Question

2p3 – xp2 –yp-14 has( p+2) as a factor and leaves a remainder - 26 when divided by ( p+3),find x and y.

Answer 1

is a factor of (x-1) and (x-2) then remainder is p(1)=0 and p(2)=0.

P(1)=1+10+p+q

         11+p+q=0-----(1)

P(-2)=-8+10(4)+(-2)p+q

          -8+40-2p+q=0

             32-2p+q=0..........(2)

Adding eq:1 and 2,

11+p+q=0

32-2p+q=0

-      +      -

-21+3p=0

3p=21

P=7

Substitute p value in equation 1

11+p+q=0

11+7+q=0

18+q=0

q=-18

P=7 and q=-18