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Since P(1) = 0, it means P has a root at x=1. However, since P'(1) = 0 too, it means that the root at x=1 is a double root.Well i get 1. , i've done it by choosing two general equations as F(x) and G(x) which are both in space S
then using those proved axiom A1 and S1 , (f+g)(1)=f(1)+g(1)=0 and same way for the derivative, but i'm not sure if it's right or not
and i can't seem to figure out what this represents geometrically :S
Can I see your working? That doesn't seem right because the cubics you're dealing with are not general at all. They must satisfy P(1)=P'(1)=0 so they will be in the general form P(x)=(ax+b)*(x-1)^2Right got it, i've done the solution but my method uses two general cubic equations,ax^3+bx^2+cx+d.. etc
is this allowed?