Factoring Trinomials: Complete the equation

Let's simplify the expression given in the left side:

$(x+3)(x+\textcolor{purple}{\boxed{?}})$ For ease of calculation we will replace the square with the question mark (indicating the missing part that needs to be completed) with the letter $\textcolor{purple}{k}$, meaning we will perform the substitution:

$$$(x+3)(x+\textcolor{purple}{\boxed{?}})=x^2+5x+6 \\ \downarrow\\ (x+3)(x+\textcolor{purple}{k})=x^2+5x+6 \\$Next, we will expand the parentheses using the expanded distribution law:

$(\textcolor{red}{a}+\textcolor{blue}{b})(c+d)=\textcolor{red}{a}c+\textcolor{red}{a}d+\textcolor{blue}{b}c+\textcolor{blue}{b}d$Let's note that in the formula template for the distribution law mentioned we assume by default that the operation between the terms inside the parentheses is addition, so we won't forget of course that the sign preceding the term is an integral part of it, and we will also apply the laws of sign multiplication and thus we can represent any expression in parentheses, which we expand using the aforementioned formula, first, as an expression where addition is performed between all terms (if necessary),

Therefore, we will first represent each of the expressions in parentheses in the multiplication on the left side as an expression where addition exists:

$(x+3)(x+\textcolor{purple}{k})=x^2+5x+6 \\ \downarrow\\ \big(x+(+3)\big)\big(x+(\textcolor{purple}{+k})\big)=x^2+5x+6 \\$Now for convenience, let's write down again the expanded distribution law mentioned earlier:

$(\textcolor{red}{a}+\textcolor{blue}{b})(c+d)=\textcolor{red}{a}c+\textcolor{red}{a}d+\textcolor{blue}{b}c+\textcolor{blue}{b}d$And we'll apply it to our problem:

$\big (\textcolor{red}{x}+\textcolor{blue}{(+3)}\big)\big(x+(+\textcolor{purple}{k})\big)=x^2+5x+6 \\ \downarrow\\ \textcolor{red}{x}\cdot x +\textcolor{red}{x}\cdot (+\textcolor{purple}{k})+\textcolor{blue}{(+3)}\cdot x +\textcolor{blue}{(+3)}\cdot (+\textcolor{purple}{k})=x^2+5x+6 \\$We'll continue and apply the laws of multiplication signs, remembering that multiplying expressions with identical signs will yield a positive result, and multiplying expressions with different signs will yield a negative result:

$\textcolor{red}{x}\cdot x +\textcolor{red}{x}\cdot (+\textcolor{purple}{k})+\textcolor{blue}{(+3)}\cdot x +\textcolor{blue}{(+3)}\cdot (+\textcolor{purple}{k})=x^2+5x+6 \\ \downarrow\\ x^2+\textcolor{purple}{k}x+3x+3\textcolor{purple}{k}=x^2+5x+6 \\$Now, we want to present the expression on the left side in a form identical to the expression on the right side, that is - as a sum of three terms with different exponents: second power (squared), first power, and zero power (i.e., the free number - not dependent on x). To do this - we will factor out the part of the expression on the left side where the terms are in the first power:

$x^2+\underline{\textcolor{purple}{k}x+3x}+3\textcolor{purple}{k}=x^2+5x+6 \\ \downarrow\\ x^2+\underline{(\textcolor{purple}{k}+3)x}+3\textcolor{purple}{k}=x^2+5x+6$

Now in order for equality to hold - we require that the coefficient of the first-power term on both sides of the equation be identical and at the same time - we require that the free term on both sides of the equation be identical as well:

$x^2+\underline{\underline{(\textcolor{purple}{k}+3)}}x+\underline{\underline{\underline{3\textcolor{purple}{k}}}}=x^2+\underline{\underline{5}}x+\underline{\underline{\underline{6}}}$In other words, we require that:

$\begin{cases} \textcolor{purple}{k}+3=5\rightarrow\boxed{\textcolor{purple}{k}=2}\\ 3\textcolor{purple}{k}=6\rightarrow\boxed{\textcolor{purple}{k}=2} \end{cases}$

Let's summarize the solution steps:

$(x+3)(x+\textcolor{purple}{\boxed{?}})=x^2+5x+6 \leftrightarrow\textcolor{red}{\boxed{\textcolor{purple}{\boxed{?}}=\textcolor{purple}{k}}} \\ \downarrow\\ (x+3)(x+\textcolor{purple}{k})=x^2+5x+6 \\ \downarrow\\ x^2+\underline{\textcolor{purple}{k}x+3x}+3\textcolor{purple}{k}=x^2+5x+6 \\ \downarrow\\ x^2+\underline{(\textcolor{purple}{k}+3)x}+3\textcolor{purple}{k}=x^2+5x+6\\ \downarrow\\ x^2+\underline{\underline{(\textcolor{purple}{k}+3)}}x+\underline{\underline{\underline{3\textcolor{purple}{k}}}}=x^2+\underline{\underline{5}}x+\underline{\underline{\underline{6}}} \\ \begin{cases} \textcolor{purple}{k}+3=5\rightarrow\boxed{\textcolor{purple}{k}=2}\\ 3\textcolor{purple}{k}=6\rightarrow\boxed{\textcolor{purple}{k}=2} \end{cases} \\ \textcolor{red}{\bm{\rightarrow}\boxed{\textcolor{purple}{\boxed{?}}=\textcolor{purple}{2}}}$Therefore, the missing expression is the number $2$meaning - the correct answer is a'.