Find the number of real solutions for the system $x^4-x^3+x^2-4x-12=0$.

First, we let $f(x)=x^4-x^3+x^2-4x-12$. $f(x)$ clearly is a quartic function and it has at most 4 real roots.

If we can factorize $f(x)$ as $f(x)=(x-a)(x-b)(x-c)(x-d)$, then $f(x)$ has 4 real roots.

Upon first glance, it looks like the rational root test is a must to figure out the root(s) for $f(x)$. But after a few trials, we realized soon that this quartic polynomial has no "simple" real roots. In other words, we have found $f(\pm 1)\ne 0,\,f(\pm 2)\ne 0,\,f(\pm 3)\ne 0,\,f(\pm 4)\ne 0,\,f(\pm 6)\ne 0,\,f(\pm 12)\ne 0$.

Therefore we need a second strategy.

We first rewrite $f(x)$ in such a way we gather [MATH]\color{yellow}\bbox[5px,purple]{x^4,\,x^2}[/MATH] in a group and then [MATH]\color{black}\bbox[5px,orange]{x^3,\,x}[/MATH] in another group:

$\begin{align*}f(x)&=x^4-x^3+x^2-4x-12\\&=(x^4+x^2)-(x^3+4x)-12\\&=x^2(x^2+1)-x(x^2+4)-12\\&=x^2(\underline{x^2+1})-x(\underline{x^2+1}+3)-12\,\,\text{(note that we want to create $x^2+1$ term)}\\&=x^2(\underline{x^2+1})-x(\underline{x^2+1})-3x-12\\&=(x^2-x)(x^2+1)-3(x+4)\end{align*}$

That doesn't seem like it would bring us to a fruitful solution.

But note that $x^4+x^2-12$ easily factors into $(x^2+4)(x^2-3)$, and $-(x^3+4x)=-x(x^2+4)$, therefore, we need to gather [MATH]\color{yellow}\bbox[5px,purple]{x^4,\,x^2,\,-12}[/MATH] in a group and then [MATH]\color{black}\bbox[5px,orange]{x^3,\,x}[/MATH] in another group:

$\begin{align*}f(x)&=x^4-x^3+x^2-4x-12\\&=(x^4+x^2-12)-(x^3+4x)\\&=(x^2+4)(x^2-3)-x(x^2+4)\\&=(x^2+4)(x^2-3-x)\\&=(x^2+4)(x^2-x-3)\end{align*}$

Up to this point, we can tell $x^4-x^3+x^2-4x-12=(x^2+4)(x^2-x-3)$, and this quartic has only two real solutions since $x^2+4=0$ has zero roots and $x^2-x-3=0$ has two distinct real roots since $(-1)^2-4(1)(-3)=13>0$.

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