If $p (x) = x^{3} - 3 x^{2} - 9 x - 9$ , find p(0), p(3), p(-3) and p(-1). What do you conclude about the zeros of p(x)? Is 0 a zero p(x)?

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Solution

$p (x) = x^{3} - 3 x^{2} - 9 x - 9 p (0) = (0)^{3} - 3 (0)^{2} - 9 (0) - 9 = - 9 p (3) = (3)^{3} - 3 (3)^{2} - 9 (3) - 9 = 27 - 27 - 27 - 9 = - 36 p (- 3) = (- 3)^{3} - 3 (- 3)^{2} - 9 (- 3) - 9 = - 27 - 27 + 27 - 9 = - 36 p (- 1) = (- 1)^{3} - 3 (- 1)^{2} - 9 (- 1) - 9 = - 1 - 3 + 9 - 9 = - 4$

as function value not come out to be zero , so no given values are the zeroes of the polynomial.

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