Follow the steps 1)-4)

1) First expand out the bracketed term:

`x-1 -(x-1)^2 = x- 1 - (x-1)(x-1) `

`= x - 1 - (x^2 - x - x + 1)`

`= x - 1 - (x^2 - 2x + 1)`

2) Now gather terms:

`x - 1 - (x-1)^2 = x - 1 - x^2 + 2x - 1 = -x^2 + 3x - 2`

3) Now, to factorise the original expression, we are looking for constants `a` and `b` such that

`-x^2 + 3x - 2 = (-x+a)(x+b)`

4) From this we can see that `(a-b) = 3` and `ab = -2`.

The idea is to spot the simple solution `a=2` and `b=-1`. This can be logically arrived at by noting that `pm 1` and `pm 2` are the only integer possibilities if `ab= -2`and that the only way these possible values of `a` and `b` can give `a-b=3` is by combining a positive 2 and a negative 1.

**The expression x - 1 - (x-1)^2 factorises to (2-x)(x-1).**

Alternatively, factor out an `x-1` right away.

`x-1-(x-1)^2=(x-1)(1-(x-1))`

`=(x-1)(1-x+1)=(x-1)(2-x).`

distribute the exponent

`x-1-(x^2+1) `

distribute the negative sign

`x-1-x^2-1 `

combine like terms

`x-1-1-x^2 `

`x-2-x^2 `

put it in factored form

`-x^2+x-2`

multiply axxc

`-1xx-2=2 `

factors of 2 that minus to 1 = 2 and 1

plug them in

`-x^2-1x+2x-2 `

factor it out

`(-x^2-1x)(2x-2) `

`-x(x-1)2(x-1) `

`(2-x)(x-1)`

x=2 x=1