How do you find the roots of 12x^3+31x^2-17x-6=0?

Use the rational roots theorem then factor to find:
##x = 2/3 ## or ## x = -1/4 ## or ## x = -3##
##f(x) = 12x^3+31x^2-17x-6##
By the rational roots theorem any rational of ##f(x)## are expressible in the form ##p/q## for integers ##p q## with ##p## a divisor of the constant term ##-6## and ##q## a divisor of the coefficient ##12## of the leading term.
That means that the only possible rational zeros are:
##+-1/12 +-1/6 +-1/4 +-1/3 +-1/2 +-2/3 +-3/4 +-1 +-3/2 +-2 +-3 +-6##
Notice that the signs of the coefficients of ##f(x)## are in the pattern ##+ + – -##. By Descartes’ Rule of Signs since there is one change of sign this cubic has exactly one positive Real zero.
So let’s look among the positive possibilities:
##f(1) = 12+31-17-6 = 20##
##f(1/2) = 12(1/8)+31(1/4)-17(1/2)-6 = (6+31-34-24)/4 = -21/4##
##f(2/3) = 12(8/27)+31(4/9)-17(2/3)-6 = (32+124-102-54)/9 = 0##
So ##x=2/3## is a zero and ##(3x-2)## a factor:
##12x^3+31x^2-17x-6 = (3x-2)(4x^2+13x+3)##
To factor the remaining quadratic use an AC method:
Look for a pair of factors of ##AC=4*3=12## with sum ##B=13##.
The pair ##12 1## works.
Use this pair to split the middle term and factor by grouping:
##4x^2+13x+3 = (4x^2+12x)+(x+3)##
##color(white)(4x^2+13x+3) = 4x(x+3)+1(x+3)##
##color(white)(4x^2+13x+3) = (4x+1)(x+3)##
So the other two zeros are:
##x = -1/4 ## and ## x = -3##