I'm trying to find a straightforward calc solution to this question. I found an algebraic solution but I don't think it's the quickest way to do it. Thanks in advance!

Find the linear function $g(x)=mx+b$ whose graph is tangent to the graph of $f(x)=x^4-6x^3+13x^2-10x+7$ at two points.

Here's my algebraic solution: 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 $h(x)≥0$ 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 $h(x)=x^4−2(x\_1+x\_2)x^3+(x^2\_1+4x\_1x\_2+x^2\_2)x^2−2(x^2\_1x\_2+x\_1x^2\_2)x+x^2\_1x^2\_2$. 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 $b=3$. I want to see a calc solution since this question appeared on a calc test.