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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 $$x = -5 , 7$$.