What are all the possible rational zeros for #f(x)=3x^3-11x^2+5x+3# and how do you find all zeros?

1 Answer
Oct 26, 2016

Answer:

#f(x)# has zeros #1#, #-1/3# and #3#

Explanation:

#f(x) = 3x^3-11x^2+5x+3#

By the rational roots theorem, any rational zero of #f(x)# is expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #3# and #q# a divisor of the coefficient #3# of the leading term.

That means that the only possible rational zeros are:

#+-1/3, +-1, +-3#

Note first that there's a shortcut in this particular example, in that the sum of the coefficients is #0#. That is #3-11+5+3 = 0#

Hence we can deduce that #f(1) = 0#, #x=1# is a zero of #f(x)# and #(x-1)# is a factor:

#3x^3-11x^2+5x+3 = (x-1)(3x^2-8x-3)#

We could just try the remaining rational possibilities that we listed, but we can factor the remaining quadratic (#3x^2-8x-3#) easily enough using an AC method.

Find a pair of factors of #AC = 3*3 = 9# which differ by #B=8#.

The pair #9, 1# works in that #9*1 = 9# and #9-1 = 8#

Use this pair to split the middle term and factor by grouping:

#3x^2-8x-3 = 3x^2-9x+x-3#

#color(white)(3x^2-8x-3) = (3x^2-9x)+(x-3)#

#color(white)(3x^2-8x-3) = 3x(x-3)+1(x-3)#

#color(white)(3x^2-8x-3) = (3x+1)(x-3)#

Hence the other two zeros of #f(x)# are #x=-1/3# and #x=3#