The number of real zeros of the polynomial p( | vedclass

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(A) To find the zeros of the polynomial $p(x) = x^3 - x$,we set $p(x) = 0$.$x^3 - x = 0$Factor out $x$:$x(x^2 - 1) = 0$Using the difference of squares

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$3$

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(A) To find the zeros of the polynomial $p(x) = x^3 - x$,we set $p(x) = 0$.$x^3 - x = 0$Factor out $x$:$x(x^2 - 1) = 0$Using the difference of squares formula $a^2 - b^2 = (a - b)(a + b)$:$x(x - 1)(x + 1) = 0$Setting each factor to zero,we get:$x = 0$,$x - 1 = 0 \implies x = 1$,and $x + 1 = 0 \implies x = -1$.Thus,the zeros are $0, 1, -1$.There are $3$ real zeros.

The number of real zeros of the polynomial $p(x) = x^3 - x$ is . . . . . . .

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