# Find a quadratic polynomial whose zeros are 2 + √3 and 2 – √3.(x2 – 2x + k) the remainder comes out to be (x + a). Find k and a.

As the remainder is x + a

x + a = (2k – 9)x + (k2 – 8k + 10)

Comparing constant and coefficient of x of LHS and RHS

1 = 2k – 9 and a = k2 – 8k + 10

2k – 9 = 1

2k = 10

k = 5

Put k = 5 in a = k2 – 8k + 10

a = 52 – 8(5) + 10

a = 25 – 40 + 10

a = 25 – 30

a = –5

Hence k = 5 and a = –5

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