What are all the possible rational zeros for #f(x)=3x^3-x^2-3x+1# and how do you find all zeros?

1 Answer
Jan 2, 2018

Answer:

#x=-1,x=1/3,x=1#

Explanation:

The pattern of signs of #f(x)# is #+ --+#. It has two changes, and Descartes' rule therefor tells us that the polynomial has either two or no roots.

The pattern of signs for #f(-x)# is #--++#, which has one change, meaning the polynomial has one negative root.

The rational roots theorem states that we can find all possible rational roots to a polynomial by dividing #+-# all factors of the last term by #+-# all factors of the first. In this case, we get:
#+-1,+-1/3#.

I'll begin by trying the negative values since I know for a fact the polynomial has a real root. Of these, we find that #f(-1)# is a root.

This means we can bring out a factor of #(x+1)# out of the polynomial using long division:
#(x+1)(3x^2-4x+1)#

We can factor this quadratic by grouping:
#(x+1)(3x^2-3x-x+1)#

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

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

We can use the zero factor principle to find that the remaining roots are #x=1# and #x=1/3#, both of which are positive, verifying our conclusions using Descartes' rule of signs.