Question

How do you use the rational roots theorem to find all possible zeros of `p(x)=x^5-4x^4+3x^3+2x-6`?

Answer 1

The only rational zero is `x=3`.

The other zeros are non-Real Complex roots.

Explanation:

The rational root theorem tells us that any rational zeros of `p(x)` are expressible in the form `m/n` for some integers `m` and `n` with `m` a divisor of the constant term `-6` and `n` a divisor of the coefficient `1` of the leading term.

That is, the only possible rational zeros are:

`+-1`, `+-2`, `+-3`, `+-6`

Trying each in turn, we find:

`p(3) = 243-324+81+6-6 = 0`

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

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

By the rational root theorem, the only remaining possible rational zeros are:

`+-1`, `+-2`

None of these is a zero of `x^4-x^3+2`, so there are no more rational zeros.

That's as far as we can go with the rational root theorem.

`color(white)()`
Epilogue

It is possible to express the solutions of `x^4-x^3+2=0` algebraically, but it gets rather messy.

If you are interested, here's the beginning of the solution...

To solve `x^4-x^3+2 = 0`, multiply by `2` and substitute `t=1/x` to get:

`0 = 4t^4-2t+2`

`= (2t^2+at+b)(2t^2-at+c)`

`= 4t^4+(2b+2c-a^2)t^2+a(c-b)t+bc`

Equating coefficients and rearranging a little:

`b+c = a^2/2`

`b-c = 2/a`

`bc = 2`

Then:

`(a^2)^2/4 = (b+c)^2 = (b-c)^2+4bc = 4/(a^2)+8`

Hence:

`(a^2)^3-32(a^2)-16 = 0`

This cubic in `a^2` has `3` Real roots, one of which is positive, giving a pair of possible Real values for `a`.

Substitute `a^2 = 8/3 sqrt(6) cos theta`

Then:

`16 = (a^2)^3-32(a^2)`

`= (8/3 sqrt(6) cos theta)^3-32(4/3 sqrt(6) cos theta)^2`

`= 256/9sqrt(6)(4 cos^3 theta - 3 cos theta)`

`= 256/9sqrt(6) cos (3theta)`

So:

`cos(3theta) = (3sqrt(6))/32`

So:

`theta = +-1/3 (arccos((3sqrt(6))/32) + 2kpi)` for `k = 0, 1, 2`

So:

`a^2 = 8/3 sqrt(6) cos(1/3 (arccos((3sqrt(6))/32) + 2kpi))`

for `k = 0, 1, 2`

This gives a positive value (`~~5.89`) for `k = 0`

So we can let

`a = sqrt(8/3 sqrt(6) cos(1/3 arccos((3sqrt(6))/32))`

Then `b = a^2/2+1/a` and `c = a^2/2-1/a`

Hence we get quadratics to solve for `t` and hence can find values for `x`.