Q)  If a, b are the zeroes of polynomial f(x) = x² – p (x + 1) – c then show that (a + 1) (b + 1) = 1 – c

Ans:

Step 1: In the given polynomial equation f(x), to find zeroes, we will start with f(x) = 0

Therefore, x² – p (x + 1) – c = 0

x² – p x – p – c = 0

x² – p x – (p + c) = 0

Step 2: Comparing it with standard quadratic equation, A x² + B x + C, we get

A = 1, B = – p; C = – p – c)

Step 3: Since, sum of the zeroes = –

∴ a + b = – = p

Since, product of the zeroes =

∴ a b = = – p – c)

By substituting value of p = (a + b) from equation (i), we get:

a b = – (a + b) – c

∴ ab = – a – b – c

∴ ab + a + b = – c

By adding 1 on both sides we get:

ab + a + b + 1 = – c + 1

∴ (a + 1)(b + 1) = 1 – c

Hence proved!

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