find the multiple roots of
` 8x^3 + 20 x^2 + 6 x-9 =0`

Answer

To find the multiple roots of the polynomial equation \( 8x^3 + 20x^2 + 6x - 9 = 0 \), we will follow these steps:

### Step 1: Define the function
Let \( f(x) = 8x^3 + 20x^2 + 6x - 9 \).

### Step 2: Find the derivative
We need to find the first derivative \( f'(x) \):
\[ f'(x) = \frac{d}{dx}(8x^3 + 20x^2 + 6x - 9) = 24x^2 + 40x + 6 \]

### Step 3: Set the derivative to zero
To find the critical points, we set the derivative equal to zero:
\[ 24x^2 + 40x + 6 = 0 \]

### Step 4: Simplify the equation
We can simplify this equation by dividing all terms by 2:
\[ 12x^2 + 20x + 3 = 0 \]

### Step 5: Factor the quadratic
Now, we will factor the quadratic equation \( 12x^2 + 20x + 3 = 0 \). We look for two numbers that multiply to \( 12 \cdot 3 = 36 \) and add to \( 20 \). The numbers \( 18 \) and \( 2 \) work:
\[ 12x^2 + 18x + 2x + 3 = 0 \]
Grouping the terms:
\[ (12x^2 + 18x) + (2x + 3) = 0 \]
Factoring by grouping:
\[ 6x(2x + 3) + 1(2x + 3) = 0 \]
Factoring out the common term:
\[ (2x + 3)(6x + 1) = 0 \]

### Step 6: Solve for x
Setting each factor to zero gives us:
1. \( 2x + 3 = 0 \) → \( x = -\frac{3}{2} \)
2. \( 6x + 1 = 0 \) → \( x = -\frac{1}{6} \)

### Step 7: Check for multiple roots
To check if \( x = -\frac{3}{2} \) is a multiple root, we substitute it back into \( f(x) \):
\[ f\left(-\frac{3}{2}\right) = 8\left(-\frac{3}{2}\right)^3 + 20\left(-\frac{3}{2}\right)^2 + 6\left(-\frac{3}{2}\right) - 9 \]
Calculating each term:
\[ = 8\left(-\frac{27}{8}\right) + 20\left(\frac{9}{4}\right) - 9 - 9 \]
\[ = -27 + 45 - 9 - 9 = 0 \]
Since \( f\left(-\frac{3}{2}\right) = 0 \), it confirms that \( x = -\frac{3}{2} \) is a root.

### Step 8: Use synthetic division to find other roots
Now we will use synthetic division to divide \( f(x) \) by \( (x + \frac{3}{2}) \) to find the remaining polynomial:
1. Divide \( 8x^3 + 20x^2 + 6x - 9 \) by \( x + \frac{3}{2} \).
2. Repeat the division to find the quotient polynomial.

After performing synthetic division twice, we will find the remaining factor:
\[ 8x - 4 = 0 \implies x = \frac{1}{2} \]

### Conclusion
The roots of the polynomial \( 8x^3 + 20x^2 + 6x - 9 = 0 \) are:
1. \( x = -\frac{3}{2} \) (multiple root)
2. \( x = -\frac{3}{2} \) (multiple root)
3. \( x = \frac{1}{2} \)