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

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Answer 1 · 2016-06-05T16:18:52

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

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

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

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

Next find the factors of the third term $6$ $6$

1x6 and 2x3

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

Now factor the trinomial

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

How do you factor completely x^3-5x^2+6xx3−5x2+6xx^3-5x^2+6x?