Question

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

Answer 1

The rational root theorem can point out the possibilities, but in this particular case there's a shortcut to the rational zeros:

`x=1` and `x=-1/3`

Explanation:

`f(x) = 3x^4-2x^3+8x^2-6x-3`

By the rational root theorem, any rational zero of `f(x)` is expressible in the form `p/q` for integers `p, q` with `p` a divior of the constant term `-3` and `q` a divisor of the coefficient `3` of the leading term.

So the only possible rational zeros are:

`+-1/3`, `+-1`, `+-3`

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Actually we can proceed more directly to find all the zeros as follows:

Note that the sum of the coefficients of `f(x)` is zero. That is:

`3-2+8-6-3 = 0`

Hence `f(1) = 0` and `(x-1)` is a factor:

`3x^4-2x^3+8x^2-6x-3 = (x-1)(3x^3+x^2+9x+3)`

The remaining cubic can be factored by grouping as follows:

`3x^3+x^2+9x+3`

`= (3x^3+x^2)+(9x+3)`

`=x^2(3x+1)+3(3x+1)`

`=(x^2+3)(3x+1)`

So another rational zero is `x=-1/3`

The remaining quadratic has no Real zeros, rational or irrational since `x^2+3 >= 3 > 0` for any Real value of `x`.

It does have Complex zeros `x = +-sqrt(3)i`