose to multiply by #20#?

I wanted a multiplier that would make the leading term into a perfect square and give the middle term enough powers of #2# so that we would not end up with fractions when divided by the square root of resulting leading term and by #2#.

The multiplier had to be a multiple of #5#, to match the existing factor #5# in the given coefficient of #x^2#. The additional #xx 2^2# helps with the middle coefficient, which would otherwise be #2(20x)(17/2)#. We could have just used #5# as a multiplier, but our arithmetic would involve fractions.