The graph of $y = x^3– 4x$ looks like,
| x | -2 | -1 | 0 | 1 | 2 |
| y | 0 | 3 | 0 | -3 | 0 |
$x^3– 4x$ = $x(x^2-4)$ i.e $x(x-2)(x+2)=0$
i.e $x=0$ and $x-2=0$ and $x+2=0$
i.e $x=0$ ; $x=2$ and $x=-2$ are the zeroes of the cubic polynomial $ x^3– 4x$.
Observe that – 2, 0 and 2 are, in fact, the x-coordinates of the only points where the graph of $y = x^3– 4x$ intersects the x-axis. Since the curve meets the x-axis in only these 3 points, their x-coordinates are the only zeroes of the polynomial.