Question

Write a polynomial function with rational coefficients so that P(x)=0 has the given root. -10i

Answer 1

Keeping in mind that complex solutions do not come all by their lonesome
they're always in pairs, so -10i has a sister, +10i, or its conjugate

drawing from that we could say that $p(x)=0 \\ \quad \\ p(x)= \begin{cases} x=-10i\implies x+10i=0\implies &(x+10i)=0 \\ \quad \\ x=+10i\implies x-10i=0\implies &(x-10i)=0 \end{cases} \\ \quad \\ thus \\ \quad \\ p(x)=0\implies (x+10i)(x-10i)=0 \\ \quad \\ --------------\\ \textit{difference of squares} \\ \quad \\ (a-b)(a+b) = a^2-b^2\qquad \qquad a^2-b^2 = (a-b)(a+b)\\ --------------\\ then \\ \quad \\ (x+10i)(x-10i)=0\implies [x^2-(10i)^2]=0 \\ \quad \\\ [x^2-(10^2\cdot i^2)]=0\implies [x^2-(100\cdot -1)]=0$

and surely you can take it from there :)