Question

How do you expand the binomial `(3x-1)^4`?

Answer 1

Using Pascal's triangle gives `81x^4-108x^3+54x^2-12x+1`.

Explanation:

To expand #(3x-1)^4, use row 4 of Pascal's triangle.
The top row is row zero, the next is row 1, the next is row 2, etc.

`color(white)(AAAAAAA)1`
`color(white)(AAAAA)1color(white)(AA)1`
`color(white)(AAA)1color(white)(AA)2color(white)(A)1`
#color(white)(A)1color(white)(AA)3color(white)(AA)3 color(white)(A)1#
`1color(white)(AA)4color(white)(AA)6color(white)(A)4color(white)(AA)1`

Row 4 is 14641. These will be used as the coefficients of each term of the expansion.

To expand #(a-b)^4, use the following process:

`1a^4b^0-4a^3b^1+6a^2b^2-4a^1b^3+1a^0b^4`

Note that the sum of the exponents is 4, because we are finding the 4th power of the binomial. The first term contains `a^4` and #b^0 #, the second `a^3` and `b^1`, the third `a^2` and `b^2`, etc.

Also, the signs alternate because we are expanding `(a-b)`. If we were expanding `(a+b)`, the signs would all be positive.

In our example `a=3x` and `b=1`.

`1(3x)^4(1)^0-4(3x)^3(1)^1+6(3x)^2(1)^2-4(3x)^1(1)^3+1(3x)^0(1)^4`

`81x^4-108x^3+54x^2-12x+1`