Question 134881
Solve the equation by finding all real roots.
x^3-7x+15x-9=0 
i have this 
1 -7 15 -9 /(divided by) 1. 
x^2-6x+9 
(x+1) (x^2-6x+9)
it says factor again and where is the soultion
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1)....1....-7....15....-9
......1....-6....9....|..0
So 1 is a Real zero
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Factorx^2-6x+9 to get (x-3)^2
So 3 is a Real zero with multiplicity two.
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Cheers,
Stan H.