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2015-04-12T18:45:24+05:30
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.                                                     
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