# How do you define a cubic function with zeros r_1, r_2, r_3 such that r_1+r_2+r_3 = 3, r_1r_2+r_2r_3+r_3r_1 = -1, r_1r_2r_3 = -3 ?

Oct 26, 2016

$f \left(x\right) = {x}^{3} - 3 {x}^{2} - x + 3$

#### Explanation:

Suppose $f \left(x\right) = {x}^{3} + a {x}^{2} + b x + c$ has zeros ${r}_{1} , {r}_{2} , {r}_{3}$

Then:

$f \left(x\right) = \left(x - {r}_{1}\right) \left(x - {r}_{2}\right) \left(x - {r}_{3}\right)$

$\textcolor{w h i t e}{f \left(x\right)} = {x}^{3} - \left({r}_{1} + {r}_{2} + {r}_{3}\right) {x}^{2} + \left({r}_{1} {r}_{2} + {r}_{2} {r}_{3} + {r}_{3} {r}_{1}\right) x - {r}_{1} {r}_{2} {r}_{3}$

We are told that:

$\left\{\begin{matrix}{r}_{1} + {r}_{2} + {r}_{3} = 3 \\ {r}_{1} {r}_{2} + {r}_{2} {r}_{3} + {r}_{3} {r}_{1} = - 1 \\ {r}_{1} {r}_{2} {r}_{3} = - 3\end{matrix}\right.$

Hence:

$f \left(x\right) = {x}^{3} - 3 {x}^{2} - x + 3$

In this particular example, we can check our result by factoring the cubic by grouping:

${x}^{3} - 3 {x}^{2} - x + 3 = \left({x}^{3} - 3 {x}^{2}\right) - \left(x - 3\right)$

$\textcolor{w h i t e}{{x}^{3} - 3 {x}^{2} - x + 3} = {x}^{2} \left(x - 3\right) - 1 \left(x - 3\right)$

$\textcolor{w h i t e}{{x}^{3} - 3 {x}^{2} - x + 3} = \left({x}^{2} - 1\right) \left(x - 3\right)$

$\textcolor{w h i t e}{{x}^{3} - 3 {x}^{2} - x + 3} = \left(x - 1\right) \left(x + 1\right) \left(x - 3\right)$

So we can write:

$\left\{\begin{matrix}{r}_{1} = 1 \\ {r}_{2} = - 1 \\ {r}_{3} = 3\end{matrix}\right.$

and we find:

$\left\{\begin{matrix}{r}_{1} + {r}_{2} + {r}_{3} = 1 - 1 + 3 = 3 \\ {r}_{1} {r}_{2} + {r}_{2} {r}_{3} + {r}_{3} {r}_{1} = \left(1\right) \left(- 1\right) + \left(- 1\right) \left(3\right) + \left(3\right) \left(1\right) = - 1 \\ {r}_{1} {r}_{2} {r}_{3} = \left(1\right) \left(- 1\right) \left(3\right) = - 3\end{matrix}\right.$