How do you factor completely $x^3-5x^2+6x$ ?

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Answer 1 · 2016-06-05T16:18:52

To factor $x^3-5x^2+6x$ completely
$x(x-2)(x-3)$

To factor $x^3-5x^2+6x$ completely

Begin by factoring out the $x$ common to each term

$x(x^2-5x+6)$

Next find the factors of the third term $6$

1x6 and 2x3

Since the second sign is $+$ the factors must add up to the middle term $-5$ . We will use $-2 and -3$

Now factor the trinomial

$x(x-2)(x-3)$

How do you factor completely x^3-5x^2+6xx^3-5x^2+6x?