How do I distribute (x-1)^5 with pascals triangle?

Question

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Answer 1

Pascals triangle tells you the coefficients of (x+y)^n, like so:

1
1 1 (x+y)^1 = 1x + 1y
1 2 1 (x+y)^2 = 1 x^2 + 2 x y + 1 x^2
1 3 3 1 (x+y)^3 = 1 x^3 + 3 x^2 y + 3 x y^2 + 1 y^3
1 4 6 4 1 (x+y)^4 = 1 x^4 + 4 x^3 y + 6 x^2 y^2 + 4 x y^3 + 1 y^4
1 5 10 10 5 1 (x+y)^5 = 1 x^5 + 5 x^4 y + 10 x^3 y^2 + 10 x^2 y^3 + 5 x y^4 + 1 y^5


Notice that the powers of x decrease and the powers of y increase. It's the same for (x-y)^n, 'cept we replace y everywhere with -y, and since (-y)^n = (-1)^n y^n and (-1)^even = 1, (-1)^odd = -1, the signs just alternate:

1
1 1 (x-y)^1 = 1x - 1y
1 2 1 (x-y)^2 = 1 x^2 - 2 x y + 1 x^2
1 3 3 1 (x-y)^3 = 1 x^3 - 3 x^2 y + 3 x y^2 - 1 y^3
1 4 6 4 1 (x-y)^4 = 1 x^4 - 4 x^3 y + 6 x^2 y^2 - 4 x y^3 + 1 y^4
1 5 10 10 5 1 1 (x-y)^5 = x^5 - 5 x^4 y + 10 x^3 y^2 - 10 x^2 y^3 + 5 x y^4 - 1 y^5

For (x-1)^5, replace y with 1:

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

Answer 2

The 5th row of Pascal's triangle is
1, 5, 10, 10, 5, 1
(x - 1)^5 =
x^5 - 5x^4 + 10x^3 - 10x^2 + 5x - 1

Answer 3

(x-1)^h
=x^5 + 5(x)^4(-1) + 10(x)^3(-1)^2 + 10(x)^2(-1)^3 + 5(x)(-1)^4 + (-1)^5
=x^5 - 5x^4 + 10x^3 - 10x^2 + 5x - 1
Basically with a minus sign the signs in the answer alternate.