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# If the zeroes of the polynomial $\left( {{x}^{4}}-6{{x}^{3}}-26{{x}^{2}}+138x-35 \right)$ are $\left( 2+\sqrt{3} \right)$ and $\left( 2-\sqrt{3} \right)$, find the other zeros.  Hint:At first from the given roots try to form a quadratic equation then from the formed quadratic equation, divide it from the given four – degree polynomial equation to get a quadratic equation. After that, factorize it to get the roots.

In the question we are given two of the zeroes of expression $\left( {{x}^{4}}-6{{x}^{3}}-26{{x}^{2}}+138x-35 \right)$ which are $\left( 2+\sqrt{3} \right)$ and $\left( 2-\sqrt{3} \right)$.
The polynomial given in the question has at most or 4 zeroes as the maximum number of zeros depending on the degree of the polynomial. If the degree of the polynomial is n then maximum zeros it can have is n.
We can also say that if the two of the zeroes are known then the expression formed by them would be a factor of the given polynomial.
Here we will use formula that is $\alpha ,\beta$ are roots of the polynomial then the expression will be ${{x}^{2}}-\left( \alpha +\beta \right)x+\alpha \beta$. Here, $\alpha$ is $\left( 2-\sqrt{3} \right)$ and $\beta$ is $\left( 2+\sqrt{3} \right)$. So, the expression will be $\left\{ {{x}^{2}}-\left( 2+\sqrt{3}+2-\sqrt{3} \right)x+\left( 2+\sqrt{3} \right)\left( 2-\sqrt{3} \right) \right\}$ or $\left\{ {{x}^{2}}-4x+\left( 4-3 \right) \right\}$ or ${{x}^{2}}-4x+1$.
So, we can write, ${{x}^{2}}-4x+1=0$
Now we can write the given expression, ${{x}^{4}}-6{{x}^{3}}-26{{x}^{2}}+138x-35$ as, $x^4-4{{x}^{3}}-2{{x}^{3}}+{{x}^{2}}+8{{x}^{2}}-35{{x}^{2}}-2x+140x-35$.
Now on rearranging we can write it as,
${{x}^{4}}-4{{x}^{3}}+{{x}^{2}}-2{{x}^{3}}+8{{x}^{2}}-2x-35{{x}^{2}}+140x-35$
Or, ${{x}^{2}}\left( {{x}^{2}}-4x+1 \right)-2x\left( {{x}^{2}}-4x+1 \right)$
$-35\left( {{x}^{2}}-4x+1 \right)$, which can be factorized and written as,
$\left( {{x}^{2}}-2x-35 \right)\left( {{x}^{2}}-4x+1 \right)$
So to find remaining roots we have to solve, ${{x}^{2}}-2x-35=0$
The equation, ${{x}^{2}}-2x-35=0$ can be written as, ${{x}^{2}}-7x+5x-35=0$.
Or, $x\left( x-7 \right)+5\left( x-7 \right)=0$
So, $\left( x+5 \right)\left( x-7 \right)=0$
Hence the values of x are -5 and 7.
Thus the remaining zeros are -5 and 7.

Note: As the degree of the polynomial given is four and also after finding one of the factor by the given two roots one can find other roots by just dividing the main given 4 – degree expression by the quadratic equation whose roots are $2+\sqrt{3}$ and $2-\sqrt{3}$.
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