Factorise :
`4a^(2)+9b^(2)+16c^(2)+12ab-24bc-16ca`

Answer

To factorize the expression \( 4a^2 + 9b^2 + 16c^2 + 12ab - 24bc - 16ca \), we can follow these steps:

### Step 1: Rearrange the expression
We start with the expression:
\[ 4a^2 + 9b^2 + 16c^2 + 12ab - 24bc - 16ca \]

### Step 2: Group terms
We can group the terms in a way that helps us identify perfect squares:
\[ (4a^2 + 12ab + 9b^2) + (16c^2 - 24bc - 16ca) \]

### Step 3: Recognize perfect squares
The first group \( 4a^2 + 12ab + 9b^2 \) can be factored as:
\[ (2a + 3b)^2 \]
The second group \( 16c^2 - 24bc - 16ca \) can be rearranged and factored as:
\[ 16c^2 - 16ca - 24bc = 16(c^2 - \frac{16}{16}ca - \frac{24}{16}bc) = 16(c^2 - ca - \frac{3}{2}bc) \]

### Step 4: Factor the second group
To factor \( c^2 - ca - \frac{3}{2}bc \), we can use the quadratic formula or look for factors. However, we can rewrite it as:
\[ c^2 - ca - \frac{3}{2}bc = (c - 3b)(c + b) \]
Thus, we have:
\[ 16(c - 3b)(c + b) \]

### Step 5: Combine the factors
Now we can combine our factors:
\[ (2a + 3b)^2 + 16(c - 3b)(c + b) \]

### Step 6: Final factorization
Putting it all together, we can express the original polynomial as:
\[ (2a + 3b - 4c)(2a + 3b + 4c) \]

Thus, the complete factorization of the expression \( 4a^2 + 9b^2 + 16c^2 + 12ab - 24bc - 16ca \) is:
\[ (2a + 3b - 4c)(2a + 3b + 4c) \]