Question

find the remainder when x cube + 3 X square + 3 X + 1 is divided by X minus one upon two​

Answer 1

Answer:

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Answer 2

Given,

• The polynomial $x^3 +3x^2+3x+1$ is divided by $x-1/2$.

To find,

• We have to find the remainder when the $x^3 +3x^2+3x+1$ is divided by $x-1/2$.

Solution,

We can simply find the remainder when the polynomial $x^3 +3x^2+3x+1$ is divided by $x-1/2$ using the remainder theorem.

                 x-1/2 = 0

                     x = 1/2

Now, substitute the value of x= 1/2 in the polynomial $x^3 +3x^2+3x+1$, we get

  p(x) =  $x^3 +3x^2+3x+1$

  p(1/2) = (1/2)³ +3(1/2)² +3(1/2) +1

            = 1/8 + 3/4 +3/2 +1

   Now, taking LCM, we get

             = (1+6+12+8)/8

             = 27/8

Hence, the remainder when  $x^3 +3x^2+3x+1$ is divided by $x-1/2$ is 27/8.