What are all the possible rational zeros for #f(x)=2x^3-8x^2+15x-27# and how do you find all zeros?

1 Answer
Sep 12, 2017

Answer:

Possible rational zeros:

#+-1/2, +-1, +-3/2, +-3, +-9/2, +-9, +-27/2, +-27#

Actual zeros:

#x = 3" "# and #" "x = 1/2(1+-sqrt(17)i)#

Explanation:

Given:

#f(x) = 2x^3-8x^2+15x-27#

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 #-27# and #q# a divisor of the coefficient #2# of the leading term.

That means that the only possible rational zeros are:

#+-1/2, +-1, +-3/2, +-3, +-9/2, +-9, +-27/2, +-27#

Using Descartes' Rule of Signs, note that the pattern of the signs of the coefficients of #f(x)# is #+ - + -#. With #3# changes of sign, we can tell that #f(x)# has #3# or #1# positive real zeros. The pattern of the signs of the coefficients of #f(-x)# is #- - - -#. With no changes, that means that #f(x)# has no negative real zeros.

After a bit of trial and error, we find:

#f(3) = 2(color(blue)(3))^3-8(color(blue)(3))^2+15(color(blue)(3))-27#

#color(white)(f(3)) = 54-72+45-27 = 0#

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

#2x^3-8x^2+15x-27 = (x-3)(2x^2-2x+9)#

Note that the discriminant of the remaining quadratic factor is negative, so it has non-real complex zeros.

We can complete the square (or use the quadratic formula) to find them.

For example:

#0 = 2(2x^2-2x+9)#

#color(white)(0) = 4x^2-4x+1+17#

#color(white)(0) = (2x-1)^2+(sqrt(17))^2#

#color(white)(0) = (2x-1)^2-(sqrt(17)i)^2#

#color(white)(0) = ((2x-1)-sqrt(17)i)((2x-1)+sqrt(17)i)#

#color(white)(0) = (2x-1-sqrt(17)i)(2x-1+sqrt(17)i)#

So:

#x = 1/2(1+-sqrt(17)i)#