Question

7. Find the zeroes of the polynomial f(x)=x^3+5x²+2x+24, if it is given that the product of its two
zeroes is 12.
4Marks​

Answer 1

→ 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 .

Answer 2

Answer:

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