Answer
$f(x)=(x+1)(4x+1)(x^{2}+2)$
Zeros: $-1,\ \ -\displaystyle \frac{1}{4},\ \ $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 $ 4,\qquad \pm 1\pm 2,\pm 4$
$\displaystyle \frac{p}{q}=\pm 1,\pm\frac{1}{2},\pm\frac{1}{4},\pm 2$
Testing with synthetic division, try $x+1$
$\left.\begin{array}{l}
-1 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr}
4 &5 &9 &10 &2 \\\hline
&-4 &-1 &-8 & -2 \\\hline
4& 1 & 8 &2 & |\ \ 0 \end{array}$
$f(x)=(x+1)(4x^{3}+x^{2}+8x+2)$
factor in pairs:
$4x^{3}+x^{2}+8x+2=x^{2}(4x+1)+2(4x+1)$
$=(4x+1)(x^{2}+2)$
$x^{2}+2$ has no real zeros and can not be factored any further.
$f(x)=(x+1)(4x+1)(x^{2}+2)$
Zeros: $-1,\ \ -\displaystyle \frac{1}{4},\ \ $with multiplicity $1$
