Q. 9 E5.0( 2 Votes )

x

Answer :

x^{4} – 6x^{3} – 26x^{2} + 138x – 35;2±√3

Given zeroes are 2 + √3 and 2 – √3

So, (x – 2 – √3)and (x – 2 + √3) are the factors of x^{4} – 6x^{3} – 26x^{2} + 138x – 35

⟹ (x – 2 – √3)(x – 2 + √3)

= x^{2} – 2x + √3 x – 2x + 4 – 2√3 – √3 x + 2√3 – 3

= x^{2} – 4x + 1 is a factor of given polynomial.

Consequently, x^{2} – 4x + 1 is also a factor of the given polynomial.

Now, let us divide x^{4} – 6x^{3} – 26x^{2} + 138x – 35 by x^{2} – 4x + 1

The division process is

Here, quotient = x^{2} – 2x – 35

= x^{2} – 7x + 5x – 35

= x(x – 7) + 5(x – 7)

= (x + 5)(x – 7)

So, the zeroes are – 5 and 7

Hence, all the zeroes of the given polynomial are – 5, 7, 2 + √3 and

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