Question

How do I find the binomial expansion of `(2x+1)^3`?

Answer 1

`1 + 6x + 12x^2 + 8x^3`

Explanation:

We must use our knowledge of the binomial expansion:

Method 1:

We can use:

`(x+1)^n = 1 + nx + (n(n-1))/(2!) x^2 + (n(n-1)(n-2))/(3!) x^3 + ...`

Substituting `n = 3` and `x` for `2x` `=>`

`(2x+1)^3 = 1 + (3*2x) + (3*2)/(2!) * (2x)^2 + (3*2*1)/(3!) * (2x)^3`

`color(red)(= 1 + 6x + 12x^2 + 8x^3`

Method 2:

We can use:

`(A+B)^n = A^n + ((n),(1))A^(n-1)B^1 + ((n),(2))A^(n-2)B^2 + ...`

Letting `A = 2x and B =1` for this circumstance:

`(2x+1)^3 = (2x)^3 + ((3),(1))(2x)^2(1) + ((3),(2))(2x)^1(1)^2 + ((3),(3))(2x)^0 (1)^3`

`= color(red)( 8x^3 + 12x^2 + 6x + 1 )`

Answer 2

`(2x+1)^3=8x^3+12x^2+6x+1`

Explanation:

we use Pascal's triangle for the coefficients

`color(white)( 0000000000000000000 )1`

`(a+b)^2rarrcolor(white)(0000000)1,2,1`

`(a+b)^3rarrcolor(white)(000000)1,3,3,1`

so we need `" "color(red)(1,3,3,1)`

the powers of the terms will sum to `3`

and startingthe first term will be `3` and descend in `1s`

thus

`(2x+1)^3=color(red)(1)(2x)^3+color(red)(3)(2x)^2(1)+color(red)(3)(2x)(1)^2+color(red)(1)*1^3`

`(2x+1)^3=8x^3+12x^2+6x+1`