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

#(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#