Question

find the zeros of the polynomial f(x)=x3-5x2-2x+24 if it is GIven that product of its two zeros is 12

Answer 1

Hey dear!!!
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==> In the question

f(x) = x³ - 5x² - 2x + 24

And we have given that the product of it's two zeroes is 12 .

So, let the zeroes of the given cubic polynomial be φ , β and γ .

From the given condition we have,

φβ = 12 ..................(1)

and also we have ,

φ + β + γ = coefficient of x²/coefficient of x³ = 5 ......................(2)

φβγ = - constant term/ coefficient of x³ = -24 ...........................(3)

Substituting the value of φβ in equation no. (3) we get,

φβγ = -24

12γ = -24

γ = -24/12

γ = -2 ......................(4)

Putting the value of γ = -2 in equation no. (2) we get

φ + β + γ = 5

φ + β + (-2) = 5

φ + β = 5 +2

φ + β = 7 ...................(5)

Now, squaring on both sides we will get ,

(φ + β)² = (7)²

We know the expression of quadratic polynomial [ (φ + β)² = (φ - β)²+ 4φβ) ]

∴ ( φ - β )² + 4*12 = 49 [∵ φβ = 12 ]

(φ - β)² + 48 = 49

( φ - β)² = 49 - 48

(φ - β)² = 1

∴ φ - β = 1 .................(6)

Now, we will add the equation no. (5) and equation no. (6) we get,

φ + β = 7
φ - β = 1
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2φ = 8

φ = 8/2

φ = 4

Putting φ = 4 in equation no. (5) we get,

φ - β = 1

4 - β = 1

- β = 1 - 4

-β = - 3 .......(multiplying by -1 on both side)

β = 3

∴ The zeroes are φ, β, γ = 4, 3, -2


Thanks !!!! ✌✌✌✌☺

[ Be Brainly ]

Answer 2

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 .

$\therefore$ β = 3 .

∴ The zeroes are α, β, γ = 4, 3, -2 .

Hence, it is solved .