Find the zeros of the quadratic polynomial $p(x)=x^{2}-3x-4$ and show it on a graph.

(N/A) Here,$p(x)=x^{2}-3x-4=(x-4)(x+1)$.
To find the zeros of $p(x)$,consider $p(x)=0$.
$\therefore (x-4)(x+1)=0$.
$\therefore x=4$ or $x=-1$.
$\therefore 4$ and $-1$ 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:

$x$	$-2$	$-1$	$0$	$3$	$4$	$5$
$p(x)=x^2-3x-4$	$6$	$0$	$-4$	$-4$	$0$	$6$

Plot these points on a graph paper. Joining all these points $(-2, 6), (-1, 0), (0, -4), (3, -4), (4, 0)$ and $(5, 6)$,we get the shape of the graph as a parabola opening upwards. We can see that this graph intersects the $X$-axis at two points $(-1, 0)$ and $(4, 0)$. Their $X$-coordinates are the zeros of this polynomial. Thus,$-1$ and $4$ are the zeros of $p(x)$.