Q:

Formulate quadratic equation

Here is an equation $$r^3-5r^2+8r-4=0$$ 
Is there a way I can formulate a quadratic equation from this?
Sorry if the question seem dumb, I am stuck and I can't figure a way out.

A:

By the Rational Root Theorem (which actually has a very simple proof, as shown in Wikipedia), if $r^3-5r^2+8r-4$ has a rational root, then it must be one of $-4,-2,-1,1,2,4$.
By checking, you'll find that $1$ is a root (actually, $2$ is also a root).
Therefore the polynomial $r^3-5r^2+8r-4$ is divisible by $r-1$. Now you can use long division (with your preferred method). Here is one way you can divide:
$$r^3-5r^2+8r-4=(r-1)r^2-4r^2+8r-4$$
$$=(r-1)r^2-(r-1)4r+4r-4$$
$$=(r-1)r^2-(r-1)4r+(r-1)4$$
$$=(r-1)\left(r^2-4r+4\right)=(r-1)(r-2)^2$$

