Obtain other zeroes of the polynomial
$$f(x) = 2x^4 + 3x^3 - 5x^2 - 9x - 3$$
if two of its zeroes are $\sqrt{3}$ and $-\sqrt{3}$.

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Step 1: $\because \sqrt{3}$ and $-\sqrt{3}$ are zeroes of $f(x)$. $	herefore (x - \sqrt{3})$ and $(x + \sqrt{3})$ are factors of $f(x)$. $	herefore x^2 - 3$ is a factor of $f(x)$
Step 2: $q(x) = \frac{2x^4 + 3x^3 - 5x^2 - 9x - 3}{x^2 - 3} = 2x^2 + 3x + 1$
Step 3: For zeroes $q(x) = 0$: $2x^2 + 3x + 1 = 0 \Rightarrow (x + 1)(2x + 1) = 0$. $x = -1, -\frac{1}{2}$
Step 4: $	herefore$ Remaining zeroes are $-1$ & $-\frac{1}{2}$

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