Answers

2015-04-30T19:52:12+05:30
Since the divisor polynomial of degree 2 (quadratic), we add a polynomial of degree 1.  M x + K.

P(x) =  x⁴ + 2 x³ - 2 x² + x - 1  + (M x + K)

   Q(x) =  x² + 2x - 3   
         - 3  has factors  3 and -1  whose sum is 2, coefficient of x.
     Q(x) =  (x + 3) ( x - 1)

   If P(x) is to be exactly divisible by  Q(x) then  P(x) must be exactly divisible without a reminder by x + 3    as well  as  x - 1.  Then  P(-3) and P(1) must both be zero.

P(x) =  x⁴ + 2 x³ - 2 x² + x - 1  + (M x + K)

P(1) =  1 + 2 - 2 + 1 - 1 + M + K = 0
                     M + K = -1    ---- (1)

P(-3) = 81 - 54 - 18  - 3 - 1 - 3 M  + K = 0
                   2 M - K = 5          --- (2)

Solve the two equations:    adding them:  3 M = 4
                 M = 4/3
                 K = - 7/3

Hence,  we can add:  4/2 x - 7/3  to the given polynomial to have the given polynomial exactly divisible by the divisor polynomial.

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