Question

How do you find the in binomial expansion of `(x-3)^5`?

Answer 1

In general the expansion of (a+b)^n is
`sum_(i=0)^n(nCi*a^(n-i)*b^i)`

where `pCq` (p Choose q)" can be evaluated as `(p!)/((p-q)!(q!))`

Alternately (and probably more simply), you could use Pascal's Triangle to optain the coefficients:

`{: (0, rarr, 1,"-","-","-", "-", "-"), (1, rarr, 1,1,"-","-", "-", "-"), (2, rarr, 1,2,1,"-", "-", "-"), (3, rarr, 1,3,3,1, "-", "-"), ( 4, rarr, 1, 4, 6, 4, 1, "-"), ( 5, rarr, 1, 5, 10, 10, 5, 1) :}`
(etc.)

Giving
`(x-3)^5`
`=(1)(x^5)(-3)^0+5(x^4)(-3)^1+10(x^3)(-3)^2+10(x^2)(-3)^3+5(x^1)(-3)^4+(1)(x^0)(-3)^5`

=`x^5-15x^4+90x^3-270x^2+405x-243`