Question

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SOLVE THIS QUES...


Find the zeros of the polynomial
$f(x) = x {}^{3} - 5 x {}^{2} - 2 x + 24$
, if it is given the product of of its two zeros is 12.


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Answer 1

$\huge \bold{hello \: mate}$

Let alpha, beta and gamma are the three zeroes of the polynomial.

We know that

Sum of the zeroes:

$\alpha + \beta + \gamma = - ( - 5) = 5 - - - - - (1)$

Product of the zeroes:

$\alpha \beta \gamma = - ( - 24) = 24$

Now, it is given that product of the two zeroes is 12, so

$12 \gamma = - 24$

$\gamma = - 2$

Now, substituting this value in (1), we get

$\alpha + \beta = 5 + 2$

$\alpha + \beta = 7 - - - - - (2)$

$( \alpha + \beta ) ^{2} = ( \alpha - \beta ) ^{2} + 4 \alpha \beta$

Substituting the obtained values,

$(7) ^{2} = ( \alpha - \beta ) ^{2} + 4(12)$

$49 = ( \alpha - \beta ) ^{2} + 48$

$49 - 48 = ( \alpha - \beta ) ^{2}$

$1 = ( \alpha - \beta ) ^{2}$

$\alpha - \beta = 1 - - - - (3)$

Adding equations (2) and (3),

$2 \alpha = 8 \\ \alpha = 4$

Substituting the above value in (2)

$\beta = 3$

Also

$\gamma = - 2$

Hence, the zeroes are 4, 3, -2.

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 .