Answer
$x\displaystyle \in\{-2,-1,1,3\}$
Work Step by Step
See Descarte's rule of sign:
It asserts that the number of real roots is at most the number of sign changes in the sequence of polynomial's coefficients (omitting the zero coefficients), and that the difference between these two numbers is always even.
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, $a_{n}$.
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$f(x)=x^4-x^{3}-7x^{2}+x+6$
a. Candidates for zeros, $\frac{p}{q}:$
$p:\qquad \pm 1,\pm2,\pm3,\pm6 $
$q:\qquad \pm 1$
$\displaystyle \frac{p}{q}:\qquad \pm 1, \pm2,\pm3,\pm6$
b. $f(x)=x^4-x^{3}-7x^{2}+x+6$, there are two sign changes therefore $f(x)$ has $2$ or $0$ positive roots.
$f(-x)=x^4+x^{3}-7x^{2}-x+6$, there are two sign changes therefore $f(x)$ has $2$ pr $0$ negative roots
c. Try for $x=1:$
$\begin{array}{lllll}
\underline{1}| & 1 & -1 & -7 & 1&6\\
& & 1 & 0 & -7&-6\\
& -- & -- & -- & --\\
& 1 & 0 & -7&-6 & |\underline{0}
\end{array}$
$1$ is a zero,
$f(x)=(x-1)(x^{3} -7x-6)$
Try for $x=-1:$
$\begin{array}{lllll}
\underline{-1}| & 1 & 0 & -7&-6\\
& & -1 & 1&6\\
& -- & -- & -- & --\\
& 1 & -1 & -6& |\underline{0}
\end{array}$
$-1$ is a zero,
$f(x)=(x-1)(x+1)(x^2-x-6)$
d. Factorize the trinomial factor $(x^{2} -x-6)$
(find two factors of $1(-6)=-6$ whose sum is $-1:$
$(-3$ and $+2$)
$x^{2} -x-6=x^{2} +2x-3x-6 \quad$...factor in pairs ...
$=x(x+2)-3(x+2)=(x+2)(x-3)$
$f(x)=(x-1)(x+1)(x+2)(x-3)$
The zeros of f satisfy $f(x)=0$
$(x+1)(x-1)(x+2)(x-3)=0$
$x\displaystyle \in\{-2,-1,1,3\}$
