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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?
Quote:
Originally Posted by zyx.gar
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 = 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.
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