Question

# Find a polynomial function of lowest degree with rational coefficients that has the given numbers as some of its zeros. 1 + i, 2

Functions
Find a polynomial function of lowest degree with rational coefficients that has the given numbers as some of its zeros.
1 + i, 2

2021-05-24
Complex zeros come in conjugate pairs so we know that 1-i is the third zero given that 1+i is a zero.
If a is a zero of a polynomial function, then x-a is one of its factor. Using 2, 1+i, and 1-i, the least possible degree of the polynomial function is 3:
$$\displaystyle{f{{\left({x}\right)}}}={a}{\left({x}-{2}\right)}{\left[{x}-{\left({1}+{i}\right)}\right]}{\left[{x}-{\left({1}-{i}\right)}\right]}$$
By expanging, $$\displaystyle{f{{\left({x}\right)}}}={a}{\left({x}-{2}\right)}{\left[{x}^{{2}}-{\left({1}-{i}\right)}{x}-{\left({1}+{i}\right)}{x}+{\left({1}+{i}\right)}{\left({1}-{i}\right)}\right]}$$
$$\displaystyle{f{{\left({x}\right)}}}={a}{\left({x}-{2}\right)}{\left[{x}^{{2}}-{x}+{i}{x}-{x}-{i}{x}+{\left({1}-{i}^{{2}}\right)}\right]}$$
$$\displaystyle{f{{\left({x}\right)}}}={a}{\left({x}-{2}\right)}{\left({x}^{{2}}-{2}{x}+{2}\right)}$$
$$\displaystyle{f{{\left({x}\right)}}}={a}{\left[{x}^{{2}}{\left({x}-{2}\right)}-{2}{x}{\left({x}-{2}\right)}+{2}{\left({x}-{2}\right)}\right]}$$
$$\displaystyle{f{{\left({x}\right)}}}={a}{\left({x}^{{3}}-{2}{x}^{{2}}+{4}{x}+{2}{x}-{4}\right)}$$
$$\displaystyle{f{{\left({x}\right)}}}={a}{\left({x}^{{3}}-{4}{x}^{{2}}+{6}{x}-{4}\right)}$$
For simplicity, let a=1: $$\displaystyle{f{{\left({x}\right)}}}={x}^{{3}}-{4}{x}^{{2}}+{6}{x}-{4}$$
2021-08-11

Answer is given below (on video)