![p(x) = {3x}^{3} - {5x}^{2} - 11x - 3 \: find \: (1) \frac{1}{ \alpha } + \frac{1}{ \beta } + \frac{1}{ \gamma } \: (2) \: \frac{1}{ \alpha \beta } + \frac{1}{ \beta \gamma } + \frac{1}{ \alpha \gamma } \: (3) \: { \alpha }^{2} \beta \gamma + \alpha { \beta }^{2} \gamma + \alpha \beta { \gamma }^{2}. p(x) = {3x}^{3} - {5x}^{2} - 11x - 3 \: find \: (1) \frac{1}{ \alpha } + \frac{1}{ \beta } + \frac{1}{ \gamma } \: (2) \: \frac{1}{ \alpha \beta } + \frac{1}{ \beta \gamma } + \frac{1}{ \alpha \gamma } \: (3) \: { \alpha }^{2} \beta \gamma + \alpha { \beta }^{2} \gamma + \alpha \beta { \gamma }^{2}.](https://tex.z-dn.net/?f=p%28x%29+%3D++%7B3x%7D%5E%7B3%7D++-++%7B5x%7D%5E%7B2%7D++-+11x+-+3+%5C%3A+find+%5C%3A+%281%29+%5Cfrac%7B1%7D%7B+%5Calpha+%7D++%2B++%5Cfrac%7B1%7D%7B+%5Cbeta+%7D++%2B++%5Cfrac%7B1%7D%7B+%5Cgamma+%7D+%5C%3A+%282%29+%5C%3A++%5Cfrac%7B1%7D%7B+%5Calpha++%5Cbeta+%7D++%2B++%5Cfrac%7B1%7D%7B+%5Cbeta++%5Cgamma+%7D++%2B++%5Cfrac%7B1%7D%7B+%5Calpha++%5Cgamma+%7D++%5C%3A+%283%29+%5C%3A+++%7B+%5Calpha+%7D%5E%7B2%7D+%5Cbeta++%5Cgamma++%2B++%5Calpha++%7B+%5Cbeta+%7D%5E%7B2%7D++%5Cgamma++%2B++%5Calpha++%5Cbeta++%7B+%5Cgamma+%7D%5E%7B2%7D.+++)
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see iam in 6 std so i dont no the asnwer
Step-by-step explanation:
but very sorry
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Answer:
Step-by-step explanation:
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
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