Question

If Alpha,beta,gama are the zeroes of the polynomial P(x)=x³-6x²-x+30 then the value of (alpha×beta+beta×gama+gama×alpha)+alpha×beta×gama​

Answer 1

$\rule{200}{1}$

*Note:- ω=gamma*

$\rule{200}{1}$

Given:-

α,β and ω are the zeroes of the polynomial, x³-6x²-x+30

To find:-

Value of αβ+βω+ωα+αβω

$\orange{\boxed{\green{\underline{\red{\mathrm{Solution:-}}}}}}$

→we know,

αβ+βω+ωα+αβω=$\frac{c}{a}$ + $\frac{-d}{a}$

→Using this↑ and polynomial we get↓

αβ+βω+ωα+αβω=$\frac{-1}{1}$ + $\frac{-30}{1}$

→αβ+βω+ωα+αβω=-1-30

→αβ+βω+ωα+αβω=-31

$\rule{200}{1}$

If the Zeros of the cubic Polynomial include αβω,then the general form of p(x) are:-

•α+β+ω = $\frac{-b}{a}$

•αβ+βω+ωα=$\frac{c}{a}$

•αβω =$\frac{-d}{a}$

$\rule{200}{1}$

Answer 2

Relation between roots and coefficients:

Let us take

f (x) = a₀xⁿ + a₁xⁿ⁻¹ + . . . + aⁿ

be a polynomial of degree n with real or complex coefficients, where a₀ ≠ 0.

If α₁, α₂, ..., αₙ be the roots of the equation f(x) = 0,

α₁ + α₂ + ... + αₙ = - a₁/a₀

α₁α₂ + α₁α₃ + ... = a₂/a₀

..... ..... ..... .....

α₁α₂ ... αₙ = (- 1)ⁿ aₙ/a₀

Now we solve the given problem:

The given polynomial is

f (x) = x³ - 6x² - x + 30

Since α, β, γ are the zeroes of f (x), we can write

α + β + γ = - (- 6/1) αβ + βγ + γα = - 1/1

or, α + β + γ = 6 ... (i) or, αβ + βγ + γα = - 1 ... (ii)

αβγ = - 30/1

or, αβγ = - 30 ... (iii)

Now, (αβ + βγ + γα) + αβγ

= (- 1) + (- 30), by (ii), (iii)

= - 1 - 30

= - 31

∴ (αβ + βγ + γα) + αβγ = - 31