Verify that 5, -2 and 1/3 are the zeroes of the cubic polynomial p(x) = (3x^3 – 10x^2 – 27x + 10) and verify ​

Question

Verify that 5, -2 and \(\frac{1}3\) are the zeroes of the cubic polynomial p(x) = (3x³ – 10x² – 27x + 10) and verify the relation between its zeroes 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 \((\frac{1}3)\) = \(\{\) 3 x \((\frac{1}3)^3\) - 10 x \((\frac{1}3)^2\) - 27 x \(\frac{1}3\) + 10 \(\}\) = ( 3 x \(\frac{1}{27}\) - 10 x \(\frac{1}9\) - 9 + 10)

= \((\frac{1}9-\frac{10}9+1)\) = \((\frac{1-10-9}9)\) = \((\frac{0}9)\) = 0

∴ 5, –2 and \(\frac{1}3\) are the zeroes of p(x).

Let α = 5, β = –2 and γ =  \(\frac{1}3\). Then we have:

(α +β + γ) = (5 − 2 + \(\frac{1}3\)) = \(\frac{10}3\) = \(\frac{-(coefficient\,of\,x^2)}{(coefficient\,of\,x^3)}\)

(αβ +βγ + γα) = (– 10 – \(\frac{2}3\) + \(\frac{5}3\)) = \(\frac{-27}3\) = \(\frac{coefficient\,of\,x}{coefficient\,of\,x^3}\)

αβγ = \(\{\) 5 x (-2) x \(\frac{1}3\)\(\}\) = \(\frac{-10}3\) = \(\frac{-(constant\,term)}{(coefficient\,of\,x^3)}\)