Question

Degree 4; zeros: -4+5i, -1 multiplicity 2

Answer 1

Answer:

$f(x) = [x^{2} + 8x + 41](x + 1)^{2}$

Step-by-step explanation:

If -4 + 5i is a root of a polynomial, then it's conjugate -4 - 5i will also be a root of the same polynomial.

Therefore, the polynomial has 4 degree, and zeroes are (- 4 + 5i), (- 4 - 5i) and -1 with multiplicity 2.

Hence, the polynomial will be

$f(x) = (x + 4 - 5i)(x + 4 + 5i)(x + 1)^{2}$

⇒ $f(x) = [x^{2} + 8x + (4 - 5i)(4 + 5i)] (x + 1)^{2}$

⇒ $f(x) = [x^{2} + 8x + 41](x + 1)^{2}$ (Answer)