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Question

(x+1)^5 Also this: (x+2)^3?

Answer 1

(x + 1)⁵

From the binomial theorem,

(x + 1)⁵ = 5c₀(x)⁵(1)⁰ + 5c₁(x)^{5 - 1}(1)¹ + 5c₂(x)^{5 - 2}(1)² + 5c₃(x)^{5 - 3}(1)³ + 5c₄(x)^{5 - 4}(1)⁴ + 5c₅(x)^{5 - 5}(1)⁵

(x + 1)⁵ = 5c₀x⁵ + 5c₁x⁴ + 5c₂x³ + 5c₃x² + 5c₄x + 5c₅

Substitute the values 5c₀ = 1, 5c₁= 5,  5c₂ = 10, 5c₃ = 10, 5c₄ = 5, and 5c₅= 1.

(x + 1)⁵ = (1)x⁵ + (5)x⁴ + (10)x³ + (10)x² + 5x + 1

(x + 1)⁵ = x⁵ + 5x⁴ + 10x³ + 10x² + 5x + 1.

Answer 2

(x + 2)³

From the binomial theorem,

(x + 2)³ = 3c₀(x)³(2)⁰ + 3c₁(x)^{3 - 1}(2)¹ + 3c₂(x)^{3 - 2}(2)² + 3c₃(x)^{3 - 3}(2)³

(x + 2)³ = 3c₀x³ + 3c₁x² (2)+ 3c₂x(4)+ 3c₃(  8)  

Substitute the values 3c₀ = 1, 3c₁= 3,  3c₂ = 3, and 3c₃ = 1.

(x + 2)³ = (1)x³ + (3)x²(2)+ 3x(4)+ (1)(8)

(x + 2)³ = x³ + 6x²+ 12x+ 8.