False
We need to factorize x2 + (a + b)x + ab, by splitting the middle term:
x2 + (a + b)x + ab = x2 + ax + bx + ab [∵, ax + bx = (a + b)x & ax × bx = abx2]
⇒ x2 + (a + b)x + ab = x(x + a) + b(x + a)
⇒ x2 + (a + b)x + ab = (x + a)(x + b)
And (x + a)(x + b) ≠ (a + b)(x + ab)
⇒ x2 + (a + b)x + ab ≠ (a + b)(x + ab)