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Certainly! To expand the function [tex]\( f(x) = (3x + 2)^4 \)[/tex], we can follow the binomial expansion theorem. Here’s the detailed step-by-step solution:
1. Identify the binomial expression and its components:
[tex]\[ f(x) = (3x + 2)^4 \][/tex]
We can express this in the form [tex]\((a + b)^n\)[/tex] where:
- [tex]\( a = 3x \)[/tex]
- [tex]\( b = 2 \)[/tex]
- [tex]\( n = 4 \)[/tex]
2. Use the binomial expansion theorem:
According to the binomial expansion theorem:
[tex]\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \][/tex]
Where [tex]\(\binom{n}{k}\)[/tex] are the binomial coefficients.
3. Apply the binomial expansion to [tex]\( (3x + 2)^4 \)[/tex]:
Expanding step-by-step we get:
[tex]\[ (3x + 2)^4 = \binom{4}{0}(3x)^4(2)^0 + \binom{4}{1}(3x)^3(2)^1 + \binom{4}{2}(3x)^2(2)^2 + \binom{4}{3}(3x)^1(2)^3 + \binom{4}{4}(3x)^0(2)^4 \][/tex]
4. Calculate and simplify each term:
- For [tex]\(\binom{4}{0}(3x)^4(2)^0\)[/tex]:
[tex]\[ \binom{4}{0} = 1, \quad (3x)^4 = 81x^4, \quad (2)^0 = 1 \][/tex]
So, [tex]\(\binom{4}{0} \cdot 81x^4 \cdot 1 = 81x^4\)[/tex]
- For [tex]\(\binom{4}{1}(3x)^3(2)^1\)[/tex]:
[tex]\[ \binom{4}{1} = 4, \quad (3x)^3 = 27x^3, \quad (2)^1 = 2 \][/tex]
So, [tex]\(\binom{4}{1} \cdot 27x^3 \cdot 2 = 4 \cdot 27x^3 \cdot 2 = 216x^3\)[/tex]
- For [tex]\(\binom{4}{2}(3x)^2(2)^2\)[/tex]:
[tex]\[ \binom{4}{2} = 6, \quad (3x)^2 = 9x^2, \quad (2)^2 = 4 \][/tex]
So, [tex]\(\binom{4}{2} \cdot 9x^2 \cdot 4 = 6 \cdot 9x^2 \cdot 4 = 216x^2\)[/tex]
- For [tex]\(\binom{4}{3}(3x)^1(2)^3\)[/tex]:
[tex]\[ \binom{4}{3} = 4, \quad (3x)^1 = 3x, \quad (2)^3 = 8 \][/tex]
So, [tex]\(\binom{4}{3} \cdot 3x \cdot 8 = 4 \cdot 3x \cdot 8 = 96x\)[/tex]
- For [tex]\(\binom{4}{4}(3x)^0(2)^4\)[/tex]:
[tex]\[ \binom{4}{4} = 1, \quad (3x)^0 = 1, \quad (2)^4 = 16 \][/tex]
So, [tex]\(\binom{4}{4} \cdot 1 \cdot 16 = 1 \cdot 1 \cdot 16 = 16\)[/tex]
5. Combine all the terms to get the expanded form:
[tex]\[ f(x) = (3x + 2)^4 = 81x^4 + 216x^3 + 216x^2 + 96x + 16 \][/tex]
Therefore, the expanded form of [tex]\( f(x) = (3x + 2)^4 \)[/tex] is:
[tex]\[ 81x^4 + 216x^3 + 216x^2 + 96x + 16 \][/tex]
1. Identify the binomial expression and its components:
[tex]\[ f(x) = (3x + 2)^4 \][/tex]
We can express this in the form [tex]\((a + b)^n\)[/tex] where:
- [tex]\( a = 3x \)[/tex]
- [tex]\( b = 2 \)[/tex]
- [tex]\( n = 4 \)[/tex]
2. Use the binomial expansion theorem:
According to the binomial expansion theorem:
[tex]\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \][/tex]
Where [tex]\(\binom{n}{k}\)[/tex] are the binomial coefficients.
3. Apply the binomial expansion to [tex]\( (3x + 2)^4 \)[/tex]:
Expanding step-by-step we get:
[tex]\[ (3x + 2)^4 = \binom{4}{0}(3x)^4(2)^0 + \binom{4}{1}(3x)^3(2)^1 + \binom{4}{2}(3x)^2(2)^2 + \binom{4}{3}(3x)^1(2)^3 + \binom{4}{4}(3x)^0(2)^4 \][/tex]
4. Calculate and simplify each term:
- For [tex]\(\binom{4}{0}(3x)^4(2)^0\)[/tex]:
[tex]\[ \binom{4}{0} = 1, \quad (3x)^4 = 81x^4, \quad (2)^0 = 1 \][/tex]
So, [tex]\(\binom{4}{0} \cdot 81x^4 \cdot 1 = 81x^4\)[/tex]
- For [tex]\(\binom{4}{1}(3x)^3(2)^1\)[/tex]:
[tex]\[ \binom{4}{1} = 4, \quad (3x)^3 = 27x^3, \quad (2)^1 = 2 \][/tex]
So, [tex]\(\binom{4}{1} \cdot 27x^3 \cdot 2 = 4 \cdot 27x^3 \cdot 2 = 216x^3\)[/tex]
- For [tex]\(\binom{4}{2}(3x)^2(2)^2\)[/tex]:
[tex]\[ \binom{4}{2} = 6, \quad (3x)^2 = 9x^2, \quad (2)^2 = 4 \][/tex]
So, [tex]\(\binom{4}{2} \cdot 9x^2 \cdot 4 = 6 \cdot 9x^2 \cdot 4 = 216x^2\)[/tex]
- For [tex]\(\binom{4}{3}(3x)^1(2)^3\)[/tex]:
[tex]\[ \binom{4}{3} = 4, \quad (3x)^1 = 3x, \quad (2)^3 = 8 \][/tex]
So, [tex]\(\binom{4}{3} \cdot 3x \cdot 8 = 4 \cdot 3x \cdot 8 = 96x\)[/tex]
- For [tex]\(\binom{4}{4}(3x)^0(2)^4\)[/tex]:
[tex]\[ \binom{4}{4} = 1, \quad (3x)^0 = 1, \quad (2)^4 = 16 \][/tex]
So, [tex]\(\binom{4}{4} \cdot 1 \cdot 16 = 1 \cdot 1 \cdot 16 = 16\)[/tex]
5. Combine all the terms to get the expanded form:
[tex]\[ f(x) = (3x + 2)^4 = 81x^4 + 216x^3 + 216x^2 + 96x + 16 \][/tex]
Therefore, the expanded form of [tex]\( f(x) = (3x + 2)^4 \)[/tex] is:
[tex]\[ 81x^4 + 216x^3 + 216x^2 + 96x + 16 \][/tex]
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