Question

Verify that 3, –2, 1 are the zeros of the cubic polynomial p(x) = (x³ – 2x² – 5x + 6) and verify the relation between it zeros and coefficients.

Answer 1

The given polynomial is p(x) = (x³ – 2x² – 5x + 6)

∴ p(3) = (3³ – 2 × 3² – 5 × 3 + 6)

= (27 – 18 – 15 + 6)

= 0

p(–2) = [(–2³) – 2 × (–2)² – 5 × (–2) + 6] 

= (–8 – 8 + 10 + 6) 

= 0

p(1) = (1³ – 2 × 1² – 5 × 1 + 6)

= (1 – 2 – 5 + 6) 

= 0 

∴ 3, –2 and 1 are the zeroes of p(x),

Let α = 3, β = –2 and γ = 1.

Then we have: 

(α + β + γ) = (3 – 2 + 1)

= 2

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

(αβ + βγ + γα) = (–6 – 2 + 3)

= `(-5)/1`

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

αβγ = {3 × (–2) × 1}

= `(-6)/1`

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