1. ## factor this completely

x^3-x^2+2

2. Originally Posted by princess_21
x^3-x^2+2
= x^3 + 1 - x^2+1
=(x+1)(x^2+1-x) - (x+1)(x-1)
=(x+1) (x^2+1-x-x+1)
=(x+1)(x^2-2x+2)...

= x^3 + 1 - x^2+1
=(x+1)(x^2+1-x) - (x+1)(x-1)
=(x+1) (x^2+1-x-x+1)
=(x+1)(x^2-2x+2)...
i didn't get the 2nd and 3rd step..

4. Originally Posted by princess_21
i didn't get the 2nd and 3rd step..
he factored by grouping.

$x^3 + 1$ <--------sum of two cubes. goes to $(x + 1)(x^2 - x + 1)$

$1 - x^2$ <--------difference of two squares. goes to $(1 - x)(1 + x)$

in the third step, he factored out the common factor, that is, $(x + 1)$

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

Expansion of
a^3 + b^3 = (a+b)(a^2+b^2-ab)
=> x^3 +1 = (x+1)(x^2+1-x) ------1
And that of
a^2-b^2 = (a+b)(a-b)
=> x^2-1 = (x+1)(x-1)--------------2

so
x^3 + 1 - x^2+1
=(x^3 +1 ) - (x^2-1)
put 1 and 2 in the equation and then take (x+1) as common

6. Originally Posted by Jhevon
he factored by grouping.

$x^3 + 1$ <--------sum of two cubes. goes to $(x + 1)(x^2 - x + 1)$

$1 - x^2$ <--------difference of two squares. goes to $(1 - x)(1 + x)$

in the third step, he factored out the common factor, that is, $(x + 1)$
how did he factored? why? the third step.
??

7. ## factor this completely

thanks.

8. I hope this would help
x^3-x^2+2 can be written as
= x^3 +x^2 - 2 x^2 -2x +2x+2
=x^2(x+1)-2x(x+1)+2(x+1)
=(x^2-2x+2)(x+1)...
...I think you got it before this reply