Find all zeros of the polynomial $P (x) = x^{3} - 3 x^{2} - 6 x + 8$ knowing that x = 1 is a zero of the polynomial.

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Solution

We can see at $x = 1, f (1) = 0$
$∴$
$(x - 1)) ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ x^{3} - {3 x}^{2} - 6 x + 8 (x^{2} - 2 x - 8 x^{3} - x^{2}- {2 x}^{2} - 6 x - {2 x}^{3} + 2 x- 8 x + 8 - 8 x + 8$

$∴ f (x) = (x - 1) (x^{2} - 2 x - 8)$

$= (x - 1) (x + 2) (x - 4)$

So all zeros of $f (n)$ are $1, - 2, 4$

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