Question

If a₁, a₂, a₃, a₄, a₅ are roots of the equation

z^{5}\overset{̶}{+}z^{4} + z^{3} + z^{2} + z + 1= 0 then $\prod_{i = 1}^{5}{(2–\alpha_{i})}$is equal to –

• A. 63
• B. 31
• C. 32
• D. 64

Answer 1

Answer: (A)

63

Answer 2

Sol. z⁵ + z⁴ + z³ + z² + z + 1 = 0

(z – a₁) (z – a₂) (z – a₃) (z – a₄) (z – a₅)

= z⁵ + z⁴ + z³ + z² + z + 1

Putting z = 2

Ž (2– a₁) (2– a₂) (2– a₃) (2– a₄) (2– a₅)

= 32 + 16 + 8 + 4 + 2 + 1

\ $\prod_{i = 1}^{5}{(2–\alpha_{i})}$ = 63