Question

Write the equation of the lowest degree polynomial with solutions x=2, -1+i root3, -1 -i root3

Answer 1

In general, a polynomial expression can be written as shown below

$\begin{gathered} f(x)=(x-c_1)(x-c_2)\cdot\cdot\cdot(x-c_n) \\ c_i\to\text{roots of the function} \\ n\to\text{ degre}e\text{ of the function} \end{gathered}$

Thus, in our case,

$g(x)=(x-2)(x-(-1+i\sqrt[]{3}))(x-(-1-i\sqrt[]{3}))$

Then, the lowest degree of the polynomial is 3.

Expanding g(x),

$\Rightarrow g(x)=(x-2)(x+1-i\sqrt[]{3})(x+1+i\sqrt[]{3})$

Notice that

$\begin{gathered} (x+1-i\sqrt[]{3})(x+1+i\sqrt[]{3})=((x+1)-i\sqrt[]{3})((x+1)+i\sqrt[]{3}) \\ =(x+1)^2-(i\sqrt[]{3})^2 \\ =(x+1)^2+3 \\ =x^2+2x+4 \end{gathered}$

Finally,

$\begin{gathered} \Rightarrow g(x)=(x-2)(x^2+2x+4)=x^3+2x^2+4x-(2x^2+4x+8)=x^3-8 \\ \Rightarrow g(x)=x^3-8 \end{gathered}$

Therefore, the answer is x^3-8