The number of distinct real roots of the equation $$3x^4 + 4x^3 - 12x^2 + 4 = 0$$ is _________.

We begin with the quartic equation

$$3x^{4}+4x^{3}-12x^{2}+4=0.$$

To determine how many real roots it possesses, we try to factor it into a product of two quadratic polynomials having real coefficients. Let us assume

$$3x^{4}+4x^{3}-12x^{2}+4=(x^{2}+ax+b)\,(3x^{2}+cx+d).$$

We now expand the right-hand side and equate coefficients term by term. Multiplying gives

$$\begin{aligned} (x^{2}+ax+b)\,(3x^{2}+cx+d) &=x^{2}(3x^{2}+cx+d)+ax(3x^{2}+cx+d)+b(3x^{2}+cx+d)\\ &=3x^{4}+cx^{3}+dx^{2}+3ax^{3}+acx^{2}+adx+3bx^{2}+bcx+bd\\ &=3x^{4}+(c+3a)x^{3}+(d+ac+3b)x^{2}+(ad+bc)x+bd. \end{aligned}$$

Comparing this with the original polynomial, we obtain the system

$$\begin{cases} c+3a=4,\\[4pt] d+ac+3b=-12,\\[4pt] ad+bc=0,\\[4pt] bd=4. \end{cases}$$

The last equation $$bd=4$$ suggests the integer possibilities $$(b,d)=(\pm1,\pm4)$$ or $$(\pm2,\pm2)$$. We test $$(b,d)=(-2,-2)$$ because it satisfies $$bd=4$$ and keeps the numbers small.

With $$b=-2$$ and $$d=-2$$, the third equation $$ad+bc=0$$ becomes

$$(-2)a+(-2)c=0 \;\Longrightarrow\; -2(a+c)=0 \;\Longrightarrow\; a+c=0 \;\Longrightarrow\; c=-a.$$

Substituting $$c=-a$$ into $$c+3a=4$$ gives

$$-a+3a=4 \;\Longrightarrow\; 2a=4 \;\Longrightarrow\; a=2,$$

and hence $$c=-2$$.

Finally, we check the second equation $$d+ac+3b=-12$$:

$$-2+(2)(-2)+3(-2)=-2-4-6=-12,$$

which is satisfied exactly. Thus our choice is consistent, and the factorisation is confirmed:

$$3x^{4}+4x^{3}-12x^{2}+4=(x^{2}+2x-2)\,(3x^{2}-2x-2).$$

Setting each quadratic factor equal to zero gives the roots.

For the first quadratic $$x^{2}+2x-2=0,$$ we apply the quadratic formula $$x=\dfrac{-2\pm\sqrt{(2)^{2}-4(1)(-2)}}{2(1)}=\dfrac{-2\pm\sqrt{4+8}}{2}=\dfrac{-2\pm\sqrt{12}}{2}=\dfrac{-2\pm2\sqrt{3}}{2}=-1\pm\sqrt{3}.$$

So we obtain two real roots $$x=-1+\sqrt{3}\quad\text{and}\quad x=-1-\sqrt{3}.$$

For the second quadratic $$3x^{2}-2x-2=0,$$ the quadratic formula gives

$$x=\dfrac{2\pm\sqrt{(-2)^{2}-4(3)(-2)}}{2\cdot3}=\dfrac{2\pm\sqrt{4+24}}{6}=\dfrac{2\pm\sqrt{28}}{6}=\dfrac{2\pm2\sqrt{7}}{6}=\dfrac{1\pm\sqrt{7}}{3}.$$

Thus we obtain two more real roots $$x=\dfrac{1+\sqrt{7}}{3}\quad\text{and}\quad x=\dfrac{1-\sqrt{7}}{3}.$$

All four solutions are real, and it is easy to see that they are distinct because the numerical values $$-1\pm\sqrt{3}$$ and $$(1\pm\sqrt{7})/3$$ are pairwise different.

Therefore the equation possesses four distinct real roots.

So, the answer is $$4$$.