3. if 2i is the root of p(x)=x^3+4x+8, then find the remaining roots of p(x)=?

Answer 1

To find the remaining roots of the polynomial, we can use the fact that complex roots of a polynomial with real coefficients come in conjugate pairs. Since 2i is a root of p(x), its conjugate -2i must also be a root.

Now, we can use polynomial division or synthetic division to divide the polynomial by (x-2i)(x+2i)=(x^2+4).

Dividing p(x) by x^2+4, we get:

p(x) = (x^2 + 4)(x - 2i)
= (x^2 + 4)(x + 2i)
= x^3 + 2ix^2 + 4x + 8i
= x^3 + 4x + 8

Therefore, the remaining roots of p(x) are the roots of x^2 + 4 = 0, which are ±2i. So the roots of p(x) are 2i, -2i, and the roots of x^2 + 4 = 0 which are ±2i.