If `(k+y)` is a factor of each of the polynomials `y^(2)+2y-15` and `y^(3)+a`, find the values of `k` and `a`.

Answer

To solve the problem, we need to find the values of \( k \) and \( a \) such that \( (k + y) \) is a factor of both polynomials \( y^2 + 2y - 15 \) and \( y^3 + a \).

### Step 1: Use the Factor Theorem
According to the Factor Theorem, if \( (k + y) \) is a factor of a polynomial \( p(y) \), then \( p(-k) = 0 \).

### Step 2: Apply to the First Polynomial
Let's first apply this to the polynomial \( y^2 + 2y - 15 \):
- We set \( y = -k \).
- Substitute \( -k \) into the polynomial:
\[ p(-k) = (-k)^2 + 2(-k) - 15 = k^2 - 2k - 15 \]
- For \( (k + y) \) to be a factor, we need:
\[ k^2 - 2k - 15 = 0 \]
This is our **Equation 1**.

### Step 3: Solve Equation 1
We will solve the quadratic equation \( k^2 - 2k - 15 = 0 \) using the middle-term splitting method:
- The factors of \(-15\) that add up to \(-2\) are \(3\) and \(-5\).
- We can rewrite the equation as:
\[ k^2 + 3k - 5k - 15 = 0 \]
- Factor by grouping:
\[ k(k + 3) - 5(k + 3) = 0 \]
- Factor out \( (k + 3) \):
\[ (k + 3)(k - 5) = 0 \]
- Setting each factor to zero gives us:
\[ k + 3 = 0 \quad \text{or} \quad k - 5 = 0 \]
Thus, \( k = -3 \) or \( k = 5 \).

### Step 4: Apply to the Second Polynomial
Next, we apply \( (k + y) \) to the second polynomial \( y^3 + a \):
- Again, set \( y = -k \):
\[ q(-k) = (-k)^3 + a = -k^3 + a \]
- For \( (k + y) \) to be a factor, we need:
\[ -k^3 + a = 0 \quad \Rightarrow \quad a = k^3 \]
This is our **Equation 2**.

### Step 5: Find Values of \( a \)
Now we will find the corresponding values of \( a \) for both values of \( k \):
1. If \( k = -3 \):
\[ a = (-3)^3 = -27 \]
2. If \( k = 5 \):
\[ a = 5^3 = 125 \]

### Final Results
Thus, the values of \( k \) and \( a \) are:
- For \( k = -3 \), \( a = -27 \)
- For \( k = 5 \), \( a = 125 \)

### Summary of Solutions
- \( (k, a) = (-3, -27) \) or \( (5, 125) \)