Posted by Belinda on Sunday, January 8, 2012 at 6:56pm.
2 x^3 -3 x^2 -6 x + 9 = 0
(2x^3 - 6x) - (3 x^2-9) = 0
2x (x^2-3) -3(x^2-3) = 0
(x^2-3)(2x-3) = 0
you take it from there.
2)
Try x = 1
3-1+4-2-4 = 0 amazing
so divide by (x-1)
(x-1)(3x^3+2x^2+6x+4) = 0
(x-1)[ 3x(x^2+2) + 2(x^2+2) ]
you take it from there
3)
(x+5)(x-3i)(x+3i)
if you have 3i, then you must have the complex conjugate -3i
Thanks so much :D
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