Since we have that 1 + 3i is one zero of p(x), then we have that its conjugate is also a root, then, we have the following complex roots for p(x):
[tex]\begin{gathered} x=1-3i \\ x=1+3i \end{gathered}[/tex]
also, notice that if we evaluate -1 on p(x), we get:
[tex]\begin{gathered} p(-1)=(-1)^3-(-1)^2+8(-1)+10=-1-1-8+10 \\ =-10+10=0 \end{gathered}[/tex]
therefore, the zeros of p(x) are:
x = 1-3i
x = 1+3i
x = -1
