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November 20th, 2016, 06:06 AM   #1
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Exclamation 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.
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November 20th, 2016, 06:21 AM   #2
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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)
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November 20th, 2016, 06:37 AM   #3
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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.
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November 20th, 2016, 08:59 AM   #4
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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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