How do you multiply `(sqrt5- sqrt2)(sqrt5+sqrt2)`?

`(sqrt(5)-sqrt(2))(sqrt(5)+sqrt(2))`

`=(sqrt(5))^2 - (sqrt(2))^2 = 5 - 2 = 3`

using the difference of squares identity:

`a^2 - b^2 = (a-b)(a+b)`

Alternatively use FOIL to pick pairs of terms to multiply and add together:

First: `sqrt(5) * sqrt(5) = 5`
Outside: `sqrt(5)*sqrt(2)`
Inside: `-sqrt(2)*sqrt(5)` = `-sqrt(5)*sqrt(2)`
Last: `-sqrt(2)*sqrt(2) = -2`

F + O + I + L = `5 + cancel(sqrt(5)sqrt(2)) - cancel(sqrt(5)sqrt(2)) - 2 = 3`

You can use "FOIL" (First-Outside-Inside-Last) as follows:

`(sqrt(5)-sqrt(2))*(sqrt(5)+sqrt(2))=sqrt(5)*sqrt(5)+sqrt(5)sqrt(2)-sqrt(5)sqrt(2)-sqrt(2)sqrt(2)`

After cancellation and use of the facts that `sqrt(5)^2=5` and `sqrt(2)^2=2`, this becomes `5-2=3`.

The reason "FOIL" works is because of abstract properties of numbers systems, such as the Distributive Property, which says that `a*(b+c)=a*b+a*c` for all numbers `a, b`, and `c`. You can use it twice when you expand `(a+b)*(c+d)` as follows:

`(a+b)*(c+d)=(a+b)*c+(a+b)*d`

`=a*c+b*c+a*d+b*d`

This can be rearranged, by the Commutative Property, to `=a*c+a*d+b*c+b*d`. (FOIL)

It's also worthwhile to memorize the difference of two squares factoring formula: `a^2-b^2=(a-b)(a+b)`. This can be applied to the problem at hand "in reverse" by letting `a=sqrt(5)` and `b=sqrt(2)`.