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
Alvin L.
Jan 2, 2018

#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.

Related questions
See all questions in Rational Zeros
Impact of this question
5487 views around the world
You can reuse this answer
Creative Commons License