Question

find all other zeros of p (x)= x^3-x^2+8x+10, given that 1+3i is a zero. ( if there is more than one zero, separate them with commas.)edit: if possible please double check answers would high appreciate it.

Answer 1

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):

$\begin{gathered} x=1-3i \\ x=1+3i \end{gathered}$

also, notice that if we evaluate -1 on p(x), we get:

$\begin{gathered} p(-1)=(-1)^3-(-1)^2+8(-1)+10=-1-1-8+10 \\ =-10+10=0 \end{gathered}$

therefore, the zeros of p(x) are:

x = 1-3i

x = 1+3i

x = -1