Find the zeroes of the polynomials.
$x^{3} - 3 x^{2} - 9 x - 5$

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Solution

Let $p (x) = x^{3} - 3 x^{2} - 9 x - 5$
The coefficient of the leading term is $1$ and the constant term is $- 5.$
Also factors of $5$ are $1$ and $5.$
So the possible integral zeroes of p(x) are $\pm 1$ and $\pm 5$
$∴ p (- 1) = (- 1)^{3} - 3 (- 1)^{2} - 9 (- 1) - 5$
$= - 1 - 3 + 9 - 5 = 9 - 9 = 0$
$\Rightarrow - 1$ is a zero of $p (x) .$
$p (5) = (5)^{3} - 3 (5)^{2} - 9 (5) - 5$
$= 125 - 75 - 45 - 5$
$= 125 - 125 = 0$
$\Rightarrow 5$ is a zero of $p (x) .$
$∴ - 1$ and $5$ are zeroes of $p (x) .$

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