Question

Suppose $p(x)$ is a monic cubic polynomial with real coefficients such that $p(3-2i)=0$ and $p(0)=-52$. determine $p(x)$ (in expanded form).

Answer 1

If $p(3-2i)=0,$ then complex number $3-2i$ is a root of cubic polynomial.

If polynomial has real coefficients, then conjugated $3+2i$ is also a root of polynomial.

Then the polynomial will be of a form

$p(x)=(x-(3-2i))(x-(3+2i))(x-a).$

Since $p(0)=-52,$ then

$-52=(0-(3-2i))(0-(3+2i))(0-a),\\ \\-52=(2i-3)(3+2i)a,\\ \\-52=(4i^2-9)a,\\ \\-52=(-4-9)a,\\ \\-13a=-52,\\ \\a=4.$

Therefore,

$p(x)=(x-(3-2i))(x-(3+2i))(x-4),\\ \\p(x)=(x^2-6x+13)(x-4),\\ \\p(x)=x^3-10x^2+37x-52.$