(x%2B1)(x%2B1)(x%2B1).jpeg "(x+1)(x+1)(x+1)(x+1)")
Expanding (x+1)(x+1)(x+1)(x+1)
This expression represents the multiplication of (x+1) by itself four times. It's a common problem in algebra and can be solved using the distributive property and some pattern recognition.
Method 1: Step-by-Step Multiplication
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Start with the first two factors: (x+1)(x+1) = x² + 2x + 1
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Multiply the result by the third factor: (x² + 2x + 1)(x+1) = x³ + 3x² + 3x + 1
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Finally, multiply the result by the fourth factor: (x³ + 3x² + 3x + 1)(x+1) = x⁴ + 4x³ + 6x² + 4x + 1
Method 2: Using the Binomial Theorem
The binomial theorem provides a shortcut for expanding expressions of the form (a+b)ⁿ. In this case, a = x, b = 1, and n = 4.
The binomial theorem states: (a + b)ⁿ = ∑ (n choose k) a^(n-k) b^k
where:
- ∑ represents the sum from k=0 to n
- (n choose k) represents the binomial coefficient, which is calculated as n!/(k!(n-k)!)
Applying this to our problem:
(x + 1)⁴ = ∑ (4 choose k) x^(4-k) 1^k
Let's expand this:
- k = 0: (4 choose 0) x⁴ 1⁰ = x⁴
- k = 1: (4 choose 1) x³ 1¹ = 4x³
- k = 2: (4 choose 2) x² 1² = 6x²
- k = 3: (4 choose 3) x¹ 1³ = 4x
- k = 4: (4 choose 4) x⁰ 1⁴ = 1
Summing these terms, we get: x⁴ + 4x³ + 6x² + 4x + 1
Conclusion
Both methods lead to the same result: (x+1)(x+1)(x+1)(x+1) = x⁴ + 4x³ + 6x² + 4x + 1. The binomial theorem provides a more concise and efficient solution, especially when dealing with higher powers.