How do you factor $7c²-28c+7$ ?

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Answer 1 · 2016-08-24T23:30:35

$7c^2-28c+7=7(c-2-sqrt(3))(c-2+sqrt(3))$

The difference of squares identity can be written:

$a^2-b^2 = (a-b)(a+b)$

We use this with $a=(c-2)$ and $b=sqrt(3)$ as follows:

$7c^2-28c+7$

$=7(c^2-4c+1)$

$=7(c^2-4c+4-3)$

$=7((c-2)^2-(sqrt(3))^2)$

$=7((c-2)-sqrt(3))((c-2)+sqrt(3))$

$=7(c-2-sqrt(3))(c-2+sqrt(3))$

How do you factor 7c²-28c+7 7c²-28c+7 ?