How do you factor #24x^2-76x+40#?

First, let's simplify this dividing the equation by #4#.

#6x^2-19x+10#

Then, we have to find its roots: #x_1=2/3# and #x_2=5/2#. But in order not to work with these fractions, we can multiply both sides by the denominator of the fractions, leaving only rational numbers:

#3x_1=2/cancel(3)*cancel(3)# which means #3x_1-2=0#
#2x_1=5/cancel(2)*cancel(2)# which means #2x_2-5=0#

Now, let's just factor these.

#(3x-2)(2x-5)#

Done.

#24x^2-76x+40 = 4*6x^2-4*19x+4*10#

#= 4*(6x^2-19x+10)#

From the rational root theorem, if #p/q# is a root of #6x^2-19x+10 = 0# written in lowest terms, then #p# must be a divisor of the constant term #10# and #q# must be a divisor of the coefficient (#6#) of the highest order term.

Notice that #6# can be factored as #2*3# and #10# can be factored as #2*5#.

Let us try #(2x-5)(3x-2)#, it seems to work...

#(2x-5)(3x-2) = 2x*3x-5*3x-2x*2-5*(-2)#

#= 6x^2 - (15+4)x + 10#

#= 6x^2 - 19x + 10#

So in summary, #24x^2-76x+40 = 4*(2x-5)(3x-1)#

#24x^2-76x+40 = 4(6x^2-19x+10)#.

To factor #6x^2-19x+10# multiply #6xx10 = 60#.

We need two numbers that multiply to give us #60# and add to get us #-19# (the number in the middle).

A little thought should convince us that both numbers must be negative. We proceed in order:

#-1xx-60# do not add up to #-19#

#-2xx-30# do not add up to #-19#

#-3xx-10# do not add up to #-19#

#-4xx-15# STOP! These do add up to #-19#.

We'll split the #-19x# into #-4x-15x# (or #-15x-4x# either will work)

#6x^2-19x+10 = 6x^2-4x-15x+10#.

Now factor by grouping:

#6x^2-19x+10 = (6x^2-4x)+(-15x+10)#.

#6x^2-19x+10 = 2x(3x-3)+(-5)(3x-2)# .

#6x^2-19x+10 = (2x-5)(3x-2)#.

So we finish with:

#24x^2-76x+40 = 4(6x^2-19x+10)#

#color(white)"ssssssssssssssss"# #= 4(2x-5)(3x-2)#