Find zeroes of polynomial f(x)=x^3-5x^2-2x+24 if it is given that product of its two zeroes is 12.
Answers
Solution :
The given polynomial is
f(x) = x³ - 5x² - 2x + 24
If α, β, γ be the zeroes of f(x), by the relation between zeroes and coefficients, we get
- α + β + γ = - (- 5)/1 = 5 .....(i)
- αβ + βγ + γα = - 2/1 = - 2 .....(ii)
- αβγ = - 24/1 = - 24 .....(iii)
ATQ, we consider αβ = 12 ....(iv)
Putting αβ = 12 in (iii), we get
12γ = - 24
or, γ = - 2
Putting γ = - 2 in (i), we get
α + β - 2 = 5
or, α + β = 7 .....(v)
Now, (α - β)² = (α + β)² - 4αβ
or, (α - β)² = 7² - 4 (12) = 49 - 48 = 1
or, α - β = ± 1 .....(vi)
Adding (v) and (vi), we get
2α = 7 ± 1 = 8, 6
or, α = 4, 3
Then, β = 7 - 4, 7 - 3
or, β = 3, 4
∴ the required zeroes of the given polynomial are 4, 3, - 2
Answer :-
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 4, 3, -2.