Question

if p (x)is equals to x cube minus x square + X + 1 find value of P (- 1)+ p(1) / 2

Answer 1

Solution :

p(x) = x³ - x² + x + 1

i ) p(-1) = (-1)³ - ( -1 )² + ( -1 ) + 1

= -1 - 1 - 1 + 1

= -2

ii ) p( 1 ) = 1³ - 1² + 1 + 1


= 1 - 1 + 1 + 1

= 2

Now ,

[ p(-1) + p(1) ]/2

= [ -2 + 2 ]/2

= 0

••••

Answer 2

p ( x ) = x³ - x² + x + 1

p ( - 1 )

Put x = - 1

==> p ( - 1 ) = ( - 1 )³ - ( - 1 )² + ( - 1 ) + 1

                 = - 1 - 1 - 1 + 1

                = - 2 .....................(1)

p ( 1 )

p ( 1 ) = 1³ - 1² + 1 + 1

         = 1 - 1 + 2

          = 2............................(2)

[ p ( - 1 ) + p ( 1 ) ] / 2

==> [ 2 - 2 ] / 2 [ from (1) and (2) ]

==> 0 / 2

==> 0

The value is 0

Hope it helps :-)

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