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(D) To find the zeros of the polynomial $p(x) = x^{3} - 9x$,we set $p(x) = 0$.$x^{3} - 9x = 0$Factor out $x$ from the expression:$x(x^{2} - 9) = 0$Usi

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$3$

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(D) To find the zeros of the polynomial $p(x) = x^{3} - 9x$,we set $p(x) = 0$.$x^{3} - 9x = 0$Factor out $x$ from the expression:$x(x^{2} - 9) = 0$Using the difference of squares formula $a^{2} - b^{2} = (a - b)(a + b)$,we can factor $x^{2} - 9$ as $(x - 3)(x + 3)$:$x(x - 3)(x + 3) = 0$Setting each factor to zero,we get:$x = 0$,$x - 3 = 0 \implies x = 3$,and $x + 3 = 0 \implies x = -3$.Thus,the zeros of the polynomial are $0, 3, -3$.Since there are $3$ distinct real values,the number of real zeros is $3$.

Find the number of real zeros of the following polynomial: $p(x) = x^{3} - 9x$.

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