Find the zeros of polynomial x3 …

Let {tex} \alpha , \beta , \gamma{/tex} be the zeroes of polynomial f(x) = x³ - 5x² - 2x + 24 such that {tex} \alpha\beta{/tex} = 12.
{tex} \alpha + \beta + \gamma = - \left( - \frac { 5 } { 1 } \right) = 5{/tex} ................ (i)
{tex} \alpha \beta + \beta \gamma + \gamma \alpha = \frac { - 2 } { 1 } = - 2{/tex}
and, {tex} \alpha \beta \gamma = - \frac { 24 } { 1 } = - 24{/tex}
Putting, {tex} \alpha \beta = 12{/tex} in {tex} \alpha \beta \gamma = - 24{/tex}, we get
{tex} 12 \gamma = - 24{/tex}
{tex} \Rightarrow \gamma = - \frac { 24 } { 12 } = - 2{/tex}
Putting {tex} \gamma= -2{/tex} in eq.(i), we get
{tex} \alpha + \beta - 2 = 5{/tex}
{tex} \Rightarrow \quad \alpha + \beta = 7{/tex}
Now, {tex} ( \alpha - \beta ) ^ { 2 } = ( \alpha + \beta ) ^ { 2 } - 4 \alpha \beta{/tex}
{tex} \Rightarrow \quad ( \alpha - \beta ) ^ { 2 } = 7 ^ { 2 } - 4 \times 12{/tex}
{tex} \Rightarrow \quad ( \alpha - \beta ) ^ { 2 } = 1{/tex}
{tex} \Rightarrow \quad \alpha - \beta = \pm 1{/tex}
Thus, we have
{tex} \alpha + \beta= 7{/tex}  and {tex} \alpha - \beta = 1 {/tex} or, {tex} \alpha + \beta= 7{/tex} and {tex} \alpha - \beta= -1{/tex}
CASE I: When {tex} \alpha + \beta= 7{/tex} and {tex} \alpha - \beta= 1{/tex}
Solving {tex} \alpha + \beta= 7{/tex} and {tex} \alpha - \beta= 1{/tex} , we get
{tex} \alpha = 4{/tex} and {tex}\beta= 3{/tex} 
CASE II: When {tex} \alpha + \beta= 7{/tex} and {tex} \alpha - \beta= -1{/tex}
Solving {tex} \alpha + \beta= 7{/tex} and {tex} \alpha - \beta= -1{/tex} , we get
{tex} \alpha = 3{/tex} and {tex}\beta= 4{/tex} .
Hence, the zeros of the given polynomial are 3, 4 and - 2 or 4,3 , -2.