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

(7, -5) Given that $2+\sqrt{3}$ and $2-\sqrt{3}$ are zeroes of the given polynomial.
Therefore,$(x-(2+\sqrt{3}))(x-(2-\sqrt{3})) = ((x-2)-\sqrt{3})((x-2)+\sqrt{3}) = (x-2)^{2} - (\sqrt{3})^{2} = x^{2}-4x+4-3 = x^{2}-4x+1$ is a factor of the given polynomial.
To find the remaining zeroes,we divide the polynomial $x^{4}-6 x^{3}-26 x^{2}+138 x-35$ by $x^{2}-4 x+1$:
$x^{4}-6 x^{3}-26 x^{2}+138 x-35 = (x^{2}-4 x+1)(x^{2}-2 x-35)$
Now,we factorize the quadratic polynomial $x^{2}-2 x-35$:
$x^{2}-2 x-35 = x^{2}-7x+5x-35 = x(x-7)+5(x-7) = (x-7)(x+5)$
Setting these factors to zero,we get $x-7=0$ or $x+5=0$,which gives $x=7$ or $x=-5$.
Hence,the other two zeroes are $7$ and $-5$.