IB Mathematics HL: When x^2+4x-b is divided by x-a the remainder is 2. Given that a, b∈R , find the smallest possible value for b.

Answer 1

The smallest possible value for b is -6

According to the Remainder Theorem:

When we divide a polynomial f(x) by x-c the remainder is f(c)

Then, for the function: x^2+4x-b  

When you substitute with x=a , fa=2 as follows:

a^2+4a-b=2

Then, b=a^2+4a-2 , completing the square 

b=(a+2)^2-6 

Then, b≥-6

Then, the smallest possible value for b is -6