Question

How do you multiply `(2x+1)(x+2)`?

Answer 1

`(2x+1)(x+2) = 2x^2+5x+2`

Explanation:

`(color(red)(2x)+colkor(blue)(1))(x+2)`
`color(white)("XXXX")` `= color(red)(2x)(x+2) + color(blue)(1)(x+2)`

`color(white)("XXXX")` `= color(red)(2x^2+4x) + color(blue)(x + 2)`

`color(white)("XXXX")` `= 2x^2+5x+2`

The following provides an even more detailed version of this, which you should ignore if the above made sense to you.

In general `color(green)((a+b)*(c) = (a*c) + (b*c)` `color(white)("XXXX")`(Distributive property)

Letting `color(green)(a)=2x` and `color(green)(b)=1` and `color(green)(c)=(x+2)`
we have:
`color(white)("XXXX")` `(2x+1)(x+2) = color(red)(2x(x+2)) + color(blue)(1(x+2))`

Also, in general, `color(orange)((p)*(q+r) = (p*q)+(p*r))` `color(white)("XXXX")`(Distributive property, again)

First letting `color(orange)(p) = 2x` and `color(orange)(q)=x` and `color(orange)(r)=2`
we have:
`color(white)("XXXX")` `color(red)(2x(x+2) = 2x*x + 2x*2)`
which simplifies as:
`color(white)("XXXX")` `color(white)("XXXX")` `color(red)(= 2x^2 +4x)`

Then letting `color(orange)(p)=1` and `color(orange)(q)=x` and `color(orange)(r)=2`
we have:
`color(white)("XXXX")` `color(blue)(1(+2) =1*x + 1*2)`
which simplifies as:
`color(white)("XXXX")` `color(white)("XXXX")` `color(blue)(= x + 2)`

So
`(2x+1)(x+2) = color(red)(2x^2+4x) + color(blue)(x+2)`

`color(white)("XXXX")` `color(white)("XXXX")` `= 2x^2 + 5x +2`