x^3-x^2+2
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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)...
Originally Posted by ADARSH = 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..
Originally Posted by princess_21 i didn't get the 2nd and 3rd step.. he factored by grouping. <--------sum of two cubes. goes to <--------difference of two squares. goes to in the third step, he factored out the common factor, that is,
Originally Posted by ADARSH = 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
Originally Posted by Jhevon he factored by grouping. <--------sum of two cubes. goes to <--------difference of two squares. goes to in the third step, he factored out the common factor, that is, how did he factored? why? the third step. ??
thanks.
Last edited by mr fantastic; January 14th 2009 at 12:24 AM.
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
Last edited by ADARSH; January 13th 2009 at 11:51 PM. Reason: ...
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