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Find all the zeroes of the polynomial $x^{4}+x^{3}-14x^{2}-2x+24,$ , if two of its zeroes are $\sqrt{2}$ and $-\sqrt{2}$ .

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Given, polynomial $P(x)=x^{4}+x^{3}-14x^{2}-2x+24$

The zeroes are $\sqrt{2}$ & $-\sqrt{2}$

We have to find there two zeroes of the $P(x)$ .

If $\sqrt{2}$ & $-\sqrt{2}$ are zeroes then $(x+\sqrt{2})$ & $(x-\sqrt{2})$ must be a factor of $P(x)$ .

$\Rightarrow$ dividing $P(x)$ by $(x-\sqrt{2})$ $(x+\sqrt{2})$ OR $(x^{2}-2)$

$\Rightarrow$ Hence $P(x)$ could be written as $\Rightarrow p(x)=(x^{2}-2)(x^{2}+x-12)$

Finding zeroes of $x^{2}+x-12$

$\Rightarrow x^{2}+(4-3)x-12$

$\Rightarrow x^{2}+4x-3x-12\Rightarrow x(x+4)-3(x+4)$

$\Rightarrow (x-3)(x+4)$

Other zeroes are $+3,-4$

Hence all zeroes of $P(x)$ are $=-\sqrt{2},\sqrt{2},3,-4$

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