Question

How do you find the number of complex, real and rational roots of `4x^3-12x+9=0`?

Answer 1

This cubic has one negative irrational Real root and two Complex ones.

Explanation:

`f(x) = 4x^3-12x+9`

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Discriminant

The discriminant `Delta` of a cubic polynomial in the form `ax^3+bx^2+cx+d` is given by the formula:

`Delta = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd`

In our example, `a=4`, `b=0`, `c=-12` and `d=9`, so we find:

`Delta = 0+27648+0-34992+0 = -7344`

Since `Delta < 0` this cubic has `1` Real zero and `2` non-Real Complex zeros, which are Complex conjugates of one another.

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Descartes' Rule of Signs

The signs of the coefficients of `f(x)` are in the pattern `+ - +`. With `2` changes of sign, `f(x)` has `0` or `2` positive Real zeros. Since we know that it does not have `2`, it must have `0` positive Real zeros. The signs of the coefficients of `f(-x)` are in the pattern `- + +`. With one change of sign, that means that `f(x)` has exactly one negative Real zero.

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Rational root theorem

Since all of the coefficients of `f(x)` are integers and the powers of `x` are in standard descending order, 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 `9` and `q` a divisor of the coefficient `4` of the leading term.

That means that the only possible rational zeros are:

`+-1/4, +-1/2, +-3/4, +-1, +-3/2, +-3, +-9/2, +-9`

Note that we already know that the only Real zero is negative, so if there is a rational zero then it is one of:

`-1/4, -1/2, -3/4, -1, -3/2, -3, -9/2, -9`

We find:

`f(-3) = -63 < 0`

`f(-3/2) = 27/2 > 0`

So the Real zero is somewhere between the possible rational zeros `-3` and `-3/2` and must be irrational.

Conclusion

`f(x)` has one Real negative irrational zero and two Complex zeros.