If $a + c + e = 0$ and $b + d = 0$ then find the zeroes of the polynomial $a x^{4} + b x^{3} + c x^{2} + d x + e = 0$ .

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Solution

Given, $a + c + e = 0$ and $b + d = 0$

$\Rightarrow c = - (a + e)$ and $d = - b$

Now, $a x^{4} + b x^{3} + c x^{2} + d x + e$

$= a x^{4} + b x^{3} + [- (a + e)] x^{2} + (- b) x + e$
$= a x^{4} - a x^{2} - e x^{2} + e + b x^{3} - b x$
$= a x^{2} (x^{2} - 1) - e (x^{2} - 1) + b x (x^{2} - 1)$
$= (x^{2} - 1) (a x^{2} - e + b x)$
$= (x + 1) (x - 1) (a x^{2} - e + b x)$ .......... $(1)$

As $(x + 1)$ and $(x - 1)$ , are the factors of $(1)$
so, it is divisible by both $(x + 1)$ and $(x - 1)$
Hence the zeroes are $- 1, 1$ and $x = \frac{- b \pm \sqrt{b^{2} + 4 a e}}{2 a}$

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