# How do you find a fourth-degree polynomial equation, with integer coefficients, for which #-sqrt11# and 2i are solutions?

##### 1 Answer

# P(x) = A(x^2-11)(x^2+4) #

Where

#### Explanation:

By the fundamental Theorem of Algebra, any polynomial of degree

# P(x) = A(x-alpha)(x-beta)(x-gamma) (x-delta) #

Where,

We are given that

# alpha = -sqrt(11) #

# beta = 2i #

We also know that complex roots must appear as complex conjugate pairs, so

# gamma = -2i #

And so we have:

# P(x) = A(x+sqrt(11))(x-2i)(x+2i) (x-delta) #

We can multiply the complex products to get real coefficients quadratic:

# (x-2i)(x+2i) = x^2-(2i)^2 #

# " " = x^2-4i^2 #

# " " = x^2+4 #

And using this we get:

# P(x) = A(x+sqrt(11))(x^2+4) (x-delta) #

We require that

# (x+sqrt(11))(x-delta) = x^2+(sqrt(11)-delta)x-sqrt(11)delta#

And if we set

# (x+sqrt(11))(x+sqrt(11)) = x^2-(sqrt(11))^2#

# " " = x^2-11 #

Hence, we have

# P(x) = A(x^2-11)(x^2+4) #

Where