SOLUTION: if the function x^3 + 2x^2 - ax - 8 is evenly divisible by (x+2), find the value of a

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Question 1192990: if the function x^3 + 2x^2 - ax - 8 is evenly divisible by (x+2), find the value of a
Found 2 solutions by ikleyn, Edwin McCravy:
Answer by ikleyn(52803) About Me  (Show Source):
You can put this solution on YOUR website!
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if the function x^3 + 2x^2 - ax - 8 is evenly divisible by (x+2), find the value of a
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According to the Remainder theorem, the number -2 is the root of the polynomial.


So, substitute this value of -2 into the polynomial.


It will give you an equation for the unknown coefficient "a"


    (-2)^3 + 2*(-2)^2 - a*(-2) - 8 = 0,

or

    -8 + 2*4 + 2a - 8 = 0,

    2a = 8

     a = 8/2 = 4.


ANSWER.  a = 4.

Solved.



Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!

             x^2 + 0x - a
x + 2)x^3 + 2x^2 - ax - 8
      x^3 + 2x^2
            0x^2 - ax 
            0x^2 + 0x
                  -ax - 8 
                  -ax -2a
                     -8+2a

The remainder should equal 0, so 

                     -8+2a = 0
                        2a = 8
                         a = 4
                         

Edwin