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

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Answer 1 · 2018-05-12T01:21:47

$(x - 3) (x + 1)$ $(x-3)(x+1)$

${3, - 1}$ ${3,-1}$

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

Factor out:

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

Factor:

$(x - 3) (x + 1)$ $(x-3)(x+1)$

${3, - 1}$ ${3,-1}$

Answer 2 · 2018-05-12T01:25:24

$(x + 1) (x - 3)$ $(x+1)(x-3)$

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

To factorise the expression we could first expand the first term. However, in this case, notice that the expression is the difference of two squares.

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

Remember the common identity: $a^{2} - b^{2} = (a + b) (a - b)$ $a^2 - b^2 = (a+b)(a-b)$

Thus, Expression $= (x - 1 + 2) (x - 1 - 2)$ $= (x-1+2)(x-1-2)$

$= (x + 1) (x - 3)$ $= (x+1)(x-3)$

How do you factor (x-1)^2 - 4(x−1)2−4(x-1)^2 - 4?