What are all the possible rational zeros for #f(x)=x^3-13x^2+23x-11# and how do you find all zeros?

1 Answer
Jan 22, 2017

Answer:

The "possible" rational zeros are: #+-1, +-11#

The actual zeros are: #1, 1, 11#

Explanation:

#f(x) = x^3-13x^2+23x-11#

By the rational roots theorem, any rational zeros of #f(x)# are expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #-11# and #q# a divisor of the coefficient #1# of the leading term.

That means that the only possible rational zeros are:

#+-1, +-11#

Note that the sum of the coefficients of #f(x)# is zero. That is:

#1-13+23-11 = 0#

Hence #f(1) = 0#, #x=1# is a zero and #(x-1)# a factor:

#x^3-13x^2+23x-11 = (x-1)(x^2-12x+11)#

The sum of the coefficients of the remaining quadratic is zero too:

#1-12+11 = 0#

So #x=1# is a zero again and #(x-1)# a factor:

#(x^2-12x+11) = (x-1)(x-11)#

So the zeros of #f(x)# are #x=1# with multiplicity #2# and #x=11#.