Let f(x) = (x^2-16)(x-3)^2+9x^2
= (x^2-16)(x^2-6x+9)+9x^2
= x^4-6x^3+2x^2+96x-144
If abs(x) >= 4 then (x^2-16) >= 0, (x-3)^2 > 0 and 9x^2 > 0, so f(x) > 0.
So the only Real roots of f(x) = 0 are in the range -4 < x < 4.
The rational root theorem tells us that any rational roots of f(x) = 0 must be of the form p/q in lowest terms, where p, q in ZZ, q > 0, p is a divisor of the constant term 144 and q is a divisor of the coefficient 1 of the leading term.
So the only possible rational roots are integer divisors of 144 strictly between -4 and 4. That gives the only possibilities as +-1, +-2, +-3. None of these work, so f(x) = 0 has no rational roots.
Next, suppose f(x) = (x^2+ax+b)(x^2+cx+d) for some integers a, b, c and d.
What can we find out about these integers?
(x^2+ax+b)(x^2+cx+d) = x^4+(a+c)x^3+(b+d+ac)x^2+(ad+bc)x+bd
Equating coefficients, we get:
(i) a+c=-6
(ii) b+d+ac=2
(iii) ad+bc=96
(iv) bd=-144
Suppose a is odd.
Then from (i) c is also odd, so ac is odd.
Then from (ii) b+d is odd, so one of b and d is odd and the other even.
Then one of ad and bc is odd and the other even.
So ad+bc is odd.
But (iii) tells us that ad+bc=96 which is even, so we have a contradiction.
Therefore a is even.
Since a is even, then from (i) we get c is even too. Thus ac is a multiple of 4.
Then from (ii), we find b+d is even but not a multiple of 4.
From (iv), at least one of b and d is even, hence they both are.
That leaves us with the following possibilities for (b, d) with b < d:
(-72, 2), (-24, 6), (-18, 8), (-8, 18), (-6, 24), (-2, 72)
..with sums:
-70, -18, -10, 10, 18, 70
..which from (ii) give us possible values for ac:
72, 20, 12, -8, -16, -68
..which in combination with (i) give us possible integer values of (a, c):
(none), (none), (none), (none), (-8, 2) / (2, -8), (none)
So (b, d) = (-6, 24) and (a, c) = (-8, 2) " or " (2, -8)
From (iii) we quickly find:
(a, c) = (2, -8)
Hence:
f(x) = (x^2+2x-6)(x^2-8x+24)
Using the quadratic formula, the first of these factors has roots:
x = -1 +-sqrt(7)
and the second only has complex roots.
