Answer :

Given that 2 + √3 and 2 - √3 are the zeroes of the given polynomial.

So (x – (2 + √3)) (x – (2 - √3)) = (x – 2 - √3) (x – 2 + √3)

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

= x^{2} – 4x + 1

The other zeroes are as follows:

x^{2} – 2x - 35 = 0

Solving the above quadratic equation.

Sum = -2

Product = -35

So the numbers which satisfy the above condition are -7 and 5

x^{2} – 7x + 5x - 35 = 0

x(x – 7) + 5(x – 7) = 0

(x + 5) (x – 7) = 0

Solving the first part,

x + 5 = 0

x = -5

Solving the second part,

x – 7 = 0

x = 7

Therefore the other zeroes of the polynomials are -5 and 7.

Rate this question :

Dividing the firsRajasthan Board Mathematics

Dividing the firsRajasthan Board Mathematics

Dividing the firsRajasthan Board Mathematics

With the followinRajasthan Board Mathematics

Find the quotientRajasthan Board Mathematics

Find the quotientRajasthan Board Mathematics

Find the quotientRajasthan Board Mathematics

With the followinRajasthan Board Mathematics

With the followinRajasthan Board Mathematics

Find the quotientRajasthan Board Mathematics