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Remainder Theorem

If roots of a...

Question

If roots of an equation $x^{n} - 1 = 0$ are $1, a_{1}, a_{2} . . . . a_{n - 1}$ , then the value of $(1 - a_{1}) (1 - a_{2}) (1 - a_{3}) . . . . . (1 - a_{n - 1})$ will be

A

n

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B

n^{2}

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C

n!

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D

0

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Solution

The correct option is A $n$
$x^{n} - 1 = (x - 1) (x - a_{1}) (x - a_{2}) . . . . . . . . . (x - a_{n - 1}) ⟹ \frac{x^{n} - 1}{x - 1} = (x - a_{1}) (x - a_{2}) . . . . . . . . . (x - a_{n - 1}) ⟹ (x - a_{1}) (x - a_{2}) . . . . . . . . . (x - a_{n - 1}) = 1 + x + x^{2} + . . . . . x^{n - 1}$
put $x = 1$ .
$⟹ (1 - a_{1}) (1 - a_{2}) . . . . . . . . . (1 - a_{n - 1}) = 1 + 1 + 1.... + 1 (n$ times $) = n$

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Similar questions

a_{1}, a_{2}, . . . . a_{n - 1}

n

th roots of unity then prove that
(i)

(1 - a_{1}) (1 - a_{2}) (1 - a_{n - 1}) . . . = n

1 + a_{1} + a_{2} + . . . . + a_{n - 1} = 0

\frac{1}{2 - a_{1}} + \frac{1}{2 - a_{2}} + . . . . \frac{1}{2 - a_{n - 1}} = \frac{(n - 2) 2^{n - 1} + 1}{2^{n} - 1}

1, a_{1}, a_{2}, . a_{n - 1}

n

roots of unity, then the value of

(1 - a_{1}) (1 - a_{2}) (1 - a_{3}) . (1 - a_{n - 1})

x^{n} - 1 = 0, n > 1, n ϵ N

1, a_{2}, . . . . a_{n}

.

(1 - a_{1}) (1 - a_{2}) . . . . (1 - a_{n})

a_{1}, a_{2}, a_{3}, \dots, a_{n}; (n \geq 2)

(n - 1) a_{1}^{2} - 2 n a_{2} < 0,

then atleast roots of the equation

x^{n} + a_{1} x^{n - 1} + a_{2} x^{n - 2} + \dots + a_{n} = 0

, are imaginary.

a_{1}, a_{2}, a_{3}, . . . . ., a_{n} (n \geq 2)

(n - 1) a_{1}^{2} = 2 n a_{2} < 0

, then show that atleast two roots of the equation

x^{n} + a_{1} x^{n - 1} + a_{2} x^{n - 2} + . . . . + a_{n} = 0

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