Solve x²+16x+64=0: Perfect Square Trinomial Decomposition

Let's try to factor using quick trinomial factoring the given expression:

$x^2+16x+64\\$

Let's look for a pair of numbers whose product is the free term in the expression, and their sum is the coefficient of the first-degree term in the expression, meaning two numbers $m,\hspace{2pt}n$ that satisfy:

$m\cdot n=64\\ m+n=16\\$From the first requirement mentioned, namely- from the multiplication, note that the product of the numbers we're looking for must yield a positive result and therefore we can conclude that both numbers have equal signs, according to multiplication rules, now we'll remember that 64 has several possible pairs of whole number factors, we won't list all possibilities here, but note that:

$64=8^2=8\cdot8$ and- $16=8+8$ Continuing, meeting the second requirement mentioned, along with the fact that the signs of the numbers we're looking for are equal to each other will lead to the conclusion that the only possibility for the two numbers we're looking for is:

$\begin{cases} m=8\\ n=8\end{cases}$

(We can check all other factor pairs of 64 to verify this is the only possibility, but once a suitable option is found - it must be the only one)

Therefore we can factor the given expression to:

$x^2+16x+64\\ \downarrow\\ (x+8)(x+8)$

And therefore clearly the suggested factorization in the problem is correct.

That is - the correct answer is answer A.

Note:

The given question could also be solved by expanding the parentheses in the suggested expression:

$(x+8)(x+8)$(using the extended distribution law or alternatively using the shortened multiplication formula for squared binomial in this case), and checking if indeed we get the given expression:

$x^2+16x+64$, however it's obviously better to try to factor the given expression- for practice purposes.