`2xya^(2) +10 y +3xa^(2) +15 ` = ______

Options

A. `(2a^(2)+5)(2y+3x)`
B. `(xa^(2)+5)(2y+3)`
C. `(ya^(2)+3)(2x+5)`
D. `(xy+5)(2a^(2)+3)`

Answer

To factor the expression \(2xy a^2 + 10y + 3xa^2 + 15\), we can follow these steps:

### Step 1: Rearrange the terms
We can rearrange the terms to group similar ones together:
\[ 2xy a^2 + 3xa^2 + 10y + 15 \]

### Step 2: Group the terms
Now we will group the first two terms and the last two terms:
\[ (2xy a^2 + 3xa^2) + (10y + 15) \]

### Step 3: Factor out the common factors
In the first group \(2xy a^2 + 3xa^2\), we can factor out \(xa^2\):
\[ xa^2(2y + 3) \]
In the second group \(10y + 15\), we can factor out \(5\):
\[ 5(2y + 3) \]

### Step 4: Combine the factored terms
Now, we can combine the factored terms:
\[ xa^2(2y + 3) + 5(2y + 3) \]

### Step 5: Factor out the common binomial
Notice that \(2y + 3\) is common in both terms, so we can factor it out:
\[ (2y + 3)(xa^2 + 5) \]

### Final Answer
Thus, the factored form of the expression \(2xy a^2 + 10y + 3xa^2 + 15\) is:
\[ (2y + 3)(xa^2 + 5) \]

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