Find the zeros of $p(x)=x^{3}-4 x$ and show them graphically.

(N/A) Here,$p(x)=x^{3}-4 x$
$=x(x^{2}-4)$
$=x(x-2)(x+2)$
To find the zeros of $p(x)$,consider $p(x)=0$.
$\therefore x(x-2)(x+2)=0$
$\therefore x=0, x=2$ or $x=-2$.
$\therefore 0, 2$ and $-2$ are the zeros of $p(x)$.
To draw the graph of this polynomial,we take some different values of $x$ and prepare the following table:

$x$	$-2$	$-1$	$0$	$1$	$2$
$p(x)=x^3-4x$	$0$	$3$	$0$	$-3$	$0$

Plot these points as shown in the graph. We can see that this graph intersects the $X$-axis at three distinct points $(-2, 0)$,$(0, 0)$ and $(2, 0)$. Their $X$-coordinates are the zeros of this polynomial. So,the zeros of $p(x)$ are $-2, 0$ and $2$.