Thread: 3rd degree polynomial (math) View Single Post
11-26-2012   #4 (permalink)
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Quote:
 Originally Posted by Degoutant Amg thanks for your fast reply! And thanks, I also kind of understood what to do : so I did this like you said: f(2) = 0 8a+4b-3(2)+2 = 0 0=8a+4b-4 f(-1)=0 f(-1)=-a+(-1)^2*b-(-1)3+2 -a+b+3+2=0 Etc -a+b-1 = 0 | +1 -a+b = 1 |+a b= 1+a Now I substitued b into 0=8a+4b-4 which will be: 8a+4(1+a)-4 0=8a+4+4a-4 0=12a|:12 a=0 Substitued a into 1+a: b=1+0 =1 The equation is: f(x)=x²+3x+2 Is this right? Because it's only 2. Degree polynomial thanks in advance!
It is good that you are checking and seeing that the answer does not make sense.

You just made an small in substitution. I have highlighted it out. Try again and see if that fixes it.

When you work out a and b again, see if f(0)=2 just to check.
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