If p(x) = x^4 - 6x^3 + 16x^2 - 25x + 10 is divided by q(x) = x^2- 2x + k, remainder = x+a, find k & a.
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f(x) = x^4 - 6x³+ 16x² - 25x + 10
Given that the remainder is x+a when f(x) is divided by x²-2x+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
Given that the remainder is x+a when f(x) is divided by x²-2x+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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