If $x^{2} - 1$ is a factor of $p x^{4} + q x^{3} + r x^{2} + s x + u$ , show that p + r + u = q + s = 0.

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Solution

$x^{2} - 1 = 0 ⟹ = (x + 1) (x - 1) = 0 ⟹ x = 0, 1$
Now $x^{2} - 1$ is a factor of $p x^{4} + q x^{3} + r x^{2} + s x + u$
Putting $x = 1$
we get,
$p (1)^{4} + q (1)^{3} + r (1)^{2} + s (1) + u = 0 ⟹ p + q + r + s + u = 0$
Putting $x = - 1$
we get,
$p (- 1)^{4} + q (- 1)^{3} + r (- 1)^{2} + s (- 1) + u = 0 ⟹ p - q + r - s + u = 0$
Hence $p + r + u = q + s = 0$

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