Question

find the value of p for which the polynomial x^3-3x^2+3x-p is exactly divisible by x-2

Answer 1

Sol : Zero of ( x - 2 ) = 2.

How is the zero of ( x - 2 ) = 2 ?

 x - 2 = 0

 x = 2.

By Remainder theorem ,

p ( x ) = x³ - 3 x² + 3 x - p

p ( 2 ) = 2³ - 3 * ( 2 )² + 3 ( 2 ) - p = 0

             8 - 3 * 4 + 6 - p = 0

             8 - 12 + 6 - p = 0

             14 - 12 - p = 0

             2 - p = 0

                  p = 2.

So , your final answer is that ( p = 2 ).