Question

What are all the possible rational zeros for `f(x)=4x^4+3x^3-2x^2+5x-12` and how do you find all zeros?

Answer 1

The "possible" rational zeros are:

`+-1/4, +-1/2, +-3/4, +-1, +-3/2, +-2, +-3, +-4, +-6, +-12`

The actual zeros are irrational and complex.

Explanation:

Given:

`f(x) = 4x^4+3x^3-2x^2+5x-12`

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Rational Roots Theorem

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 `4` of the leading term.

That means that the only possible rational zeros are:

`+-1/4, +-1/2, +-3/4, +-1, +-3/2, +-2, +-3, +-4, +-6, +-12`

That's rather a lot of possibilities to try, so let us find out more about the nature of these zeros.

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Descartes' Rule of Signs

The signs of the coefficients of `f(x)` are in the pattern `+ + - + -`. With `3` changes of sign, that means that `f(x)` has `3` or `1` real positive zero.

The signs of the coefficients of `f(-x)` are in the pattern `+ - - - -`. With `1` change of sign, that means that `f(x)` has exactly `1` real negative zero.

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Turning points

Let us first find where the turning points of `f(x)` are to help guide our search. Turning points occur where the derivative `f'(x)` of `f(x)` is zero.

`f'(x) = 16x^3+9x^2-4x+5`

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=16`, `b=9`, `c=-4` and `d=5`, so we find:

`Delta = 1296+4096-14580-172800-51840 = -233828`

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

So `f'(x)` has only one real zero and `f(x)` has only one turning point.

Therefore `f(x)` has one real negative zero and one real positive zero.

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Back to the rational possibilities...

Trying some of the possible rational zeros, we find:

`f(-2) = 10`

`f(-3/2) = -13.875`

`f(1) = -2`

`f(3/2) = 21.375`

Hence we can deduce that there is an irrational zero in `(-2, -3/2)` and one in `(1, 3/2)`.

There are no rational zeros.

graph{4x^4+3x^3-2x^2+5x-12 [-5, 5, -20, 20]}

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Where do we go from here?

to be continued...

Answer 2

Use a numerical method (Durand-Kerner) to find approximations to the irrational zeros...

Explanation:

Given:

`f(x) = 4x^4+3x^3-2x^2+5x-12`

In my first answer, we found that the zeros are not rational.

We can find approximations to the zeros using an algorithm known as Durand-Kerner.

Suppose the `4` zeros are `p, q, r, s`.

Choose initial approximations:

`p_0 = (0.4+0.9i)^0`
`q_0 = (0.4+0.9i)^1`
`r_0 = (0.4+0.9i)^2`
`s_0 = (0.4+0.9i)^3`

Then iterate using the formulas:

`p_(i+1) = p_i-f(p_i)/((p_i-q_i)(p_i-r_i)(p_i-s_i))`

`q_(i+1) = q_i-f(q_i)/((q_i-p_(i+1))(q_i-r_i)(q_i-s_i))`

`r_(i+1) = r_i-f(r_i)/((r_i-p_(i+1))(r_i-q_(i+1))(r_i-s_i))`

`s_(i+1) = s_i-f(s_i)/((s_i-p_(i+1))(s_i-q_(i+1))(s_i-r_(i+1))`

This can be done using a computer program, such as this one written in C++ ...

After a few iterations, the approximations settle down to give:

`p ~~ 1.0707`

`q ~~ -1.85404`

`r ~~ 0.0166701-1.22921i`

`s ~~ 0.0166701+1.22921i`