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)#