Binomial Theory Using Polynomial Expansion

If we expand:
 (x+a)(x+b)=x²+(a+b)x+ab
(x+a)(x+b)(x+c)=x³+(a+b+c)x²+(ab+ac+bc)x+abc
(x+a)(x+b)(x+c)(x+d)=x⁴+(a+b+c+d)x³+(ab+ac+ad+bc+bd+cd)x²+(abc+abd +cd+bcd)x+abcd

Lets say that we want to have 2 numbers or a binomial we would change the expressions above to:
(x+1)², (x+1)³, and (x+1)⁴
They equal:
(x+1)²=x²+(1+1)x+1X1=x²+2x+1
(x+1)³=x³+(1+1+1)x²+(1X1+1X1+1X1)x+1X1X1=x³+3x²+3x=1
When when we expand a binomial, the number preceding each x is just the number of combinations of a,b,c variables preceding our x in our polynomial expansion.

(x+1)⁴=x⁴+4x³+6x²+4x+1
(x+1)ⁿ=xⁿ+[nC(n-1)]xⁿ⁻¹+[nC(n-2)]xⁿ⁻²+[nC(n-3)]xⁿ⁻³...........

Now lets lets see what happens if we use a variable like y instead of 1.
(x+y)²=x²+(y+y)x+y.y=x²+2yx+y²
(x+1)³=x³+(y+y+y)x²+(y.y+y.y+y.y)x+y.y.y=x³+3yx²+3y²x+y³
(x+y)⁴=x⁴+4yx³+6x²y²+4xy³+y⁴

Our expansion becomes
(x+y)ⁿ=xⁿ+[nC(n-1)]xⁿ⁻¹y+[nC(n-2)]xⁿ⁻²y²+[nC(n-3)]xⁿ⁻³y³...........yⁿ

Lets say we have (x+y)⁶=x⁶+6x⁵y+15x⁴y²+20x³y³+15x²y⁴+6xy⁵+y⁶

If we had a number like 3 instead of y, we would substitute 3 where we have y and it would look like this:

(x+3)⁶=x⁶+6x⁵(3)+15x⁴(9)+20x³(27)+15x²(81)+6x(243)+y⁶(729)=x⁶+18x⁵+135x⁴+540x³+ 1215x²+1458x+729y⁶

As you can see there is a lot of calculation involved in expanding binomial series.  There is a way that you can just plug a number in from a table. This table is called Pascal's Triangle.  We will do post on this next.  Many people skip binomial expansion and just use Pascal's Triangle.