I'm back says the biquadratic!

$x^4-5x^3+5x^2+5x-6=0$

Notice that the sum of coefficients $=1-5+5+5-6=0$

Therefore, we can say that $x=1$ is a root of the equation. Factorize it out:

$x^4-x^3-4x^3+4x^2+x^2-x+6x-6=0\\ x^3(x-1)-4x^2(x-1)+x(x-1)+6(x-1)=0\\ (x-1)(x^3-4x^2+x+6)=0$

Now, it is given that the product of two of the real roots of the equation is $3$ . We can guess that $x=3$ is another root of the equation since $1 \times 3 = 3$ . We verify this:

$3^3-4(3)^2+3+6 = 27-36+3+6 = 0$

Factorize it out:

$(x-1)(x^3-3x^2-x^2+3x-2x+6) = 0\\ (x-1)\left(x^2(x-3)-x(x-3)-2(x-3)\right)=0\\ (x-1)(x-3)(x^2-x-2)=0\\ (x-1)(x-3)(x^2-2x+x-2)=0\\ (x-1)(x-3)\left(x(x-2)+1(x-2)\right)=0\\ (x-1)(x-3)(x-2)(x+1)=0$

The roots are: $x=1,\;2,\;3,\;-1$ . The smallest root is $x=\boxed{-1}$

Observe the title. This is the inspiration for my factorisation.

$x^4-5x^3+5x^2+5x-6=0\\x^4-5x^3+6x^2-x^2+5x-6=0\\(x^2-5x+6)(x^2-1)=0\\(x+1)(x-1)(x-2)(x-3)=0$ Clearly, the smallest root is -1.