(a) 1, 0
Let f (x) = x3 + 2x2 + 4x + b.
When divided by (x + 1), the remainder = f (–1)
Given, remainder = – 3 + 2b
∴ –3 + 2b = f (–1) = (–1)3 + 2 (–1)2 + 4 (–1) + b
⇒ –3 + 2b = –1 + 2 – 4 + b
⇒ –3 + 2b = –3 + b.
This is only possible when b = 0.
∴ f (x) = x3 + 2x2 + 4x.
Now dividing f (x) by (x + 1), we see that
∴ Quotient = x2 + ax + 3 = x2 + x + 3 ⇒ a = 1.