### Instruction: Find the real roots of
\[x^4 - 2x^3 - x + 2 = 0.\].### Response: I'm looking for the real roots of this quartic equation, which means the values of x that make the equation true.
I notice that the equation has four terms, and they are arranged in descending order of degree.
This makes me wonder if I can factor the equation by grouping pairs of terms together.
I try to group the first two terms and the last two terms, like this:
\[(x^4 - 2x^3) + (-x + 2) = 0.\]
I look for a common factor in each group.
In the first group, I can factor out x^3, and in the second group, I can factor out -1.
This gives me
\[x^3(x - 2) - 1(x - 2) = 0.\]
Now I see that both groups have a common factor of (x - 2), so I can factor that out as well.
This gives me
\[(x - 2)(x^3 - 1) = 0.\]
I have factored the equation into a product of two factors, and I know that a product is zero if and only if one of the factors is zero.
So I can split the equation into two simpler equations and solve for x in each one.
For the first factor, I have
\[x - 2 = 0.\]
This is easy to solve by adding 2 to both sides.
I get
\[x = 2.\]
So one real root of the equation is x = 2.
For the second factor, I have
\[x^3 - 1 = 0.\]
This is a bit trickier to solve, but I recognize that it is a difference of two cubes.
I remember the formula for factoring a difference of two cubes, which is
\[a^3 - b^3 = (a - b)(a^2 + ab + b^2).\]
I apply this formula to my equation, with a = x and b = 1.
I get
\[x^3 - 1 = (x - 1)(x^2 + x + 1) = 0.\]
Again, I can split the equation into two simpler equations and solve for x in each one.
For the first factor, I have
\[x - 1 = 0.\]
This is easy to solve by adding 1 to both sides.
I get
\[x = 1.\]
So another real root of the equation is x = 1.
For the second factor, I have
\[x^2 + x + 1 = 0.\]
This is a quadratic equation, and I can try to solve it by factoring, completing the square, or using the quadratic formula.
I decide to use the quadratic formula, because I don't see any obvious factors, and completing the square can be tedious.
I remember the quadratic formula, which is
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},\]
where a, b, and c are the coefficients of the quadratic equation in standard form.
I identify the coefficients in my equation.
I have a = 1, b = 1, and c = 1.
I plug these values into the quadratic formula and simplify.
I get
\[x = \frac{-1 \pm \sqrt{1^2 - 4(1)(1)}}{2(1)}\]
\[x = \frac{-1 \pm \sqrt{1 - 4}}{2}\]
\[x = \frac{-1 \pm \sqrt{-3}}{2}.\]
I notice that the expression under the square root is negative, which means that the square root is not a real number.
This means that there are no real values of x that satisfy this factor.