Factorise:
`y ^(2) + 7y - 144`

Answer

To factorise the expression \( y^2 + 7y - 144 \), we can follow these steps:

### Step 1: Identify the coefficients
The expression is in the form \( ay^2 + by + c \), where:
- \( a = 1 \) (coefficient of \( y^2 \))
- \( b = 7 \) (coefficient of \( y \))
- \( c = -144 \) (constant term)

### Step 2: Find two numbers that multiply to \( ac \) and add to \( b \)
We need to find two numbers that multiply to \( a \cdot c = 1 \cdot (-144) = -144 \) and add to \( b = 7 \).

After checking the pairs of factors of \(-144\), we find:
- The numbers \( 16 \) and \(-9\) satisfy the conditions:
- \( 16 \times (-9) = -144 \)
- \( 16 + (-9) = 7 \)

### Step 3: Rewrite the middle term using the two numbers
We can rewrite the expression \( y^2 + 7y - 144 \) as:
\[ y^2 + 16y - 9y - 144 \]

### Step 4: Group the terms
Now, we will group the terms:
\[ (y^2 + 16y) + (-9y - 144) \]

### Step 5: Factor out the common terms in each group
From the first group \( (y^2 + 16y) \), we can factor out \( y \):
\[ y(y + 16) \]
From the second group \( (-9y - 144) \), we can factor out \(-9\):
\[ -9(y + 16) \]

### Step 6: Combine the factored groups
Now we can combine the factored terms:
\[ y(y + 16) - 9(y + 16) \]
Notice that \( (y + 16) \) is common in both terms, so we can factor it out:
\[ (y + 16)(y - 9) \]

### Final Answer
Thus, the factorised form of \( y^2 + 7y - 144 \) is:
\[ (y - 9)(y + 16) \]

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