k₁ = 2, k₂ = –3

f(x) = x² + k₁x + k₂

(x – 1) is a factor of f(x) .

∴    f(1) = 0

⇒ 1 + k₁ + k₂ = 0

⇒ k₁ + k₂ = –1           ...(i)

Again,

f(–3) = 0    ⇒ (–3)² + k₁(–3) + k₂ = 0

⇒ 9 – 3k₁ + k₂ = 0   ⇒    3k₁ – k₂ = 9           ...(ii)

On adding both equations,

4k₁ = 8

⇒  k₁ = 2

From eq.(i) ,

k₁ + k₂ = –1   ⇒  2 + k₂ = –1   

⇒  k₂ = –3.