Question

if alpha and beta are the zeroes of the polynomial f(x) =x²-p(x-1)-c, then ( alpha +1) ( beta +1) is... answer 1- c​

Answer 1

Appropriate Question :-

• if alpha and beta are the zeroes of the polynomial f(x) = x² - p(x + 1) - c, then ( alpha +1) ( beta +1) is _________

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

Given that

$\rm :\longmapsto\: \alpha,\beta \: are \: zeroes \: of \: {x}^{2} - p(x + 1) - c$

can be rewritten as

$\rm :\longmapsto\: \alpha,\beta \: are \: zeroes \: of \: {x}^{2} - px - p - c$

We know that

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

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

and

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

$\rm :\implies\: \alpha \beta = \dfrac{ - p - c}{1} = - p - c$

Now, Consider

$\rm :\longmapsto\:(\alpha+1)(\beta+1)$

$\rm \:  =  \:  \: \alpha \beta + \alpha + \beta + 1$

$\rm \:  =  \:  \: - p - c + p + 1$

$\rm \:  =  \:  \: 1 - c$

Hence, Proved

Additional Information :-

$\rm :\longmapsto\: \alpha,\beta, \gamma \: are \: zeroes \: of \: a {x}^{3} + b {x}^{2} + cx + d, \: then$

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

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

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