another maths problem....?

Question

if the polynomial x^4-6x^3+16x^2-25x+10 is divided by another polynomial x^2-2x+k,the remainder comes out to be x+a,find k and a.

Answer 1

(x^2 - 2x + k)(x^2 + bx + c) + x + a = x^4 - 6x^3 + 16x^2 - 25x + 10
x^2(x^2 + bx + c) - 2x(x^2 + bx + c) + k(x^2 + bx + c) + x + a = x^4 - 6x^3 + 16x^2 - 25x + 10
x^4 + bx^3 + cx^2 -2x^3 - 2bx^2 - 2cx + kx^2 + bkx + ck + x + a = x^4 - 6x^3 + 16x^2 - 25x + 10
x^4 + (b - 2)x^3 + (c - 2b + k)x^2 + (bk - 2c + 1)x + ck + a = x^4 - 6x^3 + 16x^2 - 25x + 10

Therefore
(b - 2)x^3 = -6x^3
b - 2 = -6
b = -6 + 2 = -4

(c - 2b + k)x^2 = 16x^2
c - 2(-4) + k = 16
c + 8 + k = 16
c + k = 16 - 8
c + k = 8
c = 8 - k

(bk - 2c + 1)x = -25x
(-4)k - 2c + 1 = -25
-4k -2c = -25 - 1
-2(2k + c) = -26
2k + c = -26/-2
2k + c = 13
2k + 8 - k = 13
k + 8 = 13
k = 13 - 8
k = 5

ck + a = 10

so
b = -4
k = 5
c = 8 - k
c = 8 - 5
c = 3

ck + a = 10
(3)(5) + a = 10
15 + a = 10
a = 10 - 15 = -5

Therefore a = -5 and k = 5

Answer 2

x^4-6x^3+16x^2-25x+10
(x^2-2x+k)(x^2+bx+c)+(x +a) = x^4 +bx^3 + cx^2 -2x^3 - 2bx^2 - 2cx + kx^2 + bkx + kc + x +a
= x^4 + bx^3 - 2x^3 + cx^2 - 2bx^2 + kx^2 - 2cx + bkx + kc + x +a
= x^4 + (b - 2)x^3 + (c - 2b + k)x^2 - (2c + bk + 1)x + kc + a

Compare with
x^4-6x^3+16x^2-25x+10 = x^4 + (b - 2)x^3 + (c - 2b + k)x^2 - (2c + bk + 1)x + kc + a

(b - 2)x^3 = -6x^3
:> b = -6 +2
b = -4

(c - 2b + k)x^2 = 16x^2
c - 2(-4) + k = 16
c + k = 8

(bk - 2c + 1)x = -25x
(-4)k - 2c + 1 = -25
-4k -2c = -25 - 1
-2(2k + c) = -26
2k + c = -26/-2
2k + c = 13

ck + a = 10


c + k = 8
ck + a = 10
2k + c = 13
now solve these equation
c = 8 - k
2k + 8 - k = 13
k = 5
:> c = 8 - 5
c = 3

ck + a = 10
(3* 5) + a = 10
a = -5

:> values a = -5, c = 3, k = 5 and b = -4

Answer 3

Since the Dividend has degree 4, the quotient must have degree 2

So, (x^2-2x+k)(x^2+bx+c)+(x+a) = x^4-6x^3+16x^2-25x+10

or x^4 + (b-2)x^3 + (c-2b+k)x^2 + (-2c+kb+1)x + kc+a
= x^4-6x^3+16x^2-25x+10

b = -4

c+k+8 = 16 So, c+k=8 and c+2k=13
Thus k = 5, c =3

a = -5

Answer 4

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