Finding the Coefficient of x in Binomial Expansion

The question asks for the coefficient of the term containing 'x' in the expansion of the expression $(x + 3)^3$. This is a problem involving binomial expansion, specifically using the binomial theorem or direct multiplication.

Understanding Binomial Expansion

A binomial is an algebraic expression with two terms. Expanding a binomial raised to a power, like $(a+b)^n$, involves finding the sum of terms that result from the multiplication. The binomial theorem provides a formula for this expansion.

Applying the Binomial Theorem for $(x+3)^3$

The binomial theorem states that for any non-negative integer n,

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

where \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\) is the binomial coefficient.

In our case, we have $(x+3)^3$, so \(a=x\), \(b=3\), and \(n=3\). The expansion will have terms for \(k=0, 1, 2, 3\).

• For \(k=0\): The term is \(\binom{3}{0} x^{3-0} 3^0 = 1 \cdot x^3 \cdot 1 = x^3\).
• For \(k=1\): The term is \(\binom{3}{1} x^{3-1} 3^1 = 3 \cdot x^2 \cdot 3 = 9x^2\). This term contains \(x^2\).
• For \(k=2\): The term is \(\binom{3}{2} x^{3-2} 3^2 = 3 \cdot x^1 \cdot 9 = 27x\). This term contains \(x^1\), which is 'x'.
• For \(k=3\): The term is \(\binom{3}{3} x^{3-3} 3^3 = 1 \cdot x^0 \cdot 27 = 27\). This is the constant term.

The full expansion is the sum of these terms:

\((x+3)^3 = x^3 + 9x^2 + 27x + 27\)

Identifying the Coefficient of x

From the expanded form \(x^3 + 9x^2 + 27x + 27\), we look for the term that includes 'x' raised to the power of 1. This term is \(27x\).

The coefficient of a term is the numerical factor multiplying the variable(s). In the term \(27x\), the numerical factor is 27.

Therefore, the coefficient of x in the expansion of \((x + 3)^3\) is 27.

Alternative Method: Direct Expansion of (x+3)³

We can also expand the expression by multiplying it out directly:

\((x+3)^3 = (x+3)(x+3)(x+3)\)

First, expand \((x+3)(x+3)\):

\((x+3)(x+3) = x(x+3) + 3(x+3) = x^2 + 3x + 3x + 9 = x^2 + 6x + 9\)

Now, multiply the result by \((x+3)\):

\((x^2 + 6x + 9)(x+3) = x^2(x+3) + 6x(x+3) + 9(x+3)\)

\(= (x^3 + 3x^2) + (6x^2 + 18x) + (9x + 27)\)

\(= x^3 + 3x^2 + 6x^2 + 18x + 9x + 27\)

Combine like terms:

\(= x^3 + (3x^2 + 6x^2) + (18x + 9x) + 27\)

\(= x^3 + 9x^2 + 27x + 27\)

Determining the Coefficient of x from Direct Expansion

The direct expansion gives us \(x^3 + 9x^2 + 27x + 27\). The term with 'x' is \(27x\). The coefficient of this term is 27.

Both the binomial theorem and direct expansion methods confirm that the coefficient of x is 27.

Revision Table: Binomial Expansion Concepts

Additional Information: Pascal's Triangle and Binomial Expansion

Pascal's Triangle provides a visual way to find the binomial coefficients \(\binom{n}{k}\). Each number in Pascal's Triangle is the sum of the two numbers directly above it. The rows correspond to the power 'n' in \((a+b)^n\), starting with row 0 for \(n=0\).

• Row 0 (n=0): 1
• Row 1 (n=1): 1, 1
• Row 2 (n=2): 1, 2, 1
• Row 3 (n=3): 1, 3, 3, 1

For \((x+3)^3\), \(n=3\), so we use the coefficients from Row 3: 1, 3, 3, 1. These are \(\binom{3}{0}, \binom{3}{1}, \binom{3}{2}, \binom{3}{3}\) respectively.

The terms are formed by taking these coefficients, decreasing the power of the first term (x) starting from n down to 0, and increasing the power of the second term (3) starting from 0 up to n.

Term 1 (k=0): \(1 \cdot x^3 \cdot 3^0 = x^3\)

Term 2 (k=1): \(3 \cdot x^2 \cdot 3^1 = 9x^2\)

Term 3 (k=2): \(3 \cdot x^1 \cdot 3^2 = 27x\)

Term 4 (k=3): \(1 \cdot x^0 \cdot 3^3 = 27\)

The expansion is \(x^3 + 9x^2 + 27x + 27\). The coefficient of x is 27.