How do you find the zeros of f(x) = 4x^5 + x^4 + x^3 + x^2 - 2x - 2?

1 Answer
Jul 25, 2018

Numerically...

Explanation:

Given:

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

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 -2 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, +-1, +-2

None of these is a zero and hence we can tell that f(x) has no rational zeros.

Note that the signs of the coefficients are in the pattern + + + + - -. With one change of sign, Descartes' Rule of Signs tells us that f(x) has exactly one positive real zero.

The signs of the coefficients of f(-x) are in the pattern - + - + + -. With 4 changes of signs, that means that f(x) has 4, 2 or 0 negative real zeros.

In fact it has two complex conjugate pairs of non-real complex zeros.

Typically for a quintic polynomial none of these zeros is expressible in terms of nth roots or elementary functions. In theory they can be found using modular functions, but in practice it's probably better to stick with numerical approximations.

Using a Durand-Kerner method, I found approximate zeros:

x_1 ~~ 0.854463

x_(2,3) ~~ 0.144877+-1.00403i

x_(4,5) ~~ -0.697109+-0.287521i

Here's the C++ program I used:

enter image source here

See https://socratic.org/s/aSVeQs9C for another example and a fuller description of this method.