How do you factor #(x+1)^2-2(x+1)+1#?

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Answer 1 · 2016-09-01T23:01:14

#x xx x#

This is a disguised quadratic: it is in the form #ax^2 + bx +c#

Except that instead of #x#, the 'variable' is #(x+1)#

#Let (x+1) = b,# so that the expression becomes #b^2 - 2b +1#

The factors of 1 which add to 2 are 1 and 1.
Both signs are negative.

#(b-1)(b-1) " but " b= x+1#

#(x+1-1)(x+1-1)

#x xx x#