Answers

2016-05-06T22:13:04+05:30
f(x) = x^4 - 6x³+ 16x² - 25x + 10 
Given that the remainder is x+a when f(x) is divided by x
²-2x+k
i.e , we can write , 
x^4 - 6x³+ 16x² - 25x + 10  = (x²-2x+k) × (x²+bx+c) + (x+a)
                                           = x^4 - 2x³+kx²+bx³-2bx²+bkx+cx²-2cx+ck+x+a
                                           = x^4 + [b-2]x³ + [k+c-2b]x² + [bk-2c+1]x + [ck+a]
    Comparing the coefficients of x³ , x² , x and the constants on both sides , we get 
b - 2 =  - 6   ⇒ b= - 4
k + c - 2b = 16   ⇒ k + c = 8    ------[1]
bk - 2c + 1 = - 25  ⇒ -4k - 2c = 26 or 2k + c = 13      --------------[2]
And ck + a = 10    --------------[3]
From [1] and [2] , we get
k = 5
∴ From [1] , c = 3 and from [3] we get a = - 5

Hence k =  5 and a = - 5
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