Step 1: Verify the assertion using the quadratic formula instead of factorization.
Rather than splitting the middle term to factor the polynomial, let's plug the coefficients straight into the quadratic formula to find the zeroes of $p(y) = y^2 + 4y + 3$.
Step 2: Identify the coefficients.
Comparing with $ay^2 + by + c$: $a = 1$, $b = 4$, $c = 3$.
Step 3: Compute the discriminant.
\[ D = b^2 - 4ac = 4^2 - 4(1)(3) = 16 - 12 = 4 \]
Since $D = 4 > 0$, the equation has two distinct real roots, this already tells us the answer to the assertion before we even find the roots.
Step 4: Apply the quadratic formula to find the actual zeroes.
\[ y = \frac{-b \pm \sqrt{D}}{2a} = \frac{-4 \pm \sqrt{4}}{2(1)} = \frac{-4 \pm 2}{2} \]
This gives two values:
\[ y = \frac{-4 + 2}{2} = -1 \quad \text{and} \quad y = \frac{-4 - 2}{2} = -3 \]
So $p(y)$ has exactly two distinct zeroes, $-1$ and $-3$. Assertion (A) is true.
Step 5: Check Reason (R) using the degree of the polynomial.
A polynomial of degree $n$ can have at most $n$ real zeroes. Since $p(y)$ has degree 2, it can have at most 2 zeroes. This general statement is true, so Reason (R) is true.
Step 6: Decide if Reason (R) explains Assertion (A).
Reason (R) only tells us the ceiling on the number of zeroes, that there cannot be more than 2. It does not by itself guarantee that this particular polynomial actually reaches that ceiling with 2 distinct zeroes rather than landing on 1 repeated zero or 0 real zeroes. What actually decides that is the discriminant we computed in Step 3, which came out positive. So Reason (R) is true but is not the direct explanation of why Assertion (A) holds.
Step 7: Final answer.
Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A). This is option (B).
\[ \boxed{\text{Option (B)}} \]