Answer
$p(x)=2x^5+2x^4-18x^3-18x^2+40x+40$
Work Step by Step
Since $\sqrt{5}$ is a zero, to guarantee that the coefficients are integer it must be $-\sqrt{5}$ as another zero.
Then,
$p(x)=A(x+1)(x+2)(x-2)(x-\sqrt{5})(x+\sqrt{5})$
$p(x)=A(x+1)(x^2-4)(x^2-5)$
$p(x)=A(x+1)(x^4-9x^2+20)$
$p(x)=A(x^5+x^4-9x^3-9x^2+20x+20)$
$p(x)=Ax^5+Ax^4-9Ax^3-9Ax^2+20Ax+20A$
In order that the constant term is 40, it must be $20A=40\to A=2$.
Now, we have $p(x)=2x^5+2x^4-18x^3-18x^2+40x+40$.
