What is the binomial expansion of (x + 2)4? x4 + 4x3 + 6x2 + 4x + 1 8x3 + 24x2 + 32x x4 + 8x3 + 24x2 + 32x + 16 2x4 + 8x3 + 12x2 + 8x + 2

Answer-[tex]\boxed{\boxed{(x+2)^4=x^4+8x^3+24x^2+32x+16}}[/tex]Solution-Given expression is [tex](x+2)^4[/tex]Applying Binomial Theorem[tex]\left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i[/tex]Here, a = x, b = 2 and n = 4So,[tex]\left(x+2\right)^4=\sum _{i=0}^4\binom{4}{i}x^{\left(4-i\right)}\cdot \:2^i[/tex]Expanding the summation[tex]=\dfrac{4!}{0!\left(4-0\right)!}x^4\cdot \:2^0+\dfrac{4!}{1!\left(4-1\right)!}x^3\cdot \:2^1+\dfrac{4!}{2!\left(4-2\right)!}x^2\cdot \:2^2+\dfrac{4!}{3!\left(4-3\right)!}x^1\cdot \:2^3+\dfrac{4!}{4!\left(4-4\right)!}x^0\cdot \:2^4[/tex][tex]=\dfrac{4!}{0!\left(4\right)!}x^4\cdot \:2^0+\dfrac{4!}{1!\left(3\right)!}x^3\cdot \:2^1+\dfrac{4!}{2!\left(2\right)!}x^2\cdot \:2^2+\dfrac{4!}{3!\left(1\right)!}x^1\cdot \:2^3+\dfrac{4!}{4!\left(0\right)!}x^0\cdot \:2^4[/tex][tex]=1\cdot x^4\cdot \:1+4\cdot x^3\cdot \:2+6x^2\cdot \:4+4\cdot x\cdot \:8+1\cdot 1\cdot \:16[/tex][tex]=x^4+8x^3+24x^2+32x+16[/tex]