Question

How do you find all the real and complex roots of `2x^5-4x^4-4x^2+5=0`?

Answer 1

Use a numerical method to find approximations:

`x ~~ 2.2904`

`x ~~ 0.925274`

`x ~~ -0.80773`

`x ~~ -0.203974+-1.19116i`

Explanation:

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

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Fundamental theorem of algebra

The FTOA tells us that any non-zero polynomial of degree `> 0` has a Complex (possibly Real) zero. A corollary of this, often quoted as part of the FTOA, is that a polynomial of degree `n > 0` has exactly `n` Complex (possibly Real) zeros counting multiplicity.

In our example `f(x)` is of degree `5`, so has exactly `5` zeros counting multiplicity.

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Rational root theorem

SInce `f(x)` has integer coefficients, any rational zeros must be expressible in the for `p/q` for integers `p, q` with `p` a divisor of the constant term `5` and `q` a divisor of the coefficient `2` of the leading term.

That means that the only possible rational zeros are:

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

None of these satisfy `f(x) = 0`, so `f(x)` has no rational zeros.

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Descartes rule of signs

The signs of the coefficients of `f(x)` follow the pattern `+ - - +`. With two changes of signs, that means that `f(x)` has `0` or `2` positive Real zeros.

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

That leaves `2` or `4` non-Real, Complex zeros, which will occur in Complex conjugate pairs.

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Quintic

In common with most quintics (and higher degree polynomials), the zeros of this one cannot be expressed using elementary functions like `n`th roots, or even trigonometric functions. About the best we can do is find approximations using a numerical method such as Durand-Kerner to find zeros:

`x ~~ 2.2904`

`x ~~ 0.925274`

`x ~~ -0.80773`

`x ~~ -0.203974+-1.19116i`

See https://socratic.org/s/avVFw8eC for more details of the method and another example quintic.