If the two zeroes of the polynomial \(x^4- 6x^3 - 26x^2 -138x - 35\) are \(2 + \sqrt{3}\) and \(2 - \sqrt{3}\) find other zeroes.

Two zeroes are\(2 + \sqrt{3}\) and \(2 - \sqrt{3}\).

\(=> (x -(2 + \sqrt{3})) (x - (2 - \sqrt{3})) \)
\( = (x - 2 - \sqrt{3}) (x - 2 + \sqrt{3})\)
\(=> (x - 2)^2 - ( \sqrt{3})^2 \)
\( = x^2 - 4x + 1\)

\(x^2 - 4x + 1\) the factors of \(x^4- 6x^3 - 26x^2 -138x - 35\).

So, we'll divide the given polynomial by \(x^2 - 4x + 1\) Solving it using long division:

\(\begin{array}{rrrr|ll} x^4 -6x^3 & -26x^2 & +138x & -35 & x^2 -4x +1 \\ -x^4 + 4x^3 & -x^2 & & & x^2 -2x - 35 \\ \hline -2x^3 & -27x^2 & +138x & -35\\ \phantom{-}+2x^3 & -8x^2 & + 2x & & & \\ \hline & -35x^2 & +140x & -35 \\ & +35x^2 & -140x & +35 \\ \hline & & & 0\\ \hline \end{array}\)

\(x^4- 6x^3 - 26x^2 -138x - 35\)
=> \(( x^2 -4x +1) (x^2 -2x - 35)\)
=> \(( x^2 -4x +1) (x^2 -7x + 5x - 35)\)
=> \((x^2 -4x +1) (x(x - 7) + 5(x - 7))\)
=> \((x^2 -4x +1) (x + 5)(x - 7)\)

Thus, \(x + 5 = 0\) and \(x - 7 = 0\) are the solutions.

So, the other two zeroes are \(x = -5 , 7\).