Divide :

`(y^(2) + 6y- 16) (y+ 2)` by `(y^(2)- 4)`

Options

A. `y+8`
B. `y-2`
C. `y+4`
D. `2y+4`

Answer

To solve the problem of dividing \((y^2 + 6y - 16)(y + 2)\) by \((y^2 - 4)\), we will follow these steps:

### Step 1: Factor the expressions
First, we need to factor both the numerator and the denominator.

1. **Factor the denominator**:
\[ y^2 - 4 = (y + 2)(y - 2) \quad \text{(This is a difference of squares)} \]

2. **Factor the numerator**:
We need to factor \(y^2 + 6y - 16\). We look for two numbers that multiply to \(-16\) and add to \(6\). The numbers \(8\) and \(-2\) work:
\[ y^2 + 6y - 16 = (y + 8)(y - 2) \]

### Step 2: Rewrite the expression
Now we can rewrite the original expression using the factored forms:
\[ \frac{(y + 8)(y - 2)(y + 2)}{(y + 2)(y - 2)} \]

### Step 3: Cancel common factors
Next, we can cancel the common factors in the numerator and denominator:
\[ \frac{(y + 8) \cancel{(y - 2)} \cancel{(y + 2)}}{\cancel{(y + 2)} \cancel{(y - 2)}} = y + 8 \]

### Final Answer
Thus, the result of the division is:
\[ \boxed{y + 8} \]

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