Question

The polynomial P(x) = x^4 -8x^2-8x+15 has zeros of x=1,3,and -2+i. What is its fourth zero?

Answer 1

-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}`

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}`

Therefore, the fourth zero is:

-2 - i