Divide :
`4a^(2) + 12ab + 9b^(2) - 25c^(2)` by 2a + 3b + 5c

Answer

To divide the polynomial \(4a^2 + 12ab + 9b^2 - 25c^2\) by the binomial \(2a + 3b + 5c\), we will use polynomial long division. Here’s the step-by-step solution:

### Step 1: Set up the division
We will divide \(4a^2 + 12ab + 9b^2 - 25c^2\) by \(2a + 3b + 5c\).

### Step 2: Divide the leading term
The leading term of the dividend is \(4a^2\) and the leading term of the divisor is \(2a\). We divide these:
\[ \frac{4a^2}{2a} = 2a \]
This means the first term of our quotient is \(2a\).

### Step 3: Multiply and subtract
Now we multiply \(2a\) by the entire divisor \(2a + 3b + 5c\):
\[ 2a(2a + 3b + 5c) = 4a^2 + 6ab + 10ac \]
Next, we subtract this from the original polynomial:
\[ (4a^2 + 12ab + 9b^2 - 25c^2) - (4a^2 + 6ab + 10ac) \]
This simplifies to:
\[ (12ab - 6ab) + 9b^2 - 10ac - 25c^2 = 6ab + 9b^2 - 10ac - 25c^2 \]

### Step 4: Repeat the process
Now we take the new polynomial \(6ab + 9b^2 - 10ac - 25c^2\) and divide the leading term \(6ab\) by the leading term of the divisor \(2a\):
\[ \frac{6ab}{2a} = 3b \]
So, the next term in our quotient is \(3b\).

### Step 5: Multiply and subtract again
Multiply \(3b\) by the divisor:
\[ 3b(2a + 3b + 5c) = 6ab + 9b^2 + 15bc \]
Now, subtract this from the current polynomial:
\[ (6ab + 9b^2 - 10ac - 25c^2) - (6ab + 9b^2 + 15bc) \]
This simplifies to:
\[ (-10ac - 25c^2 - 15bc) = -10ac - 25c^2 - 15bc \]

### Step 6: Divide the leading term again
Now we take the leading term \(-10ac\) and divide by \(2a\):
\[ \frac{-10ac}{2a} = -5c \]
So, the next term in our quotient is \(-5c\).

### Step 7: Multiply and subtract one last time
Multiply \(-5c\) by the divisor:
\[ -5c(2a + 3b + 5c) = -10ac - 15bc - 25c^2 \]
Now, subtract this from the current polynomial:
\[ (-10ac - 25c^2 - 15bc) - (-10ac - 15bc - 25c^2) = 0 \]
This means there is no remainder.

### Final Result
The quotient of the division is:
\[ \boxed{2a + 3b - 5c} \]