Question 1209679: Given that x = 2 is a root of p(x) = x^4 - 3x^3 + 11x^2 - 25x + k for some constant k, find all the roots of p(x) (real and nonreal).
Enter your answer as a list of numbers separated by commas.
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! 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 $\boxed{2, 0.7923, 0.1039+2.9706i, 0.1039-2.9706i}$
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