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:

-5

Step-by-step 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 $(-2m^{3}-m+m^{2}-4)$.