2019/018) $x^2 + bx + a$ leaves same remainder when divided by x + 2 or x - a (where a ≠ -2). Show that a + b = 2.

We have
$f(x) = x^2 + bx +a$
Same remainder when divided by x + 2 or (x-a)

So $f(-2) = 4- 2b + a = f(a) =  a^2 +ab +a$
Or 4 - 2b = a^2 + ab
Or 2b+ ab = 4 - a^2
Or b(2+a) = (2-a)(2+a)
As a is not -2 so we have b = 2-a or a + b = 2

Alternatively as same remainder when divided by x+ 2 and x -a and it is degree 2 polynomial

We have $f(x) = x^2 + bx + a = (x+2)(x-a) +c $ where  c is a constant
Or  $f(x) = x^2 + bx + a = x^2+(2-a)x + c - 2a$
Comparing coefficient of x we have b= 2- a or a+b =2