Find the other zero of the quadratic polynomial `y^2 + 7y - 60` if one zero is -12

Answer

To find the other zero of the quadratic polynomial \( y^2 + 7y - 60 \) given that one zero is -12, we can follow these steps:

### Step 1: Identify the known zero
We know that one zero (let's call it \( \alpha \)) is -12.

### Step 2: Use the relationship of the sum of zeros
For a quadratic polynomial of the form \( ay^2 + by + c \), the sum of the zeros \( \alpha + \beta \) can be found using the formula:
\[ \alpha + \beta = -\frac{b}{a} \]
where \( b \) is the coefficient of \( y \) and \( a \) is the coefficient of \( y^2 \).

### Step 3: Substitute the values
In our polynomial \( y^2 + 7y - 60 \):
- \( a = 1 \)
- \( b = 7 \)

Thus, we can calculate the sum of the zeros:
\[ \alpha + \beta = -\frac{7}{1} = -7 \]

### Step 4: Substitute the known zero
Now we substitute \( \alpha = -12 \) into the equation:
\[ -12 + \beta = -7 \]

### Step 5: Solve for the other zero
To find \( \beta \), we rearrange the equation:
\[ \beta = -7 + 12 \]
\[ \beta = 5 \]

### Conclusion
The other zero of the quadratic polynomial \( y^2 + 7y - 60 \) is \( 5 \).

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