How do you use the Rational Zeros theorem to make a list of all possible rational zeros, and use the Descarte's rule of signs to list the possible positive/negative zeros of #f(x)=x^3-7x^2+x-7#?

1 Answer
Jan 2, 2018

Answer:

#x=7,x=+-i#

Explanation:

The sign pattern of #f(x)# is #+ -+ -#. This is three changes, so Descartes' rule of signs tells us that we have either #3# or #1# positive root(s) to the polynomial.

The sign pattern of #f(-x)# is #----#, which has no changes, and therefor there are no negative roots to the polynomial.

The rational roots theorem tells us that we can find all rational roots to a polynomial by dividing #+-# the factors of the last term with #+-# the factors of the first term. In this case, we get:
#+-1,+-7#

We also don't need to bother checking the negative roots, since Descartes' rule of signs told us there are no negative roots.

After checking #1# and #7#, we see that only #f(7)# is a solution. To find the remaining solutions (a cubic equation should have three solutions), we can use polynomial long division to factor:
#(x-7)(x^2+1)#

Now we can solve for when the second factor equals #0#:
#x^2+1=0#

#x^2=-1#

#x=+-sqrt(-1)#

#x=+-i#

Now that we have all the zeroes we can see that our original conclusions with Descartes' rule were correct. We only got #1# positive root, and no negative roots (Descartes' rule isn't concerned with complex roots).