Answer
a. $\pm 1, \pm 2,\pm4, \pm\frac{1}{2}, \pm\frac{1}{3}, \pm\frac{2}{3},\pm\frac{4}{3}, \pm\frac{1}{6}$
b. No positive zero.
$4$ or $2$ or $0$ negative zeros.
Work Step by Step
See The Rational Zero Theorem:
If $\frac{p}{q}$ is a zero of the polynomial $f(x) $with integer coefficients, then $p$ is a factor of the constant term, $a_{0}$, and
$q$ is a factor of the leading coefficient, $a_{n}$.
By Descartes' rule of signs, if a polynomial in one variable, $f(x) = a_{n}x^{n} + a_{n-1}x^{n-1} + a_{n-2}x^{n-2} + ...+ a_{1}x + a_{0}$ is arranged in the descending order of the exponents of the variable, then:
The number of positive real zeros of $f(x)$ is either equal to the number of sign changes in $f(x)$ or less than the number of sign changes by an even number.
The same rule applies to find the number of negative real zeros as well, but then we count the sign changes of $f(-x)$.
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$f(x)=6x^4+3x^{3}+x^{2}+3x+4$
a. Candidates for zeros, $\frac{p}{q}:$
$p:\qquad \pm 1, \pm 2, \pm4$
$q:\qquad \pm 1, \pm2, \pm3, \pm6$
$\displaystyle \frac{p}{q}:\qquad \pm 1, \pm 2,\pm4, \pm\frac{1}{2}, \pm\frac{1}{3}, \pm\frac{2}{3},\pm\frac{4}{3}, \pm\frac{1}{6}$
b. $P(x)=6x^4+3x^3+x^2+3x+4$,
There is no sign change, therefore there are no positive zeros.
$P(-x)=6x^4-3x^3+x^2-3x+4$,
There are $4$ or $2$ or $0$ negative zeros.
