Question

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?

Answer 1

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`