Question

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

Answer 1

Step-by-step explanation:

answer=-63/64

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Answer 2

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)⁵ - 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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