Question

Verify that 5, –2 and `1/3` are the zeroes of the cubic polynomial p(x) = (3x³ – 10x² – 27x + 10) and verify the relation between its zeros and coefficients.

Answer 1

p(x) = (3x³ – 10x² – 27x + 10)

p(5) = (3 × 5³ – 10 × 5² – 27 × 5 + 10) 

= (375 – 250 – 135 + 10)

= 0

p(–2) = [3 × (–2³) – 10 × (–2²) – 27 × (–2) + 10] 

= (–24 – 40 + 54 + 10)

= 0 

`p(1/3) = {3 xx (1/3)^3 - 10 xx (1/3)^2 - 27 xx 1/3 + 10}`

= `(3 xx 1/27 - 10 xx 1/9 - 9 + 10)` 

= `(1/9 - 10/9 + 1)`

= `((1 - 10 - 9)/9)`

= `(0 / 9)`

= 0 

∴ 5, –2 and `1/3` are the zeroes of p(x).

Let α = 5, β = –2 and γ = `1/3`.

Then we have:   

`(α + β + γ) = (5 - 2 + 1/3)` 

= `10/3`

= `(-("Coefficient of"  x^2))/(("Coefficient of"  x^3))` 

`(αβ + βγ + γα) = (-10 - 2/3 + 5/3)`

= `(-27)/3`

= `("Coefficient of"  x)/("Coefficient of""  x^3)`

`αβγ ={5 xx (-2) xx 1/3}`

= `(-10)/3`

= `(-("Constant term"))/(("Coefficient of"  x^3))`