What are all the possible rational zeros for #f(x)=x^3-9x^2+15x-2# and how do you find all zeros?

1 Answer
Dec 18, 2017

Answer:

#x=2# and #x=(7+-3sqrt(5))/2#

Explanation:

The rational root theorem tells us that all the possible rational zeroes will be #+-# all the factors of the coefficient of the last term divided by #+-# all the coefficients of the first term.

In this case, we get: #+-1,+-2# (because #2# is prime, and because #1# doesn't have any factors besides itself).

If we plug these into our equation we see that #x=2# is a solution.

To find the remaining zeroes, we can divide by #x-2# (using polynomial long division) to factor:
#(x^3-9x^2+15x-2)/(x-2)=x^2-7x+1#

This means that we can factor like so:
#(x-2)(x^2-7x+1)=0#

Since we already know #2# is a root, we want to look at when the other factor equals #0#:
#x^2-7x+1=0#

We can solve this by completing the square:
#(x-7/2)^2-45/4=0#

#(x-7/2)^2=45/4#

#x-7/2=+-sqrt(45/4)#

#x=7/2+-sqrt(45/4)#

#x=(7+-3sqrt(5))/2#