Foctorise :
` (a^(2) +3a -5) (a^(2) +3a +2) +6.`

Answer

To factorize the expression \( (a^2 + 3a - 5)(a^2 + 3a + 2) + 6 \), we can follow these steps:

### Step 1: Substitute for convenience
Let \( t = a^2 + 3a \). Then we can rewrite the expression as:
\[ (t - 5)(t + 2) + 6 \]

### Step 2: Expand the expression
Now, we expand \( (t - 5)(t + 2) \):
\[ t^2 + 2t - 5t - 10 + 6 = t^2 - 3t - 4 \]

### Step 3: Factor the quadratic expression
Next, we need to factor \( t^2 - 3t - 4 \). We look for two numbers that multiply to \(-4\) and add to \(-3\). These numbers are \(-4\) and \(1\).
Thus, we can factor it as:
\[ (t - 4)(t + 1) \]

### Step 4: Substitute back for \( t \)
Now, we substitute back \( t = a^2 + 3a \):
\[ (a^2 + 3a - 4)(a^2 + 3a + 1) \]

### Step 5: Factor the first term further
Now, we can factor \( a^2 + 3a - 4 \) further. We need two numbers that multiply to \(-4\) and add to \(3\). These numbers are \(4\) and \(-1\):
\[ (a + 4)(a - 1) \]

### Final Expression
Thus, the complete factorization of the original expression is:
\[ (a + 4)(a - 1)(a^2 + 3a + 1) \]