math

Find the remainder when  f(x) = 2x^3 − x^2 + x + 1  is divided by  2x + 1.

The remainder theorem states that f(x) / (x - a)  will have a remainder of  f(a).

However, this doesn't apply if the coefficient of the x-term isn't 1.

So first, let's divide both  f(x)  and  2x + 1  by 2:

         ---   This is legal because  f(x) / (x - a)  =  [ f(x) / 2 ]  /  [ (x - a) / 2) ].

     f(x) / 2  =  (2x^3 − x^2 + x + 1) / 2  =  x^3 − (1/2)x^2 + (1/2)x + (1/2)

     (2x + 1) / 2  =  x + (1/2)

Now, the coefficient of the x-term is 1.

So, to find the remainder when  x^3 − (1/2)x^2 + (1/2)x + (1/2)  is divided by  x + (1/2),  replace  x  with  -(1/2):

     --->     (-1/2)^3 − (1/2)(-1/2)^2 + (1/2)(-1/2) + (1/2)

     --->     -1/8 - 1/8 - 1/4 + 12

     --->     -1/8 - 1/8 - 2/8 + 4/8

     --->     0

Therefore, the remainder will be  0.

Simplify the following:
f(x)=(2 x^3-x^2+x+1)/(2 x+1)

The possible rational roots of 2 x^3-x^2+x+1 are x = ±1/2, x = ±1. Of these, x = -1/2 is a root. This gives 2 x+1 as all linear factors:
(((2 x+1) (2 x^3-x^2+x+1))/(2 x+1))/(2 x+1)

 | |
2 x | + | 1 | | x^2 | - | x | + | 1
2 x^3 | - | x^2 | + | x | + | 1
2 x^3 | + | x^2 | | | |
 | | -2 x^2 | + | x | |
 | | -2 x^2 | - | x | |
 | | | | 2 x | + | 1
 | | | | 2 x | + | 1
 | | | | | | 0:
(x^2-x+1 (2 x+1))/(2 x+1)

((2 x+1) (x^2-x+1))/(2 x+1) = (2 x+1)/(2 x+1)×(x^2-x+1) = x^2-x+1:
Answer: |x^2-x+1

"Find the remainder when f(x)=2x^3−x^2+x+1 is divided by 2x+1."

Here's another approach:

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