Answer
The zeros are
$1$, (which is given,)
$\displaystyle \frac{-5-\sqrt{5}}{2}$ and
$\displaystyle \frac{-5+\sqrt{5}}{2}$
Work Step by Step
x_{y}
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.
$k=1$, the x term is missing,
$\left\{\begin{array}{lllllll}
1 & | & 1 & 4 & 0 & -5 & \\
& & & 1 & 5 & 5 & \\
& & -- & -- & -- & -- & --\\
& & 1 & 5 & 5 & \fbox{$0$} &
\end{array}\right.$
$f(x)= (x-1)(x^{2}+5x+5)$
We can use the quadratic formula to find the zeros of the quotient.
$x=\displaystyle \frac{b\pm\sqrt{b^{2}-4ac}}{2a} =\frac{-5\pm\sqrt{25-4(1)(5)}}{2(1)}$
$=\displaystyle \frac{-5\pm\sqrt{5}}{2}$
The zeros are
$1$, (which is given,)
$\displaystyle \frac{-5-\sqrt{5}}{2}$ and
$\displaystyle \frac{-5+\sqrt{5}}{2}$