### Instruction: Find the real roots of
\[x^4 - 2x^3 - x + 2 = 0.\].### Response: I notice that the equation has four terms, and the first two have a common factor of $x^3$, while the last two have a common factor of $-1$.
Maybe I can factor this equation by grouping.
I write
\[x^4 - 2x^3 - x + 2 = (x^3)(x - 2) - (x - 2) = (x - 2)(x^3 - 1).\]
I recognize that $x^3 - 1$ is a difference of cubes, which has a special factorization.
I recall that
\[a^3 - b^3 = (a - b)(a^2 + ab + b^2),\]
so I apply this to $x^3 - 1$, where $a = x$ and $b = 1$.
I get
\[x^3 - 1 = (x - 1)(x^2 + x + 1).\]
I substitute this back into the original equation and obtain
\[x^4 - 2x^3 - x + 2 = (x - 2)(x - 1)(x^2 + x + 1) = 0.\]
Now I have a product of three factors equal to zero, which means that at least one of them must be zero.
So I can find the roots by setting each factor equal to zero and solving for $x$.
For the first factor, I get
\[x - 2 = 0 \implies x = 2.\]
For the second factor, I get
\[x - 1 = 0 \implies x = 1.\]
For the third factor, I have to use the quadratic formula, since it does not factor easily.
I recall that the quadratic formula is
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},\]
where $ax^2 + bx + c = 0$ is the standard form of a quadratic equation.
In this case, $a = 1$, $b = 1$, and $c = 1$.
So I plug these values into the formula and get
\[x = \frac{-1 \pm \sqrt{1 - 4}}{2} = \frac{-1 \pm \sqrt{-3}}{2}.\]
Since I have a negative number inside the square root, I know that these are complex roots, not real roots.
So I can ignore them for this problem.
Therefore, the only real roots are $x = 2$ and $x = 1$.