How do you factor $4 x^{2} - 4 x + 1$ $4x^2-4x+1$ ?

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How do you factor $4 x^{2} - 4 x + 1$ $4x^2-4x+1$ ?

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Answer 1 · 2016-05-03T13:57:00

This is a perfect square trinomial:

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

Explanation:

One way of quickly spotting this (and a few more) factorisations is the pattern: $4, 4, 1$ $4, 4, 1$ . The number $441$ $441$ is the square of $21$ $21$ , with no carried digits.

Hence we find:

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

and:

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

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How do you factor $4 x^{2} - 4 x + 1$ $4x^2-4x+1$ ?

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How do you factor $4 x^{2} - 4 x + 1$ $4x^2-4x+1$ ?