How do you use the rational root theorem to find the roots of #3x^4+2x^3-9x^2-12x-4=0#?

#f(x) = 3x^4+2x^3-9x^2-12x-4#

By the rational root theorem any rational roots of #f(x)=0# are expressible in lowest terms as #p/q# for some integers #p# and #q#, where #p# is a divisor of the constant term #4# and #q# is a divisor of the coefficient #3# of the leading term.

So the possible rational roots are:

#+-1/3#, #+-2/3#, #+-1#, #+-4/3#, #+-2#, #+-4#

Try:

#f(1/3) = 1/27+2/27-1-4-4 = 1/9-9 = -80/9#

#f(-1/3) = 1/27-2/27-1+4-4 = -28/27#

#f(2/3) = 16/27+16/27-4-8-4 = 32/27#

#f(-2/3) = 16/27 - 16/27-4+8-4 = 0#

So #x=-2/3# is a root. This 'uses up' one of the possible factors of #2# for #p# and the factor of #3# for #q#. So the only other possible rational roots are:

#+-1#, #+-2#

Try:

#f(1) = 3+2-9-12-4 = -20#

#f(-1) = 3-2-9+12-4 = 0#

#f(2) = 48+16-36-24-4 = 0#

#f(-2) = 48-16-36+24-4 = 16#

So we have identified #3# rational roots out of #4# roots, viz #-2/3#, #-1# and #2#. The fourth root must also be rational and having 'used up' the factor of #2# for #p#, it must also be #-1#.

To check, we can multiply out the corresponding factors:

#(3x+2)(x+1)(x+1)(x-2)#

#=(3x^2+5x+2)(x^2-x-2)#

#=3x^4+(5-3)x^3-(6+5-2)x^2-(10+2)x-4#

#=3x^4+2x^3-9x^2-12x-4#