Question

If α, β are the zeroes of the quadratic polynomial px² + qx + r, then find the value of α³β + β³α.

Answer 1

Given:

Quadratic polynomial px² + qx + r

α, β are the zeroes of the polynomial.

From the relationship between zeroes and coefficients:

Sum of zeroes `(α + β) = - q/p`

Product of zeroes `(αβ) = r/p`

To find: α³β + β³α

Taking αβ common:

α³β + β³α = αβ(α² + β²)

Using the identity α² + β² = (α + β)² – 2αβ:

= αβ[(α + β)² – 2αβ]

Substituting the values of (α + β) and αβ:

= `r/p [(-q/p)^2 - 2(r/p)]`

= `r/p [q^2/p^2 - (2r)/p]`

= `r/p [(q^2 - 2pr)/p^2]`

= `(r(q^2 - 2pr))/p^3`

Therefore, `α^3β + β^3α = (r(q^2 - 2pr))/p^3`.