Solve the equation `3x^(4) -10x^(3) + 4x^(2) -x-6=0` one root being `(1+sqrt(-3))/2`.

Answer

To solve the equation \(3x^4 - 10x^3 + 4x^2 - x - 6 = 0\) given that one root is \(\frac{1 + \sqrt{-3}}{2}\), we can follow these steps:

### Step 1: Identify the conjugate root
Since the given root is a complex number, its conjugate must also be a root of the polynomial. Therefore, the second root is:
\[ \frac{1 - \sqrt{-3}}{2} = \frac{1 - i\sqrt{3}}{2} \]

### Step 2: Form the quadratic factor
The roots \(\frac{1 + i\sqrt{3}}{2}\) and \(\frac{1 - i\sqrt{3}}{2}\) can be used to form a quadratic factor. The quadratic can be expressed as:
\[ (x - \frac{1 + i\sqrt{3}}{2})(x - \frac{1 - i\sqrt{3}}{2}) \]
Using the identity for the product of conjugates, we get:
\[ = \left(x - \frac{1}{2} - i\frac{\sqrt{3}}{2}\right)\left(x - \frac{1}{2} + i\frac{\sqrt{3}}{2}\right) = \left(x - \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 \]

### Step 3: Simplify the quadratic factor
Calculating the above expression:
\[ = \left(x - \frac{1}{2}\right)^2 + \frac{3}{4} \]
Expanding this gives:
\[ = x^2 - x + \frac{1}{4} + \frac{3}{4} = x^2 - x + 1 \]

### Step 4: Divide the original polynomial by the quadratic factor
Now we need to divide the original polynomial \(3x^4 - 10x^3 + 4x^2 - x - 6\) by \(x^2 - x + 1\). We can use polynomial long division for this.

1. Divide the leading term: \(3x^4 \div x^2 = 3x^2\).
2. Multiply \(3x^2\) by \(x^2 - x + 1\):
\[ 3x^2(x^2 - x + 1) = 3x^4 - 3x^3 + 3x^2 \]
3. Subtract from the original polynomial:
\[ (3x^4 - 10x^3 + 4x^2) - (3x^4 - 3x^3 + 3x^2) = -7x^3 + x^2 \]
4. Bring down the next term:
\[ -7x^3 + x^2 - x - 6 \]
5. Divide the leading term: \(-7x^3 \div x^2 = -7x\).
6. Multiply \(-7x\) by \(x^2 - x + 1\):
\[ -7x(x^2 - x + 1) = -7x^3 + 7x^2 - 7x \]
7. Subtract:
\[ (-7x^3 + x^2 - x) - (-7x^3 + 7x^2 - 7x) = -6x^2 + 6x - 6 \]
8. Bring down the last term:
\[ -6x^2 + 6x - 6 \]
9. Divide the leading term: \(-6x^2 \div x^2 = -6\).
10. Multiply \(-6\) by \(x^2 - x + 1\):
\[ -6(x^2 - x + 1) = -6x^2 + 6x - 6 \]
11. Subtract:
\[ (-6x^2 + 6x - 6) - (-6x^2 + 6x - 6) = 0 \]

### Step 5: Write the complete factorization
The division shows that:
\[ 3x^4 - 10x^3 + 4x^2 - x - 6 = (x^2 - x + 1)(3x^2 - 7x - 6) \]

### Step 6: Solve the quadratic \(3x^2 - 7x - 6 = 0\)
Now we can solve the quadratic equation \(3x^2 - 7x - 6 = 0\) using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{7 \pm \sqrt{(-7)^2 - 4 \cdot 3 \cdot (-6)}}{2 \cdot 3} \]
Calculating the discriminant:
\[ = \frac{7 \pm \sqrt{49 + 72}}{6} = \frac{7 \pm \sqrt{121}}{6} = \frac{7 \pm 11}{6} \]
Thus, we have two solutions:
1. \(x = \frac{18}{6} = 3\)
2. \(x = \frac{-4}{6} = -\frac{2}{3}\)

### Final Roots
The complete set of roots for the equation \(3x^4 - 10x^3 + 4x^2 - x - 6 = 0\) is:
1. \(x = \frac{1 + i\sqrt{3}}{2}\)
2. \(x = \frac{1 - i\sqrt{3}}{2}\)
3. \(x = 3\)
4. \(x = -\frac{2}{3}\)