Question

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

Answer 1

-5

Explanation:

Let's find the answer by dividing
`(-2m^(3)-m+m^(2)+1)` by
`(m+1)`, like this:

`(-2m^(2))*(m+1)=-2m^(3)-2m^(2)` and:

`(-2m^(3)-m+m^(2)+1)-(-2m^(3)-2m^(2))=3m^(2)-m+1` then:

`(3m)*(m+1)=3m^(2)+3m` and:

`(3m^(2)-m+1)-(3m^(2)+3m)=-4m+1` then:

`(-4)*(m+1)=-4m-4` and:

`(-4m+1)-(-4m-4)=5` notice that the remainder is 5 so we need to subtract the remainder.

Based on the previous procedure we can define:

`(-2m^(3)-m+m^(2)+1)/(m+1)=(-2m^(2)+3m-4) + 5`

In conclusion the smallest integer that can be added to the polynomial is -5, so the polynomial will be
`(-2m^(3)-m+m^(2)-4)`.