Question

verify that 3,-1,-1/3 are the zeroes of the cubic polynomial p(x)=3x2-5x2-11x-3 and then verify the relationship between zeroes and cofficients.

Answer 1

Comparing the given polynomial with ax3 + bx2 + cx + d, we get
a = 3, b = – 5, c = –11, d = – 3. Further
p(3) = 3 × 33 – (5 × 32) – (11 × 3) – 3 = 81 – 45 – 33 – 3 = 0,
p(–1) = 3 × (–1)3 – 5 × (–1)2 – 11 × (–1) – 3 = –3 – 5 + 11 – 3 = 0,
p(1/3)= 3 × [(-1/3)3] − 5 ×[(-1/3)2] − 11 × [-(1/3)]-3
= –(1/9)–(5/9)+(11/3)–3 =–(2/3) + (2/3) = 0
Therefore, 3, -1 and -(1/3) are the zeroes of 3x3 – 5x22 – 11x – 3.
So, we take α = 3, β = –1 and γ = −(1/3)
Now,
α + β + γ = 3 + (−1) + [-(1/3)] = 2 − (1/3) = 5/3 = -[-(5)/3] = -b/a,
αβ + β γ + γ α = 3 × (−1) + (−1) × [-(1/3)] + [-(1/3)] x 3 = −3 + (1/3) + −1 = -11/3 = c/a,
αβγ = 3 × (−1) × [-(1/3)] = 1 = -(-3)/3 = -d/a.                                                     

Answer 2

Comparing the given polynomial with ax3 + bx2 + cx + d, we get

a = 3, b = – 5, c = –11, d = – 3. Further

p(3) = 3 × 33 – (5 × 32) – (11 × 3) – 3 = 81 – 45 – 33 – 3 = 0,

p(–1) = 3 × (–1)3 – 5 × (–1)2 – 11 × (–1) – 3 = –3 – 5 + 11 – 3 = 0,

p(1/3)= 3 × [(-1/3)3] − 5 ×[(-1/3)2] − 11 × [-(1/3)]-3

= –(1/9)–(5/9)+(11/3)–3 =–(2/3) + (2/3) = 0

Therefore, 3, -1 and -(1/3) are the zeroes of 3x3 – 5x22 – 11x – 3.

So, we take α = 3, β = –1 and γ = −(1/3)

Now,

α + β + γ = 3 + (−1) + [-(1/3)] = 2 − (1/3) = 5/3 = -[-(5)/3] = -b/a,

αβ + β γ + γ α = 3 × (−1) + (−1) × [-(1/3)] + [-(1/3)] x 3 = −3 + (1/3) + −1 = -11/3 = c/a,

αβγ = 3 × (−1) × [-(1/3)] = 1 = -(-3)/3 = -d/a.