1. Find the zero’s of the …

f(x) = x³ - 5x² - 16x + 80
Let {tex} \alpha , \beta , \gamma{/tex} be the zeroes of polynomial f(x) such that {tex} \alpha + \beta{/tex}= 0. Then,
Sum of the zeroes ={tex} - \frac { \text { Coefficient of } x ^ { 2 } } { \text { Coefficient of } x ^ { 3 } }{/tex}
{tex} \Rightarrow \quad \alpha + \beta + \gamma = - \left( - \frac { 5 } { 1 } \right){/tex}
{tex} \Rightarrow \quad 0 + \gamma = 5{/tex} ( ∵{tex} \alpha + \beta{/tex} = 0 )
{tex} \Rightarrow \quad \gamma = 5{/tex}
Product of the zeroes = {tex} - \frac { \text { Constant term } } { \text { Coefficient of } x ^ { 3 } }{/tex}
{tex} \Rightarrow \quad \alpha \beta \gamma = - \frac { 80 } { 1 }{/tex}
{tex} \Rightarrow \quad 5 \alpha \beta = - 80{/tex}
{tex} \Rightarrow \quad \alpha \beta = - 16{/tex}
{tex} \Rightarrow \quad - \alpha ^ { 2 } = - 16{/tex}
{tex} \Rightarrow \quad \alpha = \pm 4{/tex}
Case I: When {tex} \alpha{/tex} = 4 : In this case,
{tex} \alpha + \beta = 0 {/tex}
{tex}\Rightarrow 4 + \beta = 0 {/tex}
{tex}\Rightarrow \beta = - 4{/tex}
So, the zeroes are {tex} \alpha = 4 , \beta = - 4 \text { and } \gamma = 5{/tex}
Case II: When {tex} \alpha{/tex} = -4 : In this case,
{tex} \alpha + \beta = 0{/tex}
{tex} \Rightarrow - 4 + \beta = 0 {/tex}
{tex}\Rightarrow \beta = 4{/tex}
So, the zeroes are {tex} \alpha = -4 , \beta = 4 \text { and } \gamma = 5{/tex}
Hence, in either case the zeroes are (4, -4 and 5) and (-4, 4,5).