Solve (x-8)(x+y): Binomial Expression Multiplication

Let's simplify the given expression, open the parentheses using the expanded distribution law:

$(\textcolor{red}{a}+\textcolor{blue}{b})(c+d)=\textcolor{red}{a}c+\textcolor{red}{a}d+\textcolor{blue}{b}c+\textcolor{blue}{b}d$

Note that in the formula template for the above distribution law, we take by default that the operation between the terms inside the parentheses is addition, therefore we won't forget of course that the sign preceding the term is an inseparable part of it, and we'll also apply the rules of sign multiplication and thus we can present any expression in parentheses, which we'll open using the above formula, first as an expression where addition operation exists between all terms:

$(x-8)(x+y)\\ (\textcolor{red}{x}+\textcolor{blue}{(-8)})(x+y)\\$Let's begin then with opening the parentheses:

$(\textcolor{red}{x}+\textcolor{blue}{(-8)})(x+y)\\ \textcolor{red}{x}\cdot x+\textcolor{red}{x}\cdot y+\textcolor{blue}{(-8)}\cdot x +\textcolor{blue}{(-8)}\cdot y\\ x^2+xy-8x -8y$

In calculating the above multiplications, we used the multiplication table and the laws of exponents for multiplication between terms with identical bases:

$a^m\cdot a^n=a^{m+n}$

Note that in the expression we got in the last stage there are four different terms, this is because there isn't even one pair of terms where the variables (different ones) have the same exponent, additionally the expression is already organized therefore the expression we got is the final and most simplified form:
$\textcolor{purple}{ x^2}\textcolor{green}{+xy}-8x \textcolor{orange}{-8y}\\$We highlighted the different terms using colors, and as emphasized before, we made sure that the sign preceding the term is an inseparable part of it,

We therefore concluded that the correct answer is answer A.