Answer
$P(x)=2x^5+2x^4-18x^3-18x^2+40x+40$
Work Step by Step
RECALL:
If c is a zero of a polynomial, then $x-c$ is a factor of the polynomial.
Since -2, -1, 2, and $\sqrt5$ are zeros of the polynomial, and radical zeros always come in conjugate pairs, then
$(x+2),(x+1), (x-2), (x-\sqrt5)$ and $(x+\sqrt5)$ are factors of the polynomial.
Thus, the polynomial of degree 5 with the given zeros is:
$\\P(x)=a(x+2)(x+1)(x-2)(x-\sqrt5)(x+\sqrt5)
\\P(x)=a(x^5+x^4-9x^3-9x^2+20x+20)$
where $a$ is a real number.
Since the constant term is 40, then $a$ must be:
$\\a(20)=40
\\a=\frac{40}{20}
\\a=2$
Thus, the polynomial with the given zeros whose constant term is 40 is:
$\\P(x)=2(x^5+x^4-9x^3-9x^2+20x+20)
\\P(x)=2x^5+2x^4-18x^3-18x^2+40x+40$
