Question

Expand `(2-X)^5`?

Answer 1

`(2-X)^5 = 32-80X+80X^2-40X^3+10X^4-X^5`

Explanation:

The binomial theorem tells us that:

`(a+b)^n = sum_(k=0)^n ((n),(k)) a^(n-k) b^k`

where `((n),(k)) = (n!)/((n-k)!k!)`

The binomial coefficients we want to use are:

`((5),(0)),((5),(1)),((5),(2)),((5),(3)),((5),(4)),((5),(5))`

These occur as the sixth row of Pascal's triangle:

Namely:

`1, 5, 10, 10, 5, 1`

In our example, we want `a=2` and `b=-X`, so write down powers of `2` in descending sequence from `2^5` to `2^0 = 1`:

`32, 16, 8, 4, 2, 1`

Multiply these two sequences together to get:

`32, 80, 80, 40, 10, 1`

Finally note that we need to alternate signs, due to the minus sign on `b=-X`, to get:

`(2-X)^5 = 32-80X+80X^2-40X^3+10X^4-X^5`