Question

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

Answer 1

`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.