Question

If f(p) divided by x-p and x-q have the same remainder

when f(a)=a^3+5a^2+6a+8 and p≠q,
prove that, p^2+q^2+pq+5p+5q+6=0.

Please help me... Show it step-by-step...

Answer 1

Hello,


x^2-y^2=(x+y)(x-y)
x^3-y^3=(x-y)(x²+xy+y²)

Let's use Horner's division

.........|a^3|a^2.|a^1..........|a^0
.........|1....|5....|6..............|8....
a=p...|......|p....|5p+p^2....|6p+5p^2+p^3
----------------------------------------------------------
.........|1....|5+p|6+5p+p^2|8+6p+5p^2+p^3

The remainder is 8+6p+5p^2+p^3 or 8+6q+5q^2+q^3


Thus:
8+6p+5p^2+p^3 = 8+6q+5q^2+q^3
==>p^3-q^3+5p^2-5q^2+6p-6p=0
==>(p-q)(p²+pq+q²)+5(p-q)(p+q)+6(p-q)=0
==>(p-q)[p²+pq+q²+5p+5q+6]=0 or p≠q
==>p²+pq+q²+5p+5q+6=0

And here, Mehek are there sufficients explanations?