Use Pascal's triangle to expand [tex]\((x+3)^4\)[/tex].

A. [tex]\(x^4 + 20x^3 + 90x^2 + 135x + 81\)[/tex]
B. [tex]\(x^4 + 12x^3 + 54x^2 + 108x + 81\)[/tex]
C. [tex]\(x^4 - 12x^3 + 54x^2 - 108x + 81\)[/tex]
D. [tex]\(x^4 - 20x^3 + 90x^2 - 135x + 81\)[/tex]



Answer :

To expand [tex]\((x + 3)^4\)[/tex] using Pascal's Triangle, we follow these steps:

### Step 1: Identify the coefficients from Pascal's Triangle
For the expansion of [tex]\((x + y)^n\)[/tex], the coefficients are the elements of the row of Pascal's Triangle corresponding to [tex]\(n\)[/tex]. For [tex]\(n = 4\)[/tex], the row is [tex]\([1, 4, 6, 4, 1]\)[/tex].

### Step 2: Write the general form of the Binomial Expansion
Using the coefficients from Pascal's Triangle, the general form of the binomial expansion can be written as:
[tex]\[ (x + 3)^4 = \sum_{k=0}^{4} \binom{4}{k} x^{4-k} \cdot 3^k \][/tex]

### Step 3: Substitute and simplify each term
- For [tex]\(k = 0\)[/tex]: [tex]\(\binom{4}{0} x^{4} \cdot 3^0 = 1 \cdot x^4 \cdot 1 = x^4\)[/tex]
- For [tex]\(k = 1\)[/tex]: [tex]\(\binom{4}{1} x^{3} \cdot 3^1 = 4 \cdot x^3 \cdot 3 = 12x^3\)[/tex]
- For [tex]\(k = 2\)[/tex]: [tex]\(\binom{4}{2} x^{2} \cdot 3^2 = 6 \cdot x^2 \cdot 9 = 54x^2\)[/tex]
- For [tex]\(k = 3\)[/tex]: [tex]\(\binom{4}{3} x^{1} \cdot 3^3 = 4 \cdot x \cdot 27 = 108x\)[/tex]
- For [tex]\(k = 4\)[/tex]: [tex]\(\binom{4}{4} x^{0} \cdot 3^4 = 1 \cdot 1 \cdot 81 = 81\)[/tex]

### Step 4: Combine the terms
Adding all of the simplified terms together, we get:
[tex]\[ (x + 3)^4 = x^4 + 12x^3 + 54x^2 + 108x + 81 \][/tex]

### Conclusion
The expanded form of [tex]\((x + 3)^4\)[/tex] is:
[tex]\[ x^4 + 12x^3 + 54x^2 + 108x + 81 \][/tex]

Thus, the correct answer is:
[tex]\[ \boxed{x^4 + 12 x^3 + 54 x^2 + 108 x + 81} \][/tex]

This matches option B.

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