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