Question

Find the zeros of the polynomial $f\left( x \right) = {x^3} - 5{x^2} - 2x + 24$ if it is given that the product of its two zero is 12.

Answer 1

Hint:
If $f\left( x \right) = {x^3} - 5{x^2} - 2x + 24$ let $\alpha ,\beta ,\gamma $ roots of the equation. Then $\alpha + \beta + \gamma = - \dfrac{{ - 5}}{1} = 5$,$\alpha \beta + \alpha \gamma + \beta \gamma = \dfrac{{ - 2}}{1} = - 2$and $\alpha \beta \gamma = - 24$ it is also given than $\alpha \beta = 12$ then $\gamma = - 2$ from this and three equation we can calculate the rest roots of the equation.

Complete step by step solution:
Given, $f\left( x \right) = {x^3} - 5{x^2} - 2x + 24$ and product of 2 roots are 12.
$\alpha \beta = 12$ given
We know that $\alpha \beta \gamma = - 24$
Hence $\gamma = - 2$
Also $\alpha \beta + \alpha \gamma + \beta \gamma = \dfrac{{ - 2}}{1} = - 2$ and $\alpha + \beta + \gamma = - \dfrac{{ - 5}}{1} = 5$
Substituting $\gamma = - 2$ then
$\alpha \beta - 2\alpha - 2\beta = - 2$ and $\alpha + \beta - 2 = 5$
$\alpha + \beta - 2 = 5$
$\alpha = 7 - \beta $
$\alpha \beta - 2\alpha - 2\beta = - 2$ substituting $\alpha = 7 - \beta $
$\left( {7 - \beta } \right) \beta - 2\left( {7 - \beta } \right) - 2\beta = - 2$
${\beta ^2} - 7\beta + 12 = 0$
Then ${\beta ^2} - 4\beta - 3\beta + 12 = 0$
$\left( {\beta - 4} \right)\left( {\beta - 3} \right) = 0$
$\beta = 4,3$
Then $\alpha = 3, 4$

Hence roots of the equation are 4, 3, -2.

Note:
If $f\left( x \right) = a{x^3} + b{x^2} + cx + d$ and the roots of equation is $\alpha ,\beta ,\gamma $ then $\alpha + \beta + \gamma = - \dfrac{b}{a}$ $\alpha \beta + \alpha \gamma + \beta \gamma = \dfrac{c}{a}$ and $\alpha \beta \gamma = \dfrac{{ - d}}{a}$