Question

find the zeroes of the polynomial x√2-x-p(p+1)​

Answer 1

Appropriate Question :-

Find the zeroes of the polynomial

$\rm :\longmapsto\: {x}^{2} - x - p(p + 1)$

$\large\underline{\sf{Solution-}}$

Given polynomial is represented as

$\rm :\longmapsto\:f(x) = {x}^{2} - x - p(p + 1)$

can be rewritten as

$\rm :\longmapsto\:f(x) = {x}^{2} - x(1) - p(p + 1)$

$\rm :\longmapsto\:f(x) = {x}^{2} - x(p + 1 - p) - p(p + 1)$

$\rm :\longmapsto\:f(x) = {x}^{2} - x(p + 1) + px - p(p + 1)$

$\rm :\longmapsto\:f(x) = x\bigg(x - (p + 1) \bigg) + p\bigg( x - (p + 1)\bigg)$

$\rm :\longmapsto\:f(x) = (x + p)\bigg(x - (p + 1) \bigg)$

$\rm :\longmapsto\:f(x) = (x + p)(x - p - 1)$

$\bf\implies \:Zeroes \: of \: f(x) \: = - p, \: p + 1$

Additional Information :-

$\rm :\longmapsto\: \alpha , \beta \: are \: zeroes \: of \: {ax}^{2} + bx + c \: then \:$

$\boxed{ \sf{ \: \alpha + \beta = - \: \frac{b}{a}}}$

and

$\boxed{ \sf{ \: \alpha \beta = \: \frac{c}{a}}}$

Also,

$\boxed{ \sf{ \: { \alpha }^{2} + { \beta }^{2} = {( \alpha + \beta )}^{2} - 2\alpha \beta \: }}$

$\boxed{ \sf{ \: { \alpha }^{3} + { \beta }^{3} = {( \alpha + \beta )}^{3} - 3\alpha \beta( \alpha + \beta ) \: }}$