solve the following equation , given that they have multiple roots .
`3x^4+16x^3+24x^2-16=0`

Answer

To solve the equation \(3x^4 + 16x^3 + 24x^2 - 16 = 0\) given that it has multiple roots, we can follow these steps:

### Step 1: Factor the equation
We start by rewriting the equation:
\[ 3x^4 + 16x^3 + 24x^2 - 16 = 0 \]
We can group the terms to identify possible factors. Notice that we can factor out \(x + 2\) since the equation has multiple roots.

### Step 2: Use synthetic division or polynomial long division
Since \(x + 2\) is a factor, we can perform polynomial long division or synthetic division to divide \(3x^4 + 16x^3 + 24x^2 - 16\) by \(x + 2\).

Performing synthetic division with \(x = -2\):
- Coefficients: \(3, 16, 24, 0, -16\)
- Bring down the \(3\).
- Multiply \(3\) by \(-2\) and add to \(16\): \(16 - 6 = 10\).
- Multiply \(10\) by \(-2\) and add to \(24\): \(24 - 20 = 4\).
- Multiply \(4\) by \(-2\) and add to \(0\): \(0 - 8 = -8\).
- Multiply \(-8\) by \(-2\) and add to \(-16\): \(-16 + 16 = 0\).

The result of the division is:
\[ 3x^3 + 10x^2 + 4x - 8 \]

### Step 3: Factor the cubic polynomial
Next, we need to factor \(3x^3 + 10x^2 + 4x - 8\). We can try to factor by grouping:
\[ 3x^3 + 10x^2 + 4x - 8 = (3x^3 + 6x^2) + (4x - 8) = 3x^2(x + 2) + 4(x - 2) \]
This does not seem to work directly, so we will use the Rational Root Theorem to find possible rational roots.

Testing \(x = 2\):
\[ 3(2)^3 + 10(2)^2 + 4(2) - 8 = 24 + 40 + 8 - 8 = 64 \quad \text{(not a root)} \]
Testing \(x = -2\):
\[ 3(-2)^3 + 10(-2)^2 + 4(-2) - 8 = -24 + 40 - 8 - 8 = 0 \quad \text{(is a root)} \]

### Step 4: Factor out \(x + 2\) again
Now that we know \(x + 2\) is a factor, we can divide \(3x^3 + 10x^2 + 4x - 8\) by \(x + 2\) again:
Using synthetic division again with \(x = -2\):
- Coefficients: \(3, 10, 4, -8\)
- Bring down \(3\).
- Multiply \(3\) by \(-2\) and add to \(10\): \(10 - 6 = 4\).
- Multiply \(4\) by \(-2\) and add to \(4\): \(4 - 8 = -4\).
- Multiply \(-4\) by \(-2\) and add to \(-8\): \(-8 + 8 = 0\).

The result of the division is:
\[ 3x^2 + 4x - 4 \]

### Step 5: Solve the quadratic equation
Now we need to solve \(3x^2 + 4x - 4 = 0\). We can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 3 \cdot (-4)}}{2 \cdot 3} \]
Calculating the discriminant:
\[ b^2 - 4ac = 16 + 48 = 64 \]
Thus,
\[ x = \frac{-4 \pm 8}{6} \]
Calculating the two roots:
1. \(x = \frac{4}{6} = \frac{2}{3}\)
2. \(x = \frac{-12}{6} = -2\)

### Final roots
The roots of the original equation \(3x^4 + 16x^3 + 24x^2 - 16 = 0\) are:
- \(x = -2\) (with multiplicity 3)
- \(x = \frac{2}{3}\) (with multiplicity 1)

### Summary of the solution:
The roots of the equation are \(x = -2\) (multiple root) and \(x = \frac{2}{3}\).