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

1 Answer
Aug 21, 2017

Answer:

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

Where #A# is an arbitrary integer.

Explanation:

By the fundamental Theorem of Algebra, any polynomial of degree #4# can be written in the form:

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

Where, #alpha,beta,gamma,delta# are the roots (or zeros) of the equation #P(x)=0#

We are given that #-sqrt(11)# and #2i# are solutions (presumably, although not explicitly stated, of #P(x)=0#, thus, wlog, we can write:

# alpha = -sqrt(11) #
# beta = 2i #

We also know that complex roots must appear as complex conjugate pairs, so #-2i# must be another root, so in addition:

# 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 #P(x)# has integer coefficients, rather than irrational coefficients, and this would require that the product #(x+sqrt(11))# and #(x-delta)# has integer coefficients, Now:

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

And if we set #delta=-sqrt(11)# this condition is met, and the product of the irrational coefficients are:

# (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 #A# is an arbitrary integer.