Find all the zeroes of 2x⁴ – 13x³ + 19x² + 7x – 3, if you know that two of its zeroes are 2 + √3 and 2 – √3.

Given zeroes are 2 + √3 and 2 – √3

So, (x – 2 –√3) and (x – 2 + √3) are the factors of 2x⁴ – 13x³ + 19x² + 7x – 3

⟹ (x – 2 –√3) and (x – 2 + √3)

= x² – 2x + √3x – 2x + 4 – 2√3 - √3 x + 2√3 – 3

= x² – 4x + 1 is a factor of given polynomial.

Consequently, x² – 4x + 1 is also a factor of the given polynomial.

Now, let us divide 2x⁴ – 13x³ + 19x² + 7x – 3 by x² – 4x + 1

The division process is

Here, quotient = 2x² – 5x – 3

= 2x² – 2x – 3x – 3

= 2x(x – 1) – 3(x – 1)

= (2x – 3)(x – 1)

Hence, all the zeroes of the given polynomial are , 1, 2 +√3 and 2 - √3