What is the standard form of #f(x)=(x-1)^2-(2x-1)^2 #?

1 Answer
Jim G.
Feb 20, 2016

#- 3x^2 + 2x#

Explanation:

distribute both 'pairs' of brackets using FOIL ( or by any method you may use).

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

#(2x - 1 )^2 = 4x^2 - 4x + 1#

Put these expansions back into 'original' expression.

# rArr x^2 - 2x + 1 - (4x^2 - 4x + 1 )#

# = x^2 - 2x + 1 - 4x^2 + 4x - 1 = -3x^2 + 2x#

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