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When x^4 + ax^3 + bx +c is divided by (x–1), (x+1), and (x+2), the remainders are 14, 0, and –16 respectively. Find the values of a, b, and c.

Guest Dec 2, 2017
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We have the following system of equations

 

 a    +  b   +   c   =    13    (1)

-a     - b    + c     =    -1     (2)

-8a  -2b    + c     =   -32   (3)

 

Adding  (1)  and (2) we have that  2c  = 12  ⇒   c  = 6

 

So (2) and (3) become

 

-a     -  b   =   -7   ⇒    -2a   -   2b    =    - 14     ⇒  2a  +   2b     =  14      (4)

-8a  -2b   =  -38      (5)

 

Add (4)  and (5)   and we have that

 

-6a   =   -24    ⇒   a  = 4

 

So

 

a  +  b  + c  =  13

4  +  b   + 6   = 13

b =  3

 

So    a  =  4, b = 3   and   =  6

 

 

cool cool cool

CPhill  Dec 2, 2017

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