Answer
$f(x)=(x+1)(3x+1)(x^{2}+2)$
Zeros: $-1,\ \ -\displaystyle \frac{1}{3},\ \ $with multiplicity $1$
Work Step by Step
$f(x)$ has at most $4$ real zeros, as the degree is $4$.
Rational zeros: $\displaystyle \frac{p}{q},$ in lowest terms,
$p$ must be a factor of $ 2,\qquad \pm 1,\pm 2,$
and $q$ must be a factor of $ 3,\qquad \pm 1\pm 3$
$\displaystyle \frac{p}{q}=\pm 1,\pm\frac{1}{3},\pm\frac{2}{3},\pm 2$
Testing with synthetic division, try $x+1$
$\left.\begin{array}{l}
-1 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr}
3 &4 &7 &8 &2 \\\hline
&-3 &-1 &-6 & -2 \\\hline
3& 1 & 6 &2 & |\ \ 0 \end{array}$
$f(x)=(x+1)(3x^{3}+x^{2}+6x+2)$
factor in pairs:
$3x^{3}+x^{2}+6x+2=x^{2}(3x+1)+2(3x+1)$
$=(3x+1)(x^{2}+2)$
$x^{2}+2$ has no real zeros and can not be factored any further.
$f(x)=(x+1)(3x+1)(x^{2}+2)$
Zeros: $-1,\ \ -\displaystyle \frac{1}{3},\ \ $with multiplicity $1$
