Question

in the remainder when P of x is equal to x cube minus 6 X square + 2 x minus 4 divided by q(x)=1-2x​

Answer 1

Correct Question :-

Find the remainder p(x) = x³ - 6x² + 2x - 4 divided by q(x) = 1 - 2x

Answer :-

The remainder is - 35/8 when p(x) = x³ - 6x² + 2x - 4 divided by q(x) = 1 - 2x

Solution :-

p(x) = x³ - 6x² + 2x - 4 divided by q(x) = 1 - 2x

First find the zero of 1 - 2x

To find the zero of 1 - 2x equate it to 0

1 - 2x = 0

- 2x = - 1

x = - 1/-2

x = 1/2

By Remainder theorem p(1/2) is the remainder when p(x) = x³ - 6x² + 2x - 4 divided by q(x) = 1 - 2x

p(x) = x³ - 6x² + 2x - 4

Substitute x = 1/2

p(1/2) = (1/2)³ - 6(1/2)² + 2(1/2) - 4

= 1/2³ - 6(1/2²) + 1 - 4

= 1/8 - 6(1/4) + 1 - 4

= 1/8 - 3(1/2) - 3

= 1/8 - 3/2 - 3

Taking LCM

= 1/8 - 3(4)/2(4) - 3(8)/1(8)

= 1/8 - 12/8 - 24/8

= 1/8 - 36/8

= - 35/8

Therefore the remainder is - 35/8 when p(x) = x³ - 6x² + 2x - 4 divided by q(x) = 1 - 2x

Answer 2

Answer: remainder is = -35/8

Step-by-step explanation:

p(x)= x^3 -6x^2 + 2x - 4 , divided by: q(x)=1-2x

1-2x=0

-2x=-1

x=1/2

=(1/2)^3 -6(1/2)^2 + 2(1/2) -4

= 1/8- 3/2 + 1-4

=1/8- 3/2 - 3

= (1-12-24)/8

= -35/8

hence remainder is -35/8   [we can use long division method also to solve this]

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