Question

p(x)=kx^3+3x^2-3 and q(x)=2x^3-5x+k when divided by x-4leave the same remainder in each case find the value of k

Answer 1

This is the answer i think

Answer 2

Since both leaves the same remainder, p(x) =q(x)
p(4)=q(4)
f(x)=x-4=0
x=4
k*4*4*4+3*4*4-3=2*4*4*4-5*4+k
64k+48-3=128-20+k
63k=108-45
63k=63
or,k=1.(Answer)

HOPE THIS HELPS!