If

\[x^3  4x^2 + 3x = 0\]

what real value is a solution for x?

Solution:
Factor x^2 out of the first two terms to get x^2(x  4) + 3x  12 = 0. Factor a 3 out of the last two terms to get x^2(x  4) + 3(x  4) = 0. Rewrite the equation to get (x^2 + 3)(x^2  4) = 0. Therefore, one of the solutions to the equation is x  4 = 0. Solve for x to get x = 4. The other solutions come from x^2 + 3 = 0, or x^2 = 3. This results in imaginary solutions, so the only real solution is 4.
Final answer: x = 4 is the only real solution.
Factor x^2 out of the first two terms to get x^2(x  4) + 3x  12 = 0. Factor a 3 out of the last two terms to get x^2(x  4) + 3(x  4) = 0. Rewrite the equation to get (x^2 + 3)(x^2  4) = 0. Therefore, one of the solutions to the equation is x  4 = 0. Solve for x to get x = 4. The other solutions come from x^2 + 3 = 0, or x^2 = 3. This results in imaginary solutions, so the only real solution is 4.
Final answer: x = 4 is the only real solution.