Choose the correct answer from the alternatives given.
What is the square root of $(\frac{x^{5} - 1}{x - 1}) + (x^{3} + 2 x^{2} + x)$ ?

x^{2}

+ x + 1

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x^{2} - x + 1

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x^{2} - x - 1

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x^{2} + x - 1

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Solution

The correct option is A $x^{2}$ $+ x + 1$
$: \frac{x^{3} - 1}{x - 1} + x^{3} + 2 x^{2} + x$
$: \frac{x^{5} - 1 + x^{4} + 2 x^{3} + x^{2} - x^{3} - 2 x^{2} - x}{x - 1}$
$\frac{x^{5} + x^{4} + x^{3} - x^{2} - x - 1}{x - 1} = \frac{x^{3} (x^{2} + x + 1) - 1 (x^{2} + x + 1)}{(x - 1)}$
$\frac{(x^{2} + x + 1) (x^{3} - 1)}{x - 1} = \frac{(x^{2} + x + 1) (x - 1) (x^{2} + x + 1)}{(x - 1)}$
$(x^{2} + x + 1)^{2}$
$∴$ Required square root $= \sqrt{(x^{2} + x + 1)^{2}} = (x^{2} + x + 1)$

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Question :

One of the factors of x⁵+x+1 is

(1) 2x² - x - 1

(2) x³ - x²+ 1

(3) x² - 2x² + 1

(4) x³ + x² + 1

x \neq - 1

\frac{x^{5} + x^{4} + x^{3} + x^{2}}{x^{3} + x^{2} + x + 1} = 1

\frac{x}{1 - x^{2}} - \frac{x^{3}}{1 - x^{6}} + \frac{x^{5}}{1 - x^{10}} - . . . . . . . = \frac{x}{1 + x^{2}} + \frac{x^{3}}{1 + x^{6}} + \frac{x^{5}}{1 + x^{10}} + . . . . .

x^{2} - x + 1 = 0

{(x + \frac{1}{x})}^{2} + {(x^{2} + \frac{1}{x^{2}})}^{2} + {(x^{3} + \frac{1}{x^{3}})}^{2} + {(x^{5} + \frac{1}{x^{5}})}^{2}

x \neq - 1

\frac{x^{5} + x^{4} + x^{3} + x^{2}}{x^{3} + x^{2} + x + 1}

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