What are all the possible rational zeros for #f(x)=x^3+x^2-8x-6# and how do you find all zeros?

1 Answer
Nov 15, 2017

Answer:

Possible rational zeros: #+-1, +-2, +-3, +-6#

Actual zeros: #-3# and #1+-sqrt(3)#

Explanation:

Given:

#f(x) = x^3+x^2-8x-6#

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 #6# and #q# a divisor of the coefficient #1# of the leading term.

That means that the only possible rational zeros are:

#+-1#, #+-2#, #+-3#, #+-6#

Trying each in turn, we find:

#f(-3) = -27+9+24-6 = 0#

So #x=-3# is a zero and #(x+3)# a factor:

#x^3+x^2-8x-6 = (x+3)(x^2-2x-2)#

We can factor the remaining quadratic by completing the square:

#x^2-2x-2 = x^2-2x+1-3#

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

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

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

Hence the other two zeros of our cubic function are:

#x = 1+-sqrt(3)#