Simplifying a term

[LazeFox]

I tried using the binomial formula, though, both parantheses have different x numbers. What exactly am I supposed to do here?

 

[Dr.Peterson]

View attachment 28514
I tried using the binomial formula, though, both parantheses have different x numbers. What exactly am I supposed to do here?

Just expand, that is distribute, (2x + 4y)(x + 4y), by multiplying each term of one by each term of the other; some people call this "FOIL", but I don't know what it is called in German.

 

[LazeFox]

Just expand, that is distribute, (2x + 4y)(x + 4y), by multiplying each term of one by each term of the other; some people call this "FOIL", but I don't know what it is called in German.

Ahh, I made it out to be more complicated than it actually is, thank you!

 

[HallsofIvy]

By the distributive law, (2x+ 4y)(x+ 4y)= (2x+ 4y)(x)+ (2x+ 4y)(4y)= 2x(x)+ 4y(x)+ 2x(4y)+ 4y(4y)= 2x^2+ 4xy+ 8xy+ 16y^2= 2x^2+ 12xy+ 16y^2.