Answer
2 or $0$ positive real roots exist,
2 or $0$ negative real roots exist.
Work Step by Step
Descartes's Rule of Signs (page 384)
Let $f(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{2}x^{2}+a_{1}x+a_{0}$
be a polynomial with real coefficients.
1. The number of positive real zeros of $f$ is either
$\mathrm{a}$. the same as the number of sign changes of $f(x)$
or
$\mathrm{b}$. less than the number of sign changes of $f(x)$ by a positive even integer.
If $f(x)$ has only one variation in sign, then $f$ has exactly one positive
real zero.
2. The number of NEGATIVE real zeros of $f$ is either
$\mathrm{a}$. the same as the number of sign changes of $f(-x)$
or
$\mathrm{b}$. less than the number of sign changes of $f(-x)$ by a positive even integer.
If $f(-x)$ has only one variation in sign, then $f$ has exactly one negative
real zero.
-------------
$f(x)=2x^{4}-5x^{3}-x^{2}-6x+4$
Since $f(x)$ has 2 sign variations,
(+$2x^{4}-5x^{3}$) and ($-6x+4$)
2 or $0$ positive real roots exist.
$f(-x)=2x^{4}+5x^{3}-x^{2}+6x+4$
Since $f(-x)$ has 2 sign variations,
($+5x^{3}-x^{2}$) and ($-x^{2}+6x$),
2 or $0$ negative real roots exist.
