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

1 Answer
George C.
May 30, 2016

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:

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

By the rational root theorem, any rational zero of #f(x)# is expressible in the form #p/q# for integers #p, q# with #p# a divior of the constant term #-3# and #q# a divisor of the coefficient #3# of the leading term.

So the only possible rational zeros are:

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

#color(white)()#
Actually we can proceed more directly to find all the zeros as follows:

Note that the sum of the coefficients of #f(x)# is zero. That is:

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

Hence #f(1) = 0# and #(x-1)# is a factor:

#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 #x=-1/3#

The remaining quadratic has no Real zeros, rational or irrational since #x^2+3 >= 3 > 0# for any Real value of #x#.

It does have Complex zeros #x = +-sqrt(3)i#

Related questions
See all questions in Zeros
Impact of this question
2586 views around the world
You can reuse this answer
Creative Commons License