Establish all polynomial functions, which have the property f(x) = f'(x)*f"(x), x is in the real number set.

We will take  tentatively : f(x) =
ax^n.

f'(x) = a*nx^(n-1)

f"(x)
= an(n-1)x^(n-2)

 So f(x) = f'(x)*f"(x)
gives:

a x^n = a * n x^(n-1)}{an(n-1)
x^(n-2)}

ax^n = a^2* n^2(n-1) x^(n+n-2) =
a^2*n^2(n-1)x^(2n-3)

So  equating the powers and
coefficients,

n = 2n-3.

Or
2n-n = 3

n =3.

a =
a^2n^2(n-1)

1 = a*3^2(3-1)
.

a = 1/18.

So  of f(x) =
(1/18)x^3

f'(x) = 3/18x^2 =
(1/6)x^2

f"(x) = (2/6)x =
(1/3)x

f(x) = f'(x)*f"(x) = (1/6)x^2 * (1/3)x^2 =
(1/18)x^3

Now again we assume
that

f(x) =
1/18x^3+bx^2+cx+d.

f'(x) = (3/18)x^2+2bx+c
.

f"(x) = (6/18)x+2b.

So
f'(x)*f"(x) = {(3/18)x^2 + 2bx+c}{(6/18)x+2b}

=
x^3+{(6b/18)+(12b/18)}x^2+{4b^2+6c/18)x+2bc . This should be identitically equal to
f(x). Now  equating  the coefficients of the like powers, we
get:

x^3 : both sides agree and is equal to
1

x^2: both sides agree and equal to
b.

x :  (4b^2+6c/18) =
c,

4b^2+c/3 =c

12b^2 +c =3c, 
Or 2c= 12b^2,  c = 6b^2.

Constant term: 2bc = d. So d =
2b(6b) = 12b^2.

Therefore the required polynomial
is

f(x) = (/18)x^2+bx^2+6b^2*x+12b^2 which
satisfies

f(x) = f'(x)*f"(x).