(Faulty?) Polynomial Division Problem

[DanX]

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.

 

[Dr.Peterson]

Don't use the same variables a and b in your P that they already used!

But you appear to have found that their a=0 and b=0, and P(x) = x+1. It is true that (x³+1)/(x+1) = x²-x+1 . So where is the contradiction? You were told to find what a and b have to be, and you did!

 

[Steven G]

If (x³-ax²-bx+1)/P(x) = x²-x+1, then P(x) = (x³-ax²-bx+1)/(x²-x+1).

Now do the division and make sure you do not get a remainder.

 

[JeffM]

First of all, I cannot stress enough Dr. P's point that a and b were already used so do not use them again to stand for different variables. Keep your symbols consistent and you will avoid much confusion.

However, as jomo pointed out, there is a much simpler way to go.

[MATH]\dfrac{x^3 - ax^2 - bx + 1}{P(x)} = x^2 - x + 1 \implies P(x) = \dfrac{x^3 - ax^2 - bx + 1}{x^2 - x + 1} \implies[/MATH]
[MATH]P(x) = \dfrac{x^3 - (x^2 - x) + (x^2 - x) - ax^2 - bx + 1}{x^2 - x + 1} =[/MATH]..........................................edited

[MATH]\dfrac{(x^3 - x^2 + x) + (1 - a)x^2 - (b + 1)x + 1}{x^2 - x + 1} =[/MATH]
[MATH]\dfrac{x(x^2 - x + 1) + (1 - a)x^2 - (b + 1)x + 1}{x^2 - x + 1} = x + \dfrac{(1 - a)x^2 - (b + 1)x + 1}{x^2 - x + 1}[/MATH]
Now, what do you know about equating coefficients?

 

[Dr.Peterson]

I think doing what Jomo said directly (by long division) is quite simple; it leads to a remainder of (a-b)x + a, and setting this to zero completes the work.

The multiplication approach of the OP is also reasonably straightforward. But we can make it easier:

If [MATH](x^3-ax^2-bx+1)/P(x) = x^2-x+1[/MATH], then [MATH]x^3-ax^2-bx+1 = P(x)(x^2-x+1)[/MATH]. Clearly [MATH]P(x)[/MATH] is a first-degree polynomial, so call it [MATH]px + q[/MATH].

Now if [MATH](px + q)(x^2-x+1) = x^3-ax^2-bx+1[/MATH], then the leading and constant terms of the LHS are clearly [MATH]px^3[/MATH] and [MATH]q[/MATH] respectively, so we must have [MATH]p = 1[/MATH] and [MATH]q = 1[/MATH], and [MATH]P(x) = x + 1[/MATH].

Therefore [MATH]x^3-ax^2-bx+1 = (x + 1)(x^2-x+1) = x^3 + 1[/MATH], and we conclude that [MATH]a[/MATH] and [MATH]b[/MATH] are both 0.

This is essentially the OP with a shortcut.

 

[Steven G]

First of all, I cannot stress enough Dr. P's point that a and b were already used so do not use them again to stand for different variables. Keep your symbols consistent and you will avoid much confusion.

However, as jomo pointed out, there is a much simpler way to go.

[MATH]\dfrac{x^3 - ax^2 - bx + 1}{P(x)} = x^2 - x + 1 \implies P(x) = \dfrac{x^3 - ax^2 - bx + 1}{x^2 - x + 1} \implies[/MATH]
[MATH]P(x) = \dfrac{x^3 - (x^2 - x) + (x^2 - x) - ax^2 + [COLOR=rgb(226, 80, 65)]x[/COLOR] - bx + 1}{x^2 - x + 1} =[/MATH]
[MATH]\dfrac{(x^3 - x^2 + x) + (1 - a)x^2 - (b + 1)x + 1}{x^2 - x + 1} =[/MATH]
[MATH]\dfrac{x(x^2 - x + 1) + (1 - a)x^2 - (b + 1)x + 1}{x^2 - x + 1} = x + \dfrac{(1 - a)x^2 - (b + 1)x + 1}{x^2 - x + 1}[/MATH]
Now, what do you know about equating coefficients?

To OP: JeffM had a typo inked above. That x should not be there.
I guess that Jeff goes to the corner for x minutes.

 

[JeffM]

To Jomo:

Off to the corner for x minutes.

Oh wait [MATH]x^2 - x + 1[/MATH] has no real roots so I shall have to spend imaginary time in the corner.