**Correct option (c) a+c and b+c**

**Explanation:**

x^{2 }– (a + b)x + ab + c = 0

α + β = a + b and α β = ab + c

Now (x – c – α)(x – c – β ) = c

*⇒ *(x–c)^{2} – (α + β)(x – c) + α β – c = 0

(x – c)^{2 }– (a + b)(x – c) + ab = 0

(x – c)^{2}– a(x – c) – b(x – c) + ab = 0

*⇒ *(x – c – a)(x – c – b) = 0

x = c + a and b + c