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.
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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