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If a,b are the zero of the polynomial (x)=x2p(x-1)-c,then(a+1) (b+1)

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If alpha and beeta are zeroes of polynomial x2-p(x+1)+C such that (alpha+1)

If alpha and beeta are zeroes of polynomial x2-p(x+1)+c such that (alpha+1)(beeta+1)=0. Then find the value of c.

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A)

Given that alpha and beta are the roots of the QUADRATIC EQUATION f(x) = x^2-p(x+1)-c = x^2-px-p-c = x^2 -px-(p+c),

comparing with ax^2 + bx + c, we have, a =1 , b= -p & c= -(p+c)

alpha+beta = -b/a = -(-p)/1 = p

& alpha*beta = c/a = -(p+c)/1 = -(p+c)

Therefore, (Alpha + 1)*(beta+1)

= Alpha*beta + alpha + beta + 1

= -(p+c) + p + 1

= -p-c+p+1

= 1-c

or c=1



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