Understanding the Binomial Expansion for Coefficient of x^2

The problem requires us to find the numerical factor multiplying the $x^2$ term when the expression $(2x + y)^3$ is expanded. This involves using the principles of binomial expansion.

Applying the Binomial Theorem

The binomial theorem provides a formula to expand expressions of the form $(a + b)^n$. The formula is:

$$ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k $$

For the given expression, $(2x + y)^3$, we can identify the components:

• The first term, $a = 2x$.
• The second term, $b = y$.
• The power, $n = 3$.

Identifying the Term with $x^2$

The general term in the expansion is given by $T_{k+1} = \binom{n}{k} a^{n-k} b^k$. We need to find the term where $x$ has a power of 2.

Substituting $a = 2x$, $b = y$, and $n = 3$ into the general term formula:

$$ T_{k+1} = \binom{3}{k} (2x)^{3-k} y^k $$

We need the term involving $x^2$. The power of $x$ in the general term is $3-k$. So, we set this power equal to 2:

Solving this equation for $k$:

This means the term containing $x^2$ corresponds to $k=1$ (which is the $T_2$ term).

Calculating the $x^2$ Term

Using $n=3$ and $k=1$ in the general term formula:

$$ T_{1+1} = T_2 = \binom{3}{1} (2x)^{3-1} y^1 $$

Now, we calculate each part of this term:

• Calculate the binomial coefficient $$ \binom{3}{1} $$: $$ \binom{3}{1} = \frac{3!}{1!(3-1)!} = \frac{3!}{1!2!} = \frac{3 \times 2 \times 1}{(1)(2 \times 1)} = 3 $$
• Calculate the part involving $x$: $$ (2x)^{3-1} = (2x)^2 = 2^2 \times x^2 = 4x^2 $$
• Calculate the part involving $y$: $$ y^1 = y $$

Multiply these parts together to get the full term:

$$ T_2 = 3 \times (4x^2) \times y = 12x^2y $$

Determining the Coefficient of $x^2$

The term containing $x^2$ is $12x^2y$. The coefficient of $x^2$ is the factor that multiplies $x^2$. In this term, $12y$ multiplies $x^2$.

So, the coefficient of $x^2$ is $12y$.

Verifying the Result with Options

The calculated coefficient for the $x^2$ term is $12y$. Let's compare this with the given options:

Our result, $12y$, matches the value presented in Option 3.