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 November 20th, 2016, 06:06 AM #1 Newbie   Joined: Nov 2016 From: Brazil Posts: 2 Thanks: 0 Cubic function It is known that (x-1) is the root of the function f(x) = x³ - x² + x + a. Find, if possible, all the Roots. Last edited by GabrielT; November 20th, 2016 at 06:10 AM.
 November 20th, 2016, 06:21 AM #2 Global Moderator   Joined: Dec 2006 Posts: 18,844 Thanks: 1566 That wording doesn't make sense. You probably meant that (x - 1) is a factor of the polynomial, which would imply that a = -1. x³ - x² + x - 1 = (x - 1)(x² + 1) = (x - 1)(x + i)(x - i)
 November 20th, 2016, 06:37 AM #3 Senior Member   Joined: May 2016 From: USA Posts: 997 Thanks: 409 I think you mean that 1 is a zero of the function $x^3 - x^2 + x + a.$ If that is correct, then $\exists\ p,\ q,\ r\ such\ that\ (x - 1)(px^2 + qx + r) = x^3 - x^2 + x + a.$ Do you know why? $\therefore px^2 + qx + r = \dfrac{x^3 - x^2 + x + a}{x - 1}.$ Finish it up. EFIT: Skipjack has a MUCH more direct method, but my method can be applied more generally. Last edited by JeffM1; November 20th, 2016 at 06:40 AM.
 November 20th, 2016, 08:59 AM #4 Math Team   Joined: Jul 2011 From: Texas Posts: 2,737 Thanks: 1387 synthetic division ... Code: 1] 1 -1 1 a 1 0 1 ----------------------- 1 0 1 (a+1) remainder, $a+1 = 0 \implies a = -1$ $x^3 - x^2 + x - 1 = (x-1)(x^2 + 0x + 1)$

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