Divide `(64x^(2)+48xy+9y^(2))(x+2)` by `(8x^(2)+16x+3xy+6y)`

Options

A. `8x+9y`
B. `2x+14y`
C. `8x+3y`
D. `3x-8y`

Answer

To solve the problem of dividing \((64x^{2}+48xy+9y^{2})(x+2)\) by \((8x^{2}+16x+3xy+6y)\), we can follow these steps:

### Step 1: Factor the numerator
The numerator is \((64x^{2}+48xy+9y^{2})(x+2)\). First, we need to factor \(64x^{2}+48xy+9y^{2}\).

Notice that:
\[ 64x^{2} + 48xy + 9y^{2} = (8x)^{2} + 2(8x)(3y) + (3y)^{2} \]
This can be recognized as a perfect square trinomial, which factors to:
\[ (8x + 3y)^{2} \]

Thus, the numerator becomes:
\[ (8x + 3y)^{2}(x + 2) \]

### Step 2: Factor the denominator
Now, we will factor the denominator \(8x^{2}+16x+3xy+6y\).

We can group the terms:
\[ (8x^{2} + 16x) + (3xy + 6y) \]
Factoring out common terms gives us:
\[ 8x(x + 2) + 3y(x + 2) \]
Now, we can factor out \((x + 2)\):
\[ (x + 2)(8x + 3y) \]

### Step 3: Write the division
Now we can rewrite the division:
\[ \frac{(8x + 3y)^{2}(x + 2)}{(x + 2)(8x + 3y)} \]

### Step 4: Cancel common factors
We can cancel the common factor \((x + 2)\) from the numerator and denominator:
\[ \frac{(8x + 3y)^{2}}{8x + 3y} \]
This simplifies to:
\[ 8x + 3y \]

### Final Answer
Thus, the final answer is:
\[ \boxed{8x + 3y} \]