Question

factorise x-1-(x-1)^2+ax-a​

Answer 1

Answer:

x-1-(x-1)^2+ax-a=(x-1)(2-x+a)

Step-by-step explanation:

Given:-

Given expression is x-1-(x-1)^2+ax-a

To find:-

factorise x-1-(x-1)^2+aax-a

Solution:-

Given expression is x-1-(x-1)^2+ax-a

It can be written as

=>x-1-(x-1)^2+ax-a

=>(x-1)-(x-1)(x-1)+a(x-1)

=>(x-1)[{1-(x-1)+a}]

=>(x-1)[1-x+1+a]

=>(x-1)(2-x+a)

Answer:-

x-1-(x-1)^2+ax-a=(x-1)(2-x+a)

Answer 2

Answer:

x-1-(x-1)^2+ax-a=(x-1)(2-x+a)

Step-by-step explanation:

Given:-

Given expression is x-1-(x-1)^2+ax-a

To find:-

factorise x-1-(x-1)^2+aax-a

Solution:-

Given expression is x-1-(x-1)^2+ax-a

Given expression is x-1-(x-1)^2+ax-aIt can be written as

Given expression is x-1-(x-1)^2+ax-aIt can be written as =>x-1-(x-1)^2+ax-a

Given expression is x-1-(x-1)^2+ax-aIt can be written as =>x-1-(x-1)^2+ax-a=>(x-1)-(x-1)(x-1)+a(x-1)

Given expression is x-1-(x-1)^2+ax-aIt can be written as =>x-1-(x-1)^2+ax-a=>(x-1)-(x-1)(x-1)+a(x-1)=>(x-1)[{1-(x-1)+a}]

Given expression is x-1-(x-1)^2+ax-aIt can be written as =>x-1-(x-1)^2+ax-a=>(x-1)-(x-1)(x-1)+a(x-1)=>(x-1)[{1-(x-1)+a}]=>(x-1)[1-x+1+a]

Given expression is x-1-(x-1)^2+ax-aIt can be written as =>x-1-(x-1)^2+ax-a=>(x-1)-(x-1)(x-1)+a(x-1)=>(x-1)[{1-(x-1)+a}]=>(x-1)[1-x+1+a]=>(x-1)(2-x+a)

Answer:-

x-1-(x-1)^2+ax-a=(x-1)(2-x+a)