How do you find a fourth-degree polynomial equation, with integer coefficients, for which −√11 and 2i are solutions?
1 Answer
P(x)=A(x2−11)(x2+4)
Where
Explanation:
By the fundamental Theorem of Algebra, any polynomial of degree
P(x)=A(x−α)(x−β)(x−γ)(x−δ)
Where,
We are given that
α=−√11
β=2i
We also know that complex roots must appear as complex conjugate pairs, so
γ=−2i
And so we have:
P(x)=A(x+√11)(x−2i)(x+2i)(x−δ)
We can multiply the complex products to get real coefficients quadratic:
(x−2i)(x+2i)=x2−(2i)2
=x2−4i2
=x2+4
And using this we get:
P(x)=A(x+√11)(x2+4)(x−δ)
We require that
(x+√11)(x−δ)=x2+(√11−δ)x−√11δ
And if we set
(x+√11)(x+√11)=x2−(√11)2
=x2−11
Hence, we have
P(x)=A(x2−11)(x2+4)
Where