Answer
$\displaystyle \frac{1}{2}, \frac{-2-\sqrt{2}}{2},$ and $\displaystyle \frac{-2+\sqrt{2}}{2}$
Work Step by Step
If k is a zero of f, then, by the factor theorem, (x-k) is a factor of f(x).
Synthetic division gives us a quotient q(x), with degree one less than f(x).
and we know the remainder is zero,
so $f(x)=(x-k)\cdot q(x).$
We use this first.
$k=1/2$,
$\left\{\begin{array}{llllll}
1/2 & | & 4 & 6 & -2 & -1\\
& & & 2 & 4 & 1\\
& & -- & -- & -- & --\\
& & 4 & 8 & 2 & \fbox{$0$}
\end{array}\right.$
$f(x)=(x-\displaystyle \frac{1}{2})(4x^{2}+8x+2)$
We can use the quadratic formula to find the zeros of the quotient.
$x=\displaystyle \frac{b\pm\sqrt{b^{2}-4ac}}{2a} =\frac{-8\pm\sqrt{64-4(4)(2)}}{2(4)} $
$=\displaystyle \frac{-8\pm\sqrt{32}}{8}=\frac{-8\pm 4\sqrt{2}}{8}=\frac{4(-2\pm\sqrt{2})}{8}$
$=\displaystyle \frac{-2\pm\sqrt{2}}{2}$
The other zeros (other than the given) are
$\displaystyle \frac{-2-\sqrt{2}}{2}$ and $\displaystyle \frac{-2+\sqrt{2}}{2}$
