Question

if alpha and beta are zeroes of the polynomial x^2-p(x+1)-c, then find the value of (alpha+1)(beta+1)​

Answer 1

Answer:

1 - c

Step-by-step explanation:

Given polynomial : x² - p( x + 1 ) - c

= x² - px - p - c

α, β are the zeroes

Comparing x² - px - p - c with ax² + bx + c we get,

• a = 1
• b = - p
• c = - p - c

Sum of zeroes = α + β = - b / a = - ( - p ) = p

Product of zeroes = αβ = c / a = ( - p - c ) / 1 = - p - c

( α + 1 )( β + 1 )

= α( β + 1 ) + 1( β + 1 )

= αβ + α + β + 1

= ( - p - c ) + p + 1

= - p - c + p + 1

= 1 - c

Hence the value of ( α + 1 )( β + 1 ) is ( 1 - c ).