Observe that the sum of the co-effs. of odd-powered terms of #g(x)=3-7=-4,# & that of even-powered #=2-6=-4.# Hence, #(x+1)# is a factor of #g(x).#

Now we arrange the terms of #g(x)# in such a way that #(x+1)# can be taken out as a common factor from #g(x)# as shown below :-

#g(x)=3x^3+2x^2-7x-6,#

#=3x^3+3x^2-x^2-x-6x-6,#

#=3x^2(x+1)-x(x+1)-6(x+1),#

#=(x+1)(3x^2-x-6).#

To work out the zeroes of the quadr. poly. #3x^2-x-6 = f(x), say,# we compare it with the std. qudr. poly. #ax^2+bx+c,# to give us,

#a=3, b=-1, c=-6.#

We will use the formula to find the zeroes of #f(x)#.

If #alpha, beta# are the zeroes of the std. qudr. poly., by the formula,

#alpha,beta =(-b+-sqrtDelta)/(2a), # where, #Delta=b^2-4ac=1-4*3*(-6)=1+72=73.#

Hence, #alpha,beta=(1+-sqrt73)/6.#

Altogether, the zeroes are #-1,(1+sqrt73)/6,(1-sqrt73)/6.#

Taking #sqrt73~=8.544#, the zeroes are #-1, 1.591, -1.257.#