Polynomials in 4 degree
f ( x ) = x 4 − 6 x 3 + 1 6 x 2 − 2 5 x + 1 0
When polynomial f ( x ) above is divided by x 2 − 2 x + k , the remainder is x + a , where k and a are non-zero constants. Find k + a .
The answer is 0.
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2 solutions
Let f ( x ) be as follows.
f ( x ) = ( x 2 + b x + c ) ( x 2 − 2 x + k ) + x + a = x 4 + ( b − 2 ) x 3 + ( c + k − 2 b ) x 2 + ( b k − 2 c + 1 ) x + c k + a
Equating the coefficients we have:
⎩ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎧ b − 2 = − 6 c + k − 2 b = 1 6 b k − 2 c + 1 = − 2 5 c k + a = 1 0 . . . ( 1 ) . . . ( 2 ) . . . ( 3 ) . . . ( 4 )
⎩ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎧ ( 1 ) : ( 2 ) : ( 3 ) : ( 3 a ) − ( 2 a ) : ( 2 a ) : ( 4 ) : b − 2 = − 6 c + k − 2 ( − 4 ) = 1 6 ( − 4 ) k − 2 c + 1 = − 2 5 k = 5 c + ( 5 ) = 8 ( 3 ) ( 5 ) + a = 1 0 ⟹ b = − 4 ⟹ c + k = 8 ⟹ 2 k + c = 1 3 ⟹ c = 3 ⟹ a = − 5 . . . ( 2 a ) . . . ( 3 a )
Therefore k + a = 5 − 5 = 0 .
Push through the long division (not shown) and get two equations from the coefficients of x and 1 in the remainder:
2 k − 9 = 1
1 0 − 8 k + k 2 = a
So k = 5 and a = − 5 . Thus a + k = 0 .