g(x) = x3 + ax2 + 3x +6. Given g(-1) =2, find the remainder when g(x) is divided by (3x -2).

Question

it is a remainder theoram problem plssss find d solution.......

Answer 1

g(x) = x^3 + ax^2 + 3x +6.
Given g(-1) = 2, find the remainder when g(x) is divided by (3x - 2).

-1 | 1   a     3     6
           -1  1-a   a-4
       1 a-1 4-a | a+2 = 2 => a = 0, and

g(x) = x^3 + 3x + 6

g(x)/(3x - 2) = Q(x) + R/(3x -2)
g(x) = (3x - 2)Q(x) + R
g(x) = 3(x - 2/3)Q(x) + R
g(2/3) = 3(0)Q(2/3) + R
g(2/3) = R

2/3 | 1   0  27/9    162/27
            2/3  4/9      62/27
         1 2/3 31/9 | 224/27 = Remainder & answer to problem!

check:

From synthetic division above we can write:

g(x)/(x - 2/3) = x^2 + 2x/3 + 31/9 + (224/27)/(x - 2/3)

Multiplying both sides by 1/3:

g(x)/(3x - 2) = x^2/3 + 2x/9 + 31/27 + (224/27)/(3x - 2)

g(x) =? (3x - 2)(x^2/3 + 2x/9 + 31/27 + (224/27)/(3x - 2) )

g(x) =? (3x - 2)(x^2/3 + 2x/9 + 31/27) + 224/27

g(x) =? 3x(x^2/3 + 2x/9 + 31/27)
            -2(x^2/3 + 2x/9 + 31/27)
                                     + 224/27

g(x) =? x^3 + 2x^2/3 + 31x/9
                    - 2x^2/3 -   4x/9 - 62/27
                                           + 224/27

g(x) =? x^3 + 0 + 27x/9 + 162/27

g(x) = x^3 + 3x + 6 check!

Answer 2

g ( X) =  X^3  + a X^2 + 3X + 6

 

  g( -1) =2 

 

   g( -1) = ( -1) ^3 + a ( -1)^2 + 3( -1) + 6

 

  2 = -1 + a - 3 + 6 =

 

   2 = a +2     a =0

 

   g( x ) = X^3 + 3X + 6

           

            1/3X^2+2/9 X + 31/27  ___________                  

  3X -2  l X^3 + 3X +6

              X^3±2/3 X^2

                      2/3 X^2 +3X

                      2/3 X^2 ± 4/9X

                                   31/9 X+6

                                   31/9 X ± 61/27

                                                8 8/27

   Remainder = 8 8/27 =224/27