Show that $x^4+2x^2-6x+2=0$ when $x\in\mathbb{R}$ has exactly two solutions.
I first showed that the IVT guarantees that there exists at least one zero in $(0,1)$ and at least one zero in $(1,2)$. I then was going to apply Rolle's theorem twice to show uniqueness in each interval.
My goal was to assume contradiction in each interval but I was hoping that $f'(x) > 0$, but it is not.
Where did I go wrong, and how can I fix it?
On
$$f(x) = x^4 + 2x^2-6x+2$$
$$f'(x) = 4x^3 + 4x -6$$
$$f''(x) = 12x^2 + 4 > 0 \forall_{x\in\mathbb{R}}$$
$f''$ has no zeros, so $f'$ has at most one and therefore $f$ has at most 2.
Since you've shown at least 2 exist, you are done.
On
Set $p(x)=x^4+2x^2-6x+2$. Then $p'(x)=4x^3+4x-6$ and $p''(x)=12x^2+4$, so $p'(x)$ is increasing, and it has only one root $\alpha$.
As $p'(0)=-6,\;p'(1)=2$, this root is between $0$ and $1$. Furthermore $p'(x)\to-\infty$ as $x\to-\infty$ and $p'(x)\to+\infty$ as $x\to+\infty$. This implies $p(x)$ is decreasing on $(-\infty,\alpha]$ and increasing on $\alpha,+\infty)$.
As $\lim_{x\to\pm\infty} p(x)=+\infty$ and $p(1)<0$, this again implies $p(x)$ has two roots, one $<1$, the other $>1$. The i.v.t. further shows these roots belong to $(0,1)$ and $(1,2)$ respectively.
After your findings so far, it suffices to show that the derivative $$ f'(x)=4x^3+4x-6$$ has only one real root. For this again, it is sufficient to observe that the second derivative $$ f''(x)=12x^2+4$$ is strictly positive.
In other words, your Rolle aproach should work: If there were three real roots $x_1<x_2<x_3$ of $f$, we'd have roots $\xi_1,\xi_2$ of $f'$ with $\xi_1<x_2<\xi_2$ and then a root $\eta_1$ of $f''$.