Math, asked by Anonymous, 1 year ago

if the polynomial x^4 - 6x^3 + 16x^2 - 25x + 10 is divided by x^2 - 2x + k, the remainder is x + a, find K and a.
Here ^ means raised to the power

Answers

Answered by kvnmurty
12
P(x)  =  x⁴  - 6 x³ + 16 x² - 25 x + 10
         = (x²)² - x² * 6 x + 16 * x²  - 25 x + 10

Instead of long division, i am doing as follows:
       divisor:  x²  - 2 x + k 
      so substitute  x² - 2x + k = 0    or    x² = 2 x - k  in  P(x)

 remainder  =    (2 x - k)²  - 6 x ( 2x - k )  + 16 (2x - k) - 25 x + 10
               =  4x² - 4 k x + k² - 12 x² + 6 x k + 32 x - 16 k  - 25 x + 10
             =  - 8 x² + 2 k x + 7 x + k² - 16 k + 10
            =   - 8 (2 x - k) + 2  k x + 7 x + k² - 16 k + 10
           =  (- 9  + 2 k ) * x +  k² - 8 k + 10

   as the remainder is x + a  .  compare with the above expression:
           2 k - 9 = 1      and  so  k = 5
       10 - 8 k + k² = a      =>    a = - 5

=============================

long division

      x² - 2x + k  )  x⁴ - 6 x³ + 16 x² - 25 x + 10  (  x²  - 4 x  + (8-k)
                           x⁴  - 2 x³  + k x²
                     ======================
                              - 4 x³ + (16-k) x² - 25 x
                              - 4 x³  + 8 x² - 4 k x
               ================================
                                       (8-k) x² + (4 k - 25) x  + 10
                                       8-k) x²   - 2 (8-k) x   + k (8-k)
                    ===============================
                                    (2 k  - 9) x + 10 - 8 k + k²

 we are given that  reminder is  x + a
      =>  2 k - 9 = 1      hence,    k = 5
     =>     a = 10 - 8 k + k² = - 5 
 



Anonymous: I have done the question and the above method is wrong..
Anonymous: How can you take 2k - 9 =
kvnmurty: the above answer is correct and right.
kvnmurty: what is the answer you have got , please inform
Anonymous: although the answer is same the method is un understandable
Anonymous: can you tell me how you can substitute 2k - 9 = 1
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