Question

Draw the graph of $p\left(x\right)=x^2+3x-4$ and find zeroes.
Verify the zeroes of the polynomials

Answer 1

The graph of $p\left(x\right)$ is given above.
Clearly, from the graph $x=-4,1$ are zeros of $p\left(x\right)$.
Verification:-
$p(-4)=(-4{)}^2+3\times -4-4=16-12-4=0$
$p\left(1\right)=(1{)}^2+3\times 1-4=1+3-4=0$