Question

if x^6 - 1 divided by 2x+ 1 then the absolute value of the remainder is

Answer 1

Given :  x⁶ - 1  is divided by 2x + 1

To Find : Absolute Value of Remainder

Solution:

x⁶ - 1  is divided by 2x + 1

2x + 1 = 0

=> x =  - 1/2

x⁶ - 1

= (-1/2)⁶ - 1

=  1/64 - 1

= -63/64

remainder = -63/64

Absolute value  is distance from 0 on number line /unsigned value

Absolute value of remainder = 63/64

(x⁶ - 1 )/ (2 x + 1) =

$\frac{1}{2}{x}^{5} - \frac{1}{4}{x}^{4} + \frac{1}{8}{x}^{3} - \frac{1}{16}{x}^{2} + \frac{1}{32}x - \frac{1}{64} + \frac{-{\frac{63}{64}}}{2x + 1}$

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Answer 2

Answer:

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Step-by-step explanation:

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