# factor this completely

• Jan 13th 2009, 10:02 PM
princess_21
factor this completely
x^3-x^2+2
• Jan 13th 2009, 10:15 PM
(Hi)
Quote:

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)...
• Jan 13th 2009, 10:30 PM
princess_21
Quote:

Originally Posted by ADARSH
(Hi)
= 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..
• Jan 13th 2009, 10:33 PM
Jhevon
Quote:

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, $(x + 1)$
• Jan 13th 2009, 10:37 PM
Quote:

Originally Posted by ADARSH
(Hi)
= 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)...

(Wait)
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
• Jan 13th 2009, 10:39 PM
princess_21
Quote:

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.
??
• Jan 13th 2009, 10:44 PM
princess_21
factor this completely
thanks.
• Jan 13th 2009, 10:49 PM