Roots of Polynomial Equations

Form new equation

Form New Equation

(1)   $$(x - \alpha)(x - \beta)(x - \gamma) = 0$$

If $$\alpha, \beta$$ and $$\gamma$$ are known.

(2)   $$x^3 - (\alpha + \beta + \gamma) \:.\: x^2 + (\alpha \:.\: \beta + \alpha \:.\: \gamma + \beta \:.\: \gamma) \:.\: x - (\alpha \:.\: \beta \:.\: \gamma) = 0$$

If relations of $$\alpha, \beta$$ and $$\gamma$$ are known.

Example

Find the equation which roots are −4, 3 and 5.

First method

\begin{equation*} \begin{split} & (x - \alpha)(x - \beta)(x - \gamma) = 0 \\\\ & (x + 4)(x - 3)(x - 5) = 0 \\\\ & (x^2 + x - 12)(x - 5) = 0 \\\\ & x^3 - 4x^2 - 17x + 60 = 0 \end{split} \end{equation*}

Second method

$$\alpha + \beta + \gamma$$

$$-4 + 3 + 5$$

$$4$$

$$\alpha \:.\: \beta + \alpha \:.\: \gamma + \beta \:.\: \gamma$$

$$-4 \:.\: 3 + -4 \:.\: 5 + 3 \:.\: 5$$

$$- 17$$

$$\alpha \:.\: \beta \:.\: \gamma$$

$$-4 \:.\: 3 \:.\: 5$$

$$- 60$$

Equation

\begin{equation*} \begin{split} & x^3 - (\alpha + \beta + \gamma)x^2 + (\alpha \:.\: \beta + \alpha \:.\: \gamma + \beta \:.\: \gamma)x - (\alpha \:.\: \beta \:.\: \gamma) = 0 \\\\ & \bbox[5px, border: 2px solid magenta] {x^3 - 4x^2 - 17x + 60 = 0} \end{split} \end{equation*}