Question

The given zeroes of cubic polynomial $x^3$ −𝟕𝒙 −𝟔 are -2, -1, and 3. Verify the relationship between the zeroes and the coefficients.

Answer 1

Given that,
"α = −2, β = −1, γ = 3"

On comparing the given polynomial with standard form, we get,
a =1, b = 0, c = -7, d = -6,

α + β + 𝛾 = -2 + (-1) + 3,
α + β + 𝛾 = 0,
But, α + β + 𝛾 = $\frac{-b}{a}$ = $\frac{-0}{1}$ = 0,

Now,
𝛼𝛽 + 𝛽𝛾 + 𝛾𝛼 = (-2)×(-1) + (-1)×(3) + (3)×(-2) = 2 - 3 - 6 = -7,
But, 𝛼𝛽 + 𝛽𝛾 + 𝛾𝛼 = $\frac{c}{a}$ = $\frac{-7}{1}$ = -7,

Again,
𝛼𝛽𝛾 = (-2)×(-1)×(3) = 6,
But, 𝛼𝛽𝛾 = $\frac{-d}{a}$ = $\frac{-(-6)}{1}$ = 6.

The relationship between the zeroes and coefficients of the polynomial is verified.