Remainder Theorem

When divided by (x – 3) the polynomial x³ - px² + x + 6 and 2x³- x² – (p+3) x – 6   leaves the same remainder, then find the value of p?                   ICSE 2010

Answer:

Let: F(x) = x³ - px² + x + 6

And G(x) = 2x³- x² – (p+3) x – 6   

Let the Remainder be y

Therefore, (F(x) – y) and (G(x) – y) are completely divisible by (x – 3)

Therefore, F(3) – y = 0 and G(3) – y = 0, as per the remainder theorem.

On substitution in ( F(3) – y), we get ;

27 - 9p + 3 + 6 - y = 0 

Or, 9p + y = 36    ----------------------------------(1)

On substitution in (G(3) – y) we get  ;

54 – 9 – (p + 3)3 – 6 – y = 0   

Or, 3p + y = 30    ----------------------------------- (2)

Equating (1) and (2), we get

6p = 6

Or, p = 1