Number of Real Roots of a polynomial equation

Question

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

How many real zeros are there?

I tried factoring it but couldn't find a way. I tried using Descartes rules of signs and it said that it has 1 negative root and 4 or less positive roots which is still quite vague, can this be answered without factoring? or is it really factorable?

Answer 1

I tried factoring it but couldn't find a way.

Start with $\,f(x) = x \left(x^4-x^3+x^2-4x-12\right)$. The quartic factor does not have rational roots, so look for a factorization as a product of quadratics, and you'll find that:

$$\,x^4-x^3+x^2-4x-12 = (x^2 + 4) (x^2 - x - 3)\,$$

Answer 2

Hint: Factorizing we get $$x(x^2+4)(x^2-x-3)$$

Answer 3

Hint : f(x) has three sign change using Descartes rule, hence f(x) has either 3 positive root or 1 positive root. f(-x) has one sign change hence 1 negative root.

$T=f'''(x)=60x^2-24x+6$ The equation T has negative Discriminant therefore 2 complex root. Finally the answer is 1 negative root, 2 positive roots (3real root) and 2 complex root.