Question

if alpha beta and gamma are the zeros of the polynomial 6 x cube minus 3 x square - 5 x + 1 then find the value of Alpha raised to minus 1 + beta raised to -1 and gamma raised to -1​

Answer 1

Answer:

$\large{\underline{\boxed{\sf 5}}}$

Step-by-step explanation:

Given polynomial:

6x³ - 3x² - 5x + 1

Here,

• a = 6
• b = (-3)
• c = (-5)
• d = 1

Sum of zeroes = -b/a

⇒ α + β + γ = -(-3)/6

⇒ α + β + γ = 1/2

Sum of product of zeroes = c/a

⇒ αβ + βγ + γα = -5/6

Product of zeroes = -d/a

⇒ αβγ = -1/6

Now, we have to find ;

α⁻¹ + β⁻¹ + γ⁻¹

$\implies \sf \frac{1}{ \alpha } + \frac{1}{ \beta } + \frac{1}{ \gamma } \\ \\ \implies \sf \frac{ \beta \gamma + \alpha \gamma + \alpha \beta }{ \alpha \beta \gamma } \\ \\ \implies \sf \frac{ - \frac{5}{6} }{ - \frac{1}{6} } \\ \\ \implies \sf \frac{5}{6} \times 6 \\ \\ \implies \sf 5$

Hence, the answer is 5.

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$\large{\underline{\underline{\mathfrak{\heartsuit \: Extra \: Information: \: \heartsuit}}}}$

For a cubic polynomial ax³ + bx² + cx + d, the zeroes α, β and γ, where:

• α + β + γ = - b/a
• αβ + βγ + γα = -c/a
• αβγ = d/a