# Showing a root is of an equation

• Oct 22nd 2009, 11:05 AM
zyx.gar
Showing a root is of an equation
Show that 1 is a root of the equation x^3 - 13x + 12 = 0 I can usually do this by dividing the factor in to the equation but I can't seem to work this one out mainly because when I do it I end up having to take x^2 away from 13x which as far as I know you can't do. Any help on this one?
• Oct 22nd 2009, 11:35 AM
Amer
Quote:

Originally Posted by zyx.gar
Show that 1 is a root of the equation x^3 - 13x + 12 = 0 I can usually do this by dividing the factor in to the equation but I can't seem to work this one out mainly because when I do it I end up having to take x^2 away from 13x which as far as I know you can't do. Any help on this one?

1 is he root of equation x^3-13x+12 if

(1)^3 - 13(1) +12 = 0 and this is right

and

$x^3-13x+12 = (x-1)(x^2+x-12)$
• Oct 23rd 2009, 03:10 AM
pacman
x^3 - 13x + 12 = 0,

factor by grouping:

x^3 +4x^2 - 4x^2 - 13x + 12 = 0,

x^3 +4x^2 - 4x^2 - 13x - 3x + 3x + 12 = 0, group them

(x^3 + 4x^2) + (- 4x^2 - 13x - 3x) + (3x + 12) = 0,

(x^3 + 4x^2) + (- 4x^2 - 16x) + (3x + 12) = 0,

x^2(x + 4) - 4x(x + 4) + 3(x + 4) = 0, factoring (x + 4),

(x + 4)(x^2 - 4x + 3) = 0, now you left with a quadratic (x^2 - 4x + 3),

this quadratic x^2 - 4x + 3 = 0 is easier to factor.

. . . .