(c) –9, 20, –12
Given, f(x) = x3 + lx2 + mx + n.
f(1) = f(2) = 0 ⇒ (x – 1) and (x – 2) are factors of f(x).
Since, f(x) is polynomial of degree 3, it shall have three linear factors. So, let the third factor be (x – k).
Then, f(x) = (x – 1) (x – 2) (x – k)
⇒ f(x) = x3 + lx2 + mx + n = (x – 1) (x – 2) (x – k)
Given, f(4) = f(0)
⇒ (4 – 1) (4 – 2) (4 – k) = (–1) (–2) (–k)
⇒ 24 – 6k = – 2k ⇒ 4k = 24 ⇒ k = 6
∴ f(x) = (x – 1) (x – 2) (x – 6) = (x2 – 3x + 2) (x – 6)
= x3 – 9x2 + 20x – 12
∴ x3 + lx2 + mx + n = x3 – 9x2 + 20x – 12
⇒ l = – 9, m = 20, n = – 12.