Question 693647
Any rational zeros would be of the form {{{p/q}}},
with {{{p}}} being a factor of the constant ({{{-9}}}),
and {{{q}}} being a factor of the leading coefficient ({{{2}}})
Factors of {{{9}}} are 1, 3, and 9.
Factors of {{{2}}} are 1, and 2.
Possible rational zeros are -1/2, -3/2, -9/2, -1, -3, -9, 1/2, 3/2, 9/2, 1, 3, and 9.
Descartes rule of signs helps narrow down your choices.
{{{P(x)=2x^3-13x^2+24x-9}}} has 3 changes of sign going from one coefficient to the next. That means that it has 3 or 1 positive zeros. (It could be two different zeros if one is a double zero).
{{{P(-x)=-2x^3-13x^2-24x-9}}} has 0 changes of sign, meaning that there are 0 negative zeros.
Now we say that possible rational zeros are 1/2, 3/2, 9/2, 1, 3, and 9.
Trying the possible zeros, using synthetic division, we find out that
{{{2x^3-13x^2+24x-9=(2x-1)(x^2-6x+9)}}}
Then we recognize that {{{x^2-6x+9=(x-3)^2}}} and have the complete factorization,
{{{2x^3-13x^2+24x-9=(2x-1)(x-9)^2}}},
which tells us that the zeros are {{{highlight(3)}}} (with multiplicity 2, a double zero), and {{{highlight(1/2)}}}