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

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

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Answer 1 · 2016-02-20T21:19:57

$- 3 x^{2} + 2 x$ $- 3x^2 + 2x$

Explanation:

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

${(x - 1)}^{2} = x^{2} - 2 x + 1$ $(x - 1 )^2 = x^2 - 2x + 1$

${(2 x - 1)}^{2} = 4 x^{2} - 4 x + 1$ $(2x - 1 )^2 = 4x^2 - 4x + 1$

Put these expansions back into 'original' expression.

$\Rightarrow x^{2} - 2 x + 1 - (4 x^{2} - 4 x + 1)$ $rArr x^2 - 2x + 1 - (4x^2 - 4x + 1 )$

$= x^{2} - 2 x + 1 - 4 x^{2} + 4 x - 1 = - 3 x^{2} + 2 x$ $= x^2 - 2x + 1 - 4x^2 + 4x - 1 = -3x^2 + 2x$

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

1 Answer

Explanation:

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