Find a polynomial with roots from another

I apologize if the title is not complete enough.

Let $f(x) = x^3 - 7x - 3$ and $g(x) = x^3 - 4x -5$ be polynomials with complex coefficients.

I need to find a polynomial of degree $3$ with roots $g(x_1)$, $g(x_2)$, and $g(x_3)$, where $x_1$, $x_2$ and $x_3$ are roots of $f$. The coefficient in front of $x^3$ must be $1$.

I can't find roots of $f$. Any suggestions?

• Are you familiar with (elementary) symmetric polynomials? If not, then this exercise is a bit cruel IMHO. – Jyrki Lahtonen Aug 29 '16 at 20:28
• A rather similar question: (math.stackexchange.com/q/1415302) – Jean Marie Aug 29 '16 at 21:14

Suppose $\;a\;$ is a root of $\;f\;$ , so

$$0=f(a)=a^3-7a-3\implies a^3=7a+3\,,\,\,\text{so}\;$$

$$\; g(a)=a^3-4a+5=7a+3-4a+5=3a+8$$

Thus, if the roots of $\;f\;$ are $\;a,b,c\;$ , then the wanted polynomial is

$$h(x)=(x-3a-8)(x-3b-8)(x-3c-8)=$$

$$=x^3-(3a+8+3b+8+3c+8)x^2+(9ab-24(a+b)....$$

Now, if you know the relations between coefficients of a polynomial and its roots, then for example:

$$a+b+c=0\implies3a+8+3b+8+3c+8=24$$

and also

$$ab+ac+bc=-7\;,\;\;abc=3$$

You may want to google "Vieta's Formulas"

• I somehow missed that $a\mapsto 3a+8$ relation, and it became hairy. Good job, DonAntonio! – Jyrki Lahtonen Aug 29 '16 at 20:49