How do you expand `(2x+5)^7`?

`(2x+5)^7` simply means `(2x+5)` times itself 7 times.
So `(2x+5)^3` would be `(2x+5)(2x+5)(2x+5)`
`(2x+5)^5` would be `(2x+5)(2x+5)(2x+5)(2x+5)(2x+5)`

This works regardless of whats in the brackets.
`(3x^3-5xm×38)^2` is `(3x^3-5xm×38)(3x^3-5xm×38)`

And without brackets `5^5` is 5×5×5×5×5
`38×12^3` is 38(12×12×12)

And so forth.

`"using the "color(blue)"binomial theorem"`

`• (x+y)^n=sum_(r=0)^n((n),(r))x^(n-r)y^r`

`"where " ((n),(r))=(n!)/(r!(n-r)!)`

`"here " x=2x" and " y=5`

`rArr(2x+5)^7`

`=((7),(0))(2x)^7 .5^0+((7),(1))(2x)^6 .5^1+((7),(2))(2x)^5 .5^2`

`color(white)(=)+((7),(3))(2x)^4 .5^3+((7),(4))(2x)^3 .5^4+((7),(5))(2x)^2 .5^5`

`color(white)(=)+((7),(6))(2x)^1 .5^6+((7),(7))(2x)^0 .5^7`

`"we can obtain the binomial coefficients using the "`
`"appropriate row of "color(blue)"Pascal's triangle"`

`"for n = 7 the row of coefficients is"`

`1color(white)(x)7color(white)(x)21color(white)(x)35color(white)(x)35color(white)(x)21color(white)(x)7color(white)(x)1`

`=1. 128x^7+7.5.64x^6+21.25.32x^5+35.125.16x^4`

`color(white)(=)+35.625.8x^3+21.3125.4x^2+7.15625.2x+78125`

`=128x^7+2240x^6+16800x^5+70000x^4+175000x^3`

`color(white)(=)+262500x^2+218750x+78125`