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

Jul 25, 2018

Numerically...

#### Explanation:

Given:

$f \left(x\right) = 4 {x}^{5} + {x}^{4} + {x}^{3} + {x}^{2} - 2 x - 2$

By the rational roots theorem, any rational zeros of $f \left(x\right)$ are expressible in the form $\frac{p}{q}$ 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:

$\pm \frac{1}{4} , \pm \frac{1}{2} , \pm 1 , \pm 2$

None of these is a zero and hence we can tell that $f \left(x\right)$ 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 $f \left(x\right)$ has exactly one positive real zero.

The signs of the coefficients of $f \left(- x\right)$ are in the pattern $- + - + + -$. With $4$ changes of signs, that means that $f \left(x\right)$ 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 $n$th 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:

${x}_{1} \approx 0.854463$

${x}_{2 , 3} \approx 0.144877 \pm 1.00403 i$

${x}_{4 , 5} \approx - 0.697109 \pm 0.287521 i$

Here's the C++ program I used:

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