Question

find the qubic polynomial with the sum, sum of the product of its zeros taken two at a time and the product of its zeros as 2,-7,-14 respectively​

Answer 1

EXPLANATION.

To find a cubic polynomial.

As we know that,

Sum of the zeroes of the cubic polynomial.

⇒ α + β + γ = - b/a.

⇒ α + β + γ = 2. - - - - - (1).

Products of the zeroes of the cubic polynomial two at a time.

⇒ αβ + βγ + γα = c/a.

⇒ αβ + βγ + γα = - 7. - - - - - (2).

Products of the zeroes of the cubic polynomial.

⇒ αβγ = - d/a.

⇒ αβγ = - 14. - - - - - (3).

As we know that,

Formula of a cubic polynomial.

⇒ x³ - (α + β + γ)x² + (αβ + βγ + γα)x - αβγ.

Put the values in the equation, we get.

⇒ x³ - (2)x² + (-7)x - (-14).

⇒ x³ - 2x² - 7x + 14.

                                                                                                                       

MORE INFORMATION.

Higher Degree Equations.

The equation.

f(x) = α₀xⁿ + α₁xⁿ⁻¹ + . . . . . + αₙ₋₁x + αₙ = 0. - - - - - (1).

Where the coefficient α₀, α₁. . . . . αₙ ∈ C and α₀ ≠ 0 is called an equation of nth degree, which has exactly n roots α₁, α₂ . . . . . αₙ ∈ C, Then we can write.

p(x) = α₀(x - α₁)(x - α₂) . . . . . (x - αₙ) = α₀[xⁿ - (∑α₁)xⁿ⁻¹ + (∑α₁α₂)xⁿ⁻² - . . . . . + (-1)ⁿα₁α₂ . . . . . αₙ]. - - - - - (2).

Comparing equation (1) and (2), we get.

∑α₁ = α₁ + α₂ + . . . . . + αₙ = -α₁/α₀.

∑α₁α₂ = α₁α₂ + . . . . . + αₙ₋₁αₙ = α₂/α₀.

And So on . . . . .

α₁α₂ . . . . . αₙ = (-1)ⁿαₙ/α₀.

Answer 2

Step-by-step explanation:

$\tt\alpha + \beta + \gamma = 2 \\ \tt \alpha \beta + \beta \gamma + \gamma \alpha = - 7 \\ \tt \: \alpha \beta \gamma = - 14 \\ \tt \: Formula \: used \\ \leadsto \: \tt {x}^{3} - ( \alpha + \beta + \gamma ) {x }^{2} +( \alpha \beta + \beta \gamma + \gamma \alpha )x - \alpha \beta \gamma \\ \leadsto \: \tt{x}^{3} - 2 {x }^{2} - 7x + 14$