# How do you find all rational roots for #3x^4 - 2x^3 + 8x^2 - 6x - 3 = 0#?

##### 1 Answer

The rational root theorem can point out the possibilities, but in this particular case there's a shortcut to the rational zeros:

#x=1# and#x=-1/3#

#### Explanation:

By the rational root theorem, any rational zero of

So the only possible rational zeros are:

#+-1/3# ,#+-1# ,#+-3#

Actually we can proceed more directly to find all the zeros as follows:

Note that the sum of the coefficients of

#3-2+8-6-3 = 0#

Hence

#3x^4-2x^3+8x^2-6x-3 = (x-1)(3x^3+x^2+9x+3)#

The remaining cubic can be factored by grouping as follows:

#3x^3+x^2+9x+3#

#= (3x^3+x^2)+(9x+3)#

#=x^2(3x+1)+3(3x+1)#

#=(x^2+3)(3x+1)#

So another rational zero is

The remaining quadratic has no Real zeros, rational or irrational since

It does have Complex zeros