Find the LCM of `3y +12, y^2 –16 `and `y^4 - 64y`

Answer

To find the LCM of the polynomials \(3y + 12\), \(y^2 - 16\), and \(y^4 - 64y\), we will follow these steps:

### Step 1: Factor each polynomial

1. **Factor \(3y + 12\)**:
\[ 3y + 12 = 3(y + 4) \]

2. **Factor \(y^2 - 16\)**:
This is a difference of squares:
\[ y^2 - 16 = (y - 4)(y + 4) \]

3. **Factor \(y^4 - 64y\)**:
First, factor out \(y\):
\[ y^4 - 64y = y(y^3 - 64) \]
Then, factor \(y^3 - 64\) as a difference of cubes:
\[ y^3 - 64 = (y - 4)(y^2 + 4y + 16) \]
So, we have:
\[ y^4 - 64y = y(y - 4)(y^2 + 4y + 16) \]

### Step 2: Write down the factored forms

Now we have:
- \(3y + 12 = 3(y + 4)\)
- \(y^2 - 16 = (y - 4)(y + 4)\)
- \(y^4 - 64y = y(y - 4)(y^2 + 4y + 16)\)

### Step 3: Identify the unique factors

The unique factors from all three polynomials are:
- \(3\)
- \(y + 4\)
- \(y - 4\)
- \(y\)
- \(y^2 + 4y + 16\)

### Step 4: Form the LCM

The LCM is the product of the highest powers of all unique factors:
\[ \text{LCM} = 3 \cdot (y + 4) \cdot (y - 4) \cdot y \cdot (y^2 + 4y + 16) \]

### Step 5: Write the final answer

Thus, the LCM of the given polynomials is:
\[ \text{LCM} = 3y(y + 4)(y - 4)(y^2 + 4y + 16) \]

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