Answer:
-2 - i
Step-by-step explanation:
When the polynomial has a zero with the form a + bi, it also has a zero with the form a - bi. So, the fourth zero of the polynomial should be:
-2 - i
And we can prove that it is a zero replacing x by (-2 - i) on the initial equation as:
![\begin{gathered} P(x)=(-2-i)^4-8(-2-i_{})^2-8(-2-i)+15 \\ P(x)=(4+4i+i^2)^2-8(4+4i+i^2)+(16+8i)+15 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/zpsetpdpip9n6djgv7eafu7n5rws2vcblu.png)
Where i²= -1, so P(x) is:
![\begin{gathered} P(x)=(4i+3)^2-8(4i+3)+16+8i+15 \\ P(x)=16i^2+24i+9-32i-24+16+8i+15 \\ P(x)=-16^{}+24i+9-32i-24+16+8i+15 \\ P(x)=0 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/f6kksnk2ony7uidepjwy2ppzaqm4qp84am.png)
Therefore, the fourth zero is:
-2 - i