## Calculus (3rd Edition)

$p(x)=x^2+2x-3$.
We have $$p(x)=x^2+ax+b, \quad p'(x)= 2x+a.$$ Since, $p(1)=0$ and $p'(1)=4$, we get the equations $$a+b=-1, \quad a=2$$ Hence we have the solution $a=2, b=-3$. Then, $p(x)=x^2+2x-3$.