Question

How do you find the roots for `f(x)=16x^4+8x^3+4x^2-2x-15` using the fundamental theorem of algebra?

Answer 1

The FTOA tells us that there are `4` zeros, but not how to find them.

Explanation:

Fundamental Theorem of Algebra

The fundamental theorem of algebra (FTOA) tells us that any non-constant polynomial with Complex (possibly Real) coefficients will have a zero in the Complex numbers.

A straightforward corollary of that, often stated as part of the FTOA is that a polynomial of degree `n > 0` has exactly `n` Complex (possibly Real) zeros counting multiplicity.

Given:

`f(x) = 16x^4+8x^3+4x^2-2x-15`

Note that `f(x)` is of degree `4`. So by the FTOA we deduce that it has `4` Complex (possibly Real) zeros counting multiplicity.

That is all the FTOA tells us. It does not tell us how to find them.

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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 `-15` and `q` a divisor of the coefficient `16` of the leading term.

That means that the only possible rational zeros are:

`+-1/16, +-1/8, +-3/16, +-1/4, +-5/16, +-3/8, +-1/2, +-5/8, +-3/4, +-15/16, +-1, +-5/4, +-3/2, +-15/8, +-5/3, +-3, +-5, +-15`

That is rather a lot of possibilities to try, but we can cut it down radically as follows:

Note that:

`f(x) = 16x^4+8x^3+4x^2-2x-15=(2x)^4+(2x)^3+(2x)^2-(2x)-15`

So the only possible rational values of `2x` making `f(x) = 0` are `+-1, +-3, +-5, +-15`. Hence the only possible rational zeros are:

`+-1/2, +-3/2, +-5/2, +-15/2`

None of these work. So `f(x)` has no rational zeros.

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Descartes' Rule of Signs

The coefficients of `f(x)` have signs in the pattern `+ + + - -`. With one change of sign, that means that `f(x)` has exactly `1` positive Real zero.

The coefficients of `f(-x)` have signs in the pattern `+ - + + -`. With `3` changes of sign, that means that `f(x)` has exactly `1` or `3` negative Real zeros.

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Where do we go from here?

It is possible to solve `f(x) = 0` algebraically, but it gets very long and messy. See https://socratic.org/s/axciCVkR for a typical quartic. Alternatively you can use a numerical method such as Durand-Kerner to find approximations:

`x_1 ~~ -1.0194`

`x_2 ~~ 0.85875`

`x_(3,4) ~~ -0.16967+-1.02084i`