What are all the possible rational zeros for #f(x)=5x^3-9x^2-12x-2# and how do you find all zeros?

1 Answer
Jun 17, 2017

Answer:

The zeros of #f(x)# are:

#-1/5" "# and #" "1+-sqrt(3)#

Explanation:

Given:

#f(x) = 5x^3-9x^2-12x-2#

By the rational root 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 #-2# and #q# a divisor of the coefficient #5# of the leading term.

That means that the only possible rational zeros are:

#+-1/5, +-2/5, +-1, +-2#

First check #f(+-1)#...

#f(1) = 5-9-12-2 = -18#

#f(-1) = -5-9+12-2 = -4#

We can deduce that not all of the zeros are rational, since there are not enough factors to go around.

Trying the other possibilities, we eventually find:

#f(-1/5) = -5(1/125)+9(1/25)+12(1/5)-2#

#color(white)(f(-1/5)) = (-1-9+60-50)/25 = 0#

So #x=-1/5# is a zero and #(5x+1)# a factor:

#5x^3-9x^2-12x-2 = (5x+1)(x^2-2x-2)#

#color(white)(5x^3-9x^2-12x-2) = (5x+1)(x^2-2x+1-3)#

#color(white)(5x^3-9x^2-12x-2) = (5x+1)((x-1)^2-(sqrt(3))^2)#

#color(white)(5x^3-9x^2-12x-2) = (5x+1)(x-1-sqrt(3))(x-1+sqrt(3))#

So the zeros are:

#-1/5" "# and #" "1+-sqrt(3)#