How do you find a fourth-degree polynomial equation, with integer coefficients, for which 11 and 2i are solutions?

1 Answer
Aug 21, 2017

P(x)=A(x211)(x2+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(xα)(xβ)(xγ)(xδ)

Where, α,β,γ,δ are the roots (or zeros) of the equation P(x)=0

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

α=11
β=2i

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

γ=2i

And so we have:

P(x)=A(x+11)(x2i)(x+2i)(xδ)

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

(x2i)(x+2i)=x2(2i)2
=x24i2
=x2+4

And using this we get:

P(x)=A(x+11)(x2+4)(xδ)

We require that P(x) has integer coefficients, rather than irrational coefficients, and this would require that the product (x+11) and (xδ) has integer coefficients, Now:

(x+11)(xδ)=x2+(11δ)x11δ

And if we set δ=11 this condition is met, and the product of the irrational coefficients are:

(x+11)(x+11)=x2(11)2
=x211

Hence, we have

P(x)=A(x211)(x2+4)

Where A is an arbitrary integer.