(x+y)^0=1 of course because anything to the power if 0 equal 1 (x+y)^1= x+y anything to a power of 1 is just itself.
(x+y)^2= (x+y)(x+y) NOT x^2+y^2.
So expand (x+y)(x+y)=x^2+xy+yx+y^2 or x^2+2xy+y^2.
(x+y)^3=(x+y)(x+y)(x+y) now expanding it is getting quite long. Of course we could do this using the distribution property, but there must be an easier way of expanding binomials with out doing all the steps that is takes to expand something like (x+y)^10.
As surprising as it might be, there is an easier way of find what a binomial equals to a larger power. Using combinations we can find the coefficients of each term. Lets look at an example. The (x+y)^3 was the one I didn’t finish. Let’s look at it now. Using the equation in combination, we can insert the power that we are using and for each term to find the coefficient of each term. Ex:
This process can be even easier. Blaise Pascal, a famous French mathematical among other things put together a triangle made up of numbers where each number represents the coefficient of each term (when expanded) in a binomial to a certain power. He named the triangle “triangle du arithmétique” or in English, “The Arithmetical Triangle”. Now the triangle is called “Pascal’s triangle named after the creator, Blaise Pascal. We can find this triangle by using combinations. Ex:
Moving back to the binomial theorem, you can use any binomial and find what it equals using Pascal’s triangle or combinations. We can us any two random numbers, a number and a variable, variables with coefficients, subtraction instead of adding etc. Lets look at some examples.