How do you multiply $(2x+1)(2x+1)$?

Hint: In this question, we have to multiply 2 binomial equations. Thus, for solving this problem, we use the FOIL method. In the FOIL method, F means we multiply the First terms together. O means the outermost term that is when we multiply the Outermost terms when binomials are placed side by side. I mean when we multiply the Innermost terms together and L means when we multiply the Last term together in each binomial. After that sum up the partial products of the two binomials beginning from the first, outer, inner, and then the last, to get our required answer.

Complete step by step answer:
According to the question, we have to solve $(2x+1)(2x+1)$.
Now we use the FOIL method, which is
We multiply the first terms together, we get
\[(a+b)(a+b)=a.(a)={{a}^{2}}\] ---- (1)
Now, multiply the outermost terms of both the binomial equations, we get
$(a+b)(a+b)=a.(b)=ab$ ---- (2)
Again multiply the innermost terms of the binomial equation, we get
$(a+b)(a+b)=b.(a)=ba$ ----- (3)
Eventually, multiply the last terms of both the binomial equation, to get
$(a+b)(a+b)=b.(b)={{b}^{2}}$ ----- (4)
Therefore, add all the partial products (1), (2), (3), and (4), we get
\[(a+b)(a+b)={{a}^{2}}+ab+ba+{{b}^{2}}\]
On further simplification, we get
\[(a+b)(a+b)={{a}^{2}}+2ab+{{b}^{2}}\]
Similarly, we solve $(2x+1)(2x+1)$ , where a=2x and b=1, we get
$\text{Equation: }(2x+1)(2x+1)$
We multiply the first terms together, we get
\[\Rightarrow (2x+1)(2x+1)=2x.(2x)=4{{x}^{2}}\] ---- (5)
Now, multiply the outermost terms of both the binomial equations, we get
\[\Rightarrow (2x+1)(2x+1)=2x.(1)=2x\] \[\] ---- (6)
Again multiply the innermost terms of the binomial equation, we get
\[\Rightarrow (2x+1)(2x+1)=1.(2x)=2x\] ----- (7)
Finally, multiply the last terms of both the binomial equation, to get
\[\Rightarrow (2x+1)(2x+1)=1.(1)=1\] ----- (8)
Therefore, add all the partial products (5), (6), (7), and (8), we get
\[\Rightarrow (2x+1)(2x+1)=4{{x}^{2}}+2x+2x+1\]
On further simplification, we get
\[\Rightarrow (2x+1)(2x+1)=4{{x}^{2}}+4x+1\] which is our required answer.

Note:
Always keep in mind the definition of the FOIL method, otherwise when you add all the partial products in the end you will get the wrong answer. One of the alternative methods we can solve this problem is by using the distributive property $(a+b)(a+b)=a(a+b)+b(a+b)$ . In all steps, we follow the distributive property to get our required answer.
An alternative method:
Equation: $(2x+1)(2x+1)$
Now, apply distributive property $(a+b)(a+b)=a(a+b)+b(a+b)$ , we get
$\Rightarrow (2x+1)(2x+1)=2x(2x+1)+1(2x+1)$
Again, distributive property $(a+b)(a+b)=a(a+b)+b(a+b)$ , we get
$\Rightarrow 2x(2x+1)+1(2x+1)=2x(2x)=2x(1)+1(2x)+1(1)$
On further simplifications, we get
$\Rightarrow 2x(2x)=2x(1)+1(2x)+1(1)=4{{x}^{2}}+2x+2x+1$
Therefore, we get
$\Rightarrow 4{{x}^{2}}+2x+2x+1=4{{x}^{2}}+4x+1$ which is our required answer.