Let f(x) = x² + 6x²-x+ 5 then ,
number to be added be P
then,
f(x) = x² + 6x²-x+ 5 +P
According to the qn,
(x+3) is exactly divisible by zero then,
R=0
comparing .. we get a= -3
now by remainder theorm
R=f(a)
0=f(-3)
0=(-3)² + 6(-3)²-(-3)+ 5 + P
0= 9 + 54 + 3 + 5 + P
-71=P
therefore, -71 should be added.
Hope you understand