What are all the possible rational zeros for #f(x)=3x^5-6x^4+2x^2-6x+12# and how do you find all zeros?

1 Answer
Sep 1, 2017

Answer:

Assuming:

#f(x) = 3x^5-6x^4-x^3+2x^2-6x+12#

Possible rational zeros:

#+-1/3,+-2/3,+-1,+-4/3,+-2,+-3,+-4,+-6,+-12#

Actual zeros:

#2," "+-sqrt(1/6(1+sqrt(73)))," "+-sqrt(1/6(sqrt(73)-1))i#

Explanation:

I think there is a missing term #-x^3# in the question as:

  • The given quintic has no solutions expressible using radicals, let alone rational roots.

  • There is no #x^3# term - so one might well have been missed.

  • If the missing term is #-x^3# then the quintic is solvable using precalculus level methods and has one rational zero.

So, suppose:

#f(x) = 3x^5-6x^4-x^3+2x^2-6x+12#

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 #12# and #q# a divisor of the coefficient #3# of the leading term.

That means that the only possible rational zeros are:

#+-1/3, +-2/3, +-1, +-4/3, +-2, +-3, +-4, +-6, +-12#

We find:

#f(2) = 3(color(blue)(2))^5-6(color(blue)(2))^4-(color(blue)(2))^3+2(color(blue)(2))^2-6(color(blue)(2))+12#

#color(white)(f(2)) = 96-96-8+8-12+12 = 0#

So #x=2# is a zero and #(x-2)# a factor:

#3x^5-6x^4-x^3+2x^2-6x+12 = (x-2)(3x^4-x^2-6)#

Then:

#12(3x^4-x^2-6) = 36x^4-12x^2-72#

#color(white)(12(3x^4-x^2-6)) = (6x^2)^2-2(6x^2)+1-73#

#color(white)(12(3x^4-x^2-6)) = (6x^2-1)^2-(sqrt(73))^2#

#color(white)(12(3x^4-x^2-6)) = ((6x^2-1)-sqrt(73))((6x^2-1)+sqrt(73))#

#color(white)(12(3x^4-x^2-6)) = (6x^2-1-sqrt(73))(6x^2-1+sqrt(73))#

Hence:

#x = +-sqrt(1/6(1+sqrt(73)))" "# or #" "x = +-sqrt(1/6(sqrt(73)-1))i#