I've been going through my textbook, trying to go from start to finish, doing pretty much all the hardest exercises so that I can complete it and move to next grade math book (unsure whether to count this as Algebra I or Intermediary Algebra), and I've been doing fine until I found this exercise. It's a mess:
"Find the values of parameters a, b, knowing that there exists a binomial P(x), so that (x3-ax2-bx+1)/P(x) = x2-x+1"
So I gave it a try. I'll write "x3-ax2-bx+1" as A(x), "x2-x+1" as B(x). First I'd like to say I'm assuming that after division there's no remainder, it should divide cleanly, therefore P(x) would be a clean factor of A(x), therefore P(x)*B(x)=A(x).
Then I thought that the a, b variables would correspond to the generic formula for a binomial, which would be P(x)=ax+b, noted in the book a couple pages prior. Then what I decided to do is actually multiply P(x) by B(x) and I got: (ax+b)(x2-x+1)=ax3-ax2+bx2+ax-bx+b.
So that'd mean that b remains to be 1, since it's the constant at the end which matches the constant in F(x). Right? And since a is the only one with an x3 term, that'd mean that a also needs to be equal to 1. Replacing a,b with 1 you get: A(x) = x3-x2+x2+x-x+1, P(x)=x+1. This'd mean that in A(x), the 2nd and 1st degree terms cancel out -- A(x) = x3+1. If you decided to divide A(x) by P(x) now you'd get B(x), but...There's a problem with that. That implies that, from our first representation of A(x), a,b now both have to be 0. A contradiction, atleast from what I understand. If you have any idea of how to solve this, please tell.
"Find the values of parameters a, b, knowing that there exists a binomial P(x), so that (x3-ax2-bx+1)/P(x) = x2-x+1"
So I gave it a try. I'll write "x3-ax2-bx+1" as A(x), "x2-x+1" as B(x). First I'd like to say I'm assuming that after division there's no remainder, it should divide cleanly, therefore P(x) would be a clean factor of A(x), therefore P(x)*B(x)=A(x).
Then I thought that the a, b variables would correspond to the generic formula for a binomial, which would be P(x)=ax+b, noted in the book a couple pages prior. Then what I decided to do is actually multiply P(x) by B(x) and I got: (ax+b)(x2-x+1)=ax3-ax2+bx2+ax-bx+b.
So that'd mean that b remains to be 1, since it's the constant at the end which matches the constant in F(x). Right? And since a is the only one with an x3 term, that'd mean that a also needs to be equal to 1. Replacing a,b with 1 you get: A(x) = x3-x2+x2+x-x+1, P(x)=x+1. This'd mean that in A(x), the 2nd and 1st degree terms cancel out -- A(x) = x3+1. If you decided to divide A(x) by P(x) now you'd get B(x), but...There's a problem with that. That implies that, from our first representation of A(x), a,b now both have to be 0. A contradiction, atleast from what I understand. If you have any idea of how to solve this, please tell.