Question 1209679
Here's how to find all the roots of p(x):

1. **Find k:**
   Since x = 2 is a root, p(2) = 0.
   p(2) = 2⁴ - 3(2)³ + 11(2)² - 25(2) + k = 0
   16 - 24 + 44 - 50 + k = 0
   -14 + k = 0
   k = 14

2. **Rewrite p(x):**
   p(x) = x⁴ - 3x³ + 11x² - 25x + 14

3. **Polynomial division:**
   Since x = 2 is a root, (x - 2) is a factor. Divide p(x) by (x - 2):

```
x^3 - x^2 + 9x - 7
------------------
x - 2 | x^4 - 3x^3 + 11x^2 - 25x + 14
      x^4 - 2x^3
      ----------
          -x^3 + 11x^2
          -x^3 + 2x^2
          ----------
               9x^2 - 25x
               9x^2 - 18x
               ----------
                   -7x + 14
                   -7x + 14
                   ----------
                        0
```

So, p(x) = (x - 2)(x³ - x² + 9x - 7)

4. **Solve the cubic:**
   Now we need to find the roots of the cubic x³ - x² + 9x - 7 = 0.
   We can try rational root theorem, but it doesn't seem to give integer roots.

   Let's use the rational root theorem to test possible rational roots.  Factors of 7 are ±1, ±7.
   If x = 1, 1 - 1 + 9 - 7 = 2 ≠ 0
   If x = -1, -1 - 1 - 9 - 7 = -18 ≠ 0
   If x = 7, 7³ - 7² + 9(7) - 7 = 343 - 49 + 63 - 7 = 350 ≠ 0
   If x = -7, (-7)³ - (-7)² + 9(-7) - 7 = -343 - 49 - 63 - 7 = -462 ≠ 0

   Since the rational root theorem doesn't give us integer roots, the roots are likely irrational or complex.

5. **Use a calculator or software:**
   Using a calculator or software to find the roots of the cubic x³ - x² + 9x - 7 = 0 gives the following approximate roots:
   x ≈ 0.7923
   x ≈ 0.1039 + 2.9706i
   x ≈ 0.1039 - 2.9706i

Therefore, the roots of p(x) are approximately 2, 0.7923, 0.1039 + 2.9706i, and 0.1039 - 2.9706i.

Final Answer: The final answer is $\boxed{2, 0.7923, 0.1039+2.9706i, 0.1039-2.9706i}$