Explanation:
Let's say we have (m*x+d)², with m and d representing constant values (a number). If we expand, we can see that
(m*x+d)² = (m*x+d) * (m*x+d) = m²*x²+2*m*x*d + d². Matching that up with
ax²+bx+c, the value multiplied by x² in our factored perfect square is m² (so m²=a), the value multiplied by x is 2md ( so 2md = b), and the constant is d².
Going back to the problem, we want to see how the values of a and c correspond with a perfect square trinomial. In our perfect square of (m*x+d)², our resulting trinomial has m² = a and d² = c. This points to the fact that a must be a square of something (for example, if a = 1, √1=1, so this works) as well as c if they are part of a perfect square trinomial. If a and c are both squares of other numbers, then it is possible that ax²+bx+c is a perfect square trinomial. If they are not, then it is not possible