Finding a tangent line that intersects a quartic at two points

Let $(x_1, f(x_1))$ and $(x_2, f(x_2)$ be the points where $g(x)$ is tangent to $f(x)$ . Let's form the function $h(x)=f(x)-g(x)$ , so $h(x)=x^4-6x^3+13x^2-(10+m)x+7-b$ . The zeros of $h(x)$ are $x=x_1$ and $x=x_2$ and we also know that $h(x) \geq 0$ for all $x$ . What's nice is that $h(x)$ can also be written as $h(x)=(x-x_1)^2(x-x_2)^2$ . We can expand this to get $h(x)=x^4-2(x_1+x_2)x^3+(x_1^2+4x_1x_2+x_2^2)x^2-2(x_1^2x_2+x_1x_2^2)x+x_1^2x_2^2$ . By comparing the two forms of h(x) we can get that $x_1=1$ and $x_2=2$ (or vice versa) and therefore $m=2$ and $b=3$ .

Whenever we talk about the tangents of polynomial curves, on solving the curve and the tangent we get a perfect square factor at the point of tangency. Thus, x^4-6x^3+13x^2-10x+7-(mx+c) must factor into product of 2 perfect squares (since it's stated tangent at 2 points). In other words we can also say that the biquadratic must be the perfect square of a quadratic. On simplifying it a bit we get, (x^2-3x+2)^2+2x+3-(mx+c). Thus m=2 and c=3. We actually also got the points of tangency by solving the quadratic formed above, which gives us the roots x=1 and x=2.

I have written a note to the claim I made in the first statement. Viewers may check it out.