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

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Answer 1 · 2015-05-22T13:49:42

We can factor out $x$ $x$ from all the elements that contain it:

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

And then factor the quadratic inside the parenthesis, which, using Bhaskara, will give us the following roots:

$(2+-sqrtcancel((4-4(1)(1))))/2$
$2/2=1$

$x_1=x_2=1$

Thus, the factors are both $(x-1)$

Finally: $x((x-1)(x-1))+2$

Answer 2 · 2015-05-22T14:12:22

Factor by grouping:

$x^3-2x^2-x+2$

$=(x^3-2x^2)+(-x+2)$ (Notice the inserted addition before the second pair.)

$=x^2(x-2)+(-1)(x-2)$

$=(x^2-1)(x-2)$ Now the first binomial is a difference of squares, so

$=(x+1)(x-1)(x-2)$ .

$x^3-2x^2-x+2=(x+1)(x-1)(x-2)$

How do you factor x^3 - 2x^2 - x + 2 x3−2x2−x+2x^3 - 2x^2 - x + 2 ?