Question

The smallest integer that can be added to -2m3-m+m2+1 to make it completely divisible by m+1 is

Answer 1

Answer with explanation:

P(m)= -2 m³- m + m²+1

= -2 m³+ m²-m +1--------Dividend

Q(m) = m +1-------Remainder

We have to find smallest Integer added to P(m), so that it can be divisible by m+1.

→Q(m)=0

Put,→ m+1=0

→m = -1

⇒P(-1)=-2 × (-1)³+(-1)² -(-1)+1

=-2 × (-1) +1 +1+1

=2+3

→5 ≠ 0

→Additive Inverse of 5= -5

→Because→ (5) +(-5)=0

Hence smallest positive Integer added to P(m) so that it can be completely divisible by Q(m) is -5.

Answer 2

-5

Explanation:

By synthetic division, polynomial long division or evaluating the function at x=-1, you find the remainder from division by m+1 to be 5. In order to make the remainder be zero, -5 must be added to the the expression.