If #4x^3-12x^2-37x-15 = 0# has an integer root then it has a corresponding factor of the form #(x+a)# with #a# a divisor of the constant term #-15#. That gives possibilities of #+-1#, #+-3#, #+-5# or #+-15#. If we try #x=5# we find:
#4x^3-12x^2-37x-15#
#= (4xx125)-(12xx25)-(37xx5)-15#
#= 500-300-185-15 = 0#
So #x=5# is a root of #4x^3-12x^2-37x-15 = 0# and #(x-5)# must be a factor of #4x^3-12x^2-37x-15#.
Next use synthetic division to find:
#4x^3-12x^2-37x-15 = (x-5)(4x^2+8x+3)#
The remaining quadratic factor #4x^2+8x+3# is of the form #ax^2+bx+c# with #a=4#, #b=8# and #c=3#.
This has discriminant
#Delta = b^2 - 4ac = 8^2 - (4xx4xx3) = 64 - 48 = 16 = 4^2#
...a nice positive square number, so #4x^2+8x+3 = 0# has 2 distinct rational roots, given by the formula
#x = (-b+-sqrt(Delta))/(2a) = (-8+-4)/8 = (-2+-1)/2#
Multiplying both sides by #2# we get
#2x = -2+-1#
Hence #(2x+1)# and #(2x+3)# are factors of #4x^2+8x+3#
Putting this together:
#4x^3-12x^2-37x-15 = (x-5)(2x+1)(2x+3)#
