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If two zeroes of the polynomial $x 4 − 6 x 3 − 26 x 2 + 138 x − 35$ are $2 ± 3,$ the other zeroes are

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7, 8

7, 3

7, 5

5, 6

answer is C.

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Detailed Solution

Given a polynomial

x 4 − 6 x 3 − 26 x 2 + 138 x − 35

.
As,

2 − 3

and

2 + 3

are the zeros of polynomial.
Therefore,

x − 2 + 3

and

x − 2 − 3

are the factors of the given polynomial.

x − 2 + 3 x − 2 − 3

= x − 2 − 3 x − 2 + 3

= x − 2 2 − 32

= x 2 − 4 x + 4 − 3

= x 2 − 4 x + 1

Using long division method, dividing the polynomial,
Question Image

The quotient is,

Now, midterm splitting,

x 2 − 7 x + 5 x − 35 = 0

x (x − 7) + 5 (x − 7) = 0

(x − 7) (x − 5) = 0

x = 7,5 are other zeros of given polynomial.
Hence, option 3 is correct.

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