Answer :-
→ The zeroes are α, β, γ = 4, 3, -2 .
Step-by-step explanation :-
Given :-
→ f(x) = x³ - 5x² - 2x + 24
→ The product of its two zeroes is 12 .
To Find :-
→ Zeroes of polynomial [ α, β, γ ] .
Solution :-
→ Let the zeroes of the given cubic polynomial be α , β and γ .
From the given condition we have,
∵ αβ = 12 ..................(1) .
and also we have an identity ,
∵ α + β + γ = - coefficient of x²/coefficient of x³ = -(-5)/1 = 5 ................(2).
∵ αβγ = - constant term/ coefficient of x³ = -24 ......................(3) .
Putting the value of αβ in equation (3), we get
∵ αβγ = -24 .
⇒ 12γ = -24 .
⇒ γ = -24/12 .
∴ γ = -2 ......................(4) .
Putting the value of γ = -2 in equation (2), we get
∵ α + β + γ = 5 .
⇒ α + β + (-2) = 5 .
⇒ α + β = 5 + 2 .
⇒ α + β = 7 ................(5) .
Now,
→ Squaring on both sides, we get
∵ (α + β)² = (7)²
We know the identity [ (α + β)² = (α - β)²+ 4αβ) ]
∴ ( α - β )² + 4 × 12 = 49. [∵ αβ = 12 ]
⇒ (α - β)² + 48 = 49 .
⇒ ( α - β)² = 49 - 48 .
⇒ (α - β)² = 1 .
∴ α - β = 1 ...............(6) .
Now, add in equation (5) and (6), we get
α + β = 7
α - β = 1
+. - .....+
----------------
⇒ 2α = 8 .
⇒ α = 8/2 .
∴ α = 4 .
Putting α = 4 in equation (5), we get
∵ α - β = 1 .
⇒ 4 - β = 1 .
⇒ - β = 1 - 4 .
⇒ -β = - 3 .
β = 3 .
∴ The zeroes are α, β, γ = 4, 3, -2 .
