Fewpal
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Verify that 3, –1, -1/3 are the zeroes of the cubic polynomial p(x) = 3x3 – 5x2 – 11x – 3, and then verify the relationship between the zeroes and the coefficients.

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Solution : 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 = 0

Therefore, 3, -1 and -1/3 are the zeroes of 3x3 - 5x2 -11x -3
So, we take α = 3, β = -1, γ = -1/3,

Now,

α + β + γ = 5/3 = -(-5)/3 = -b/a,
αβ + βγ + γα = -11/3 = c/a,
αβγ = 1 = -(-3)/3 = -d/a

by (10 points)
So there are 2 ways of doing this question? 1 by putting values and 2 by coefficient method ?

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