Question

How do you find all rational roots for `8y^4 - 6y^3 + 17y^2 - 12y + 2 = 0`?

Answer 1

The rational roots are `1/4`, `1/2`.

The remaining two roots are `+-sqrt(2)i`

Explanation:

`f(y)=8y^4-6y^3+17y^2-12y+2`

By the rational root theorem, any rational zeros of `f(y)` must be 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 `8` of the leading term.

That means that the only possible rational zeros are:

`+-1/8, +-1/4, +-1/2, +-1, +-2`

In addition, note that `f(-y) = 8y^4+6y^3+17y^2+12y+2` has all positive coefficients. Hence `f(y)` has no negative zeros.

So the only possible rational zeros of `f(y)` are:

`1/8, 1/4, 1/2, 1, 2`

We find:

`f(1/4) = 8(1/4)^4-6(1/4)^3+17(1/4)^2-12(1/4)+2`

`=(1-3+34-96+64)/32 = 0`

`f(1/2) = 8(1/2)^4-6(1/2)^3+17(1/2)^2-12(1/2)+2`

`=(2-3+17-24+8)/4 = 0`

So `y=1/4` and `y=1/2` are zeros and `(4y-1)` and `(2y-1)` are factors:

`8y^4-6y^3+17y^2-12y+2`

`=(4y-1)(2y^3-y^2+4y-2)`

`=(4y-1)(2y-1)(y^2+2)`

`y^2+2 >= 2 > 0` for all Real values of `y`, so there are no more Real, let alone rational, zeros.

The last two zeros are `+-sqrt(2)i` since:

`(y-sqrt(2)i)(y+sqrt(2)i) = y^2-(sqrt(2)i)^2 = y^2+2`