Question

if alpha and beta are the zeroes of the polynomial x2-px+q then find the value of alpha3 beta2+alpha2 beta23

Answer 1

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

$\rm :\longmapsto\: \alpha \: and \: \beta \: are \: zeroes \: of \: {x}^{2} - px + q$

$\large\underline{\sf{To\:Find - }}$

$\rm :\longmapsto\: { \alpha }^{3} { \beta }^{2} + { \alpha }^{2} { \beta }^{3}$

8

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

Given that

$\rm :\longmapsto\: \alpha \: and \: \beta \: are \: zeroes \: of \: {x}^{2} - px + q$

We know,

$\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

$\rm :\implies\: \alpha + \beta = - \: \dfrac{( - p)}{1} = p - - (1)$

And

$\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}$

$\rm :\implies\: \alpha \beta = \dfrac{q}{1} = q - - - (2)$

Now,

Consider,

$\rm :\longmapsto\: { \alpha }^{3} { \beta }^{2} + { \alpha }^{2} { \beta }^{3}$

$\rm \: \: = \: \: { \alpha }^{2} { \beta }^{2}( \alpha + \beta)$

$\rm \: \: = \: \: {(\alpha \beta) }^{2}( \alpha + \beta)$

$\rm \: \: = \: \: {q}^{2}p$

Hence,

$\rm :\longmapsto\: { \alpha }^{3} { \beta }^{2} + { \alpha }^{2} { \beta }^{3} = {pq}^{2}$

Additional Information :-

$\red{\rm :\longmapsto\: \alpha, \: \beta, \: \gamma \: are \: zeroes \: of \: {ax}^{3} + {bx}^{2} + cx + d \: then}$

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

$\boxed{ \sf{ \: \alpha \beta + \beta \gamma + \gamma \alpha = \: \dfrac{c}{a}}}$

$\boxed{ \sf{ \: \alpha \beta \gamma = - \: \dfrac{d}{a}}}$