Polynomial division unclear

[Ognjen]

In my workbook, the problem wants me to find all solutions for x, if P(x) / Q(x) is zero. P(x) = x^3 - 5x^2 + 2x + 8; Q(x) = x + 1.

So when you divide the two polynomials ( by long division ) you get x^2 - 6x + 8. Since this quadratic expression we got has to be equal to 0 ( by the formulation of the problem ), we apply the quadratic formula and get 2 solutions: 4 and 2.

HOWEVER, the official solution says there is another solution, that we get when we multiply the P(x) / Q(x) = x^2 - 6x + 8 by Q(x), then make Q(x) zero, so that ( x - 1 ) = 0 and thus x_3 = -1. How come for polynomials the rule ''you can't divide by zero'' seems to be ignored ? Expressions like:

[math]\frac{x^2 + 1}{x^2 - 1} = 1[/math]
cannot be defined for x^2 = 1 or x = +- 1 no matter what ( even if we multiply the equation by ( x^2 - 1 ) to get rid of the discomfort of having to divide by zero ). So why does it appear here ? How can it suddenly become plausible to divide by zero ?

I would appreciate any answers as this is a seemingly simple yet essential question for understanding the core of polynomials.

 

[topsquark]

Let's make sure the question is absolutely clear. You are given two polynomials P(x) and Q(x). You are to divide them and find the remainder in terms of x. Then you are to set the remainder to 0 and solve for x.

If that's the case then you are done. P(x) and Q(x) have a zero in common: x = -1. But as you mention that zero is not going make the remainder zero for the reason you stated... The function P(x)/Q(x) does not exist at x = -1.

-Dan

 

[Steven G]

Two initial comments.
1st) x-1 ≠ 0 when x=-1. Rather x-1=0 when x= 1
2nd) If something works for numbers, then it works for variables. If something works for variables, then it works for numbers. If something doesn't work for numbers, then it doesn't work for variables. If something doesn't work for variables, then it doesn't work for numbers.
That is, dividing by a polynomial or a real number both have the same reals.

(x^2 + 1)/(x^2 - 1) = 1. The only way a fraction can possibly be 1 is if the numerator and denominator are equal. You can't add and subtract 1 from the same number (x^2 in this case) and expect equality.

Now when does a fraction equal 0. Answer---when the numerator is 0 and the denominator is not 0. No need to do the long division (but you can).

So when does P(x) = x^3 - 5x^2 + 2x + 8 = 0? And do any of these values make Q(x) = 0?

 

[Ognjen]

So when does P(x) = x^3 - 5x^2 + 2x + 8 = 0? And do any of these values make Q(x) = 0?

I do the long division so that I can have an AX^2 + BX + C expression that I can apply the quadratic equation formula upon.

As I said, its result is x^2 - 6x + 8, which according to the text of the problem has to equal 0. Thus, 2 solutions are 2 roots of the quadratic equation x^2 - 6x + 8. None of them make Q(x) = 0.

The problem I have is that the official solution says that this problem has not 2 but 3 solutions, 2 of which are solutions of the quadratic, and one of which is the result of equating Q(x) to 0, as if making denominator zero makes the division equal to zero in any way...

They justify this by saying P(x) \ Q(x) = x^2 - 6x + 8 => P(x) = (x^2 - 6x + 8)*Q(x).

EDIT: I think I mispresented the text of the problem. I'm supposed to divide 2 polynomials, then using the solution of the division, find solutions for P(x) = 0. Still, I don't think the path to solving this diverges in any way.

 

[topsquark]

I do the long division so that I can have an AX^2 + BX + C expression that I can apply the quadratic equation formula upon.

As I said, its result is x^2 - 6x + 8, which according to the text of the problem has to equal 0. Thus, 2 solutions are 2 roots of the quadratic equation x^2 - 6x + 8. None of them make Q(x) = 0.

The problem I have is that the official solution says that this problem has not 2 but 3 solutions, 2 of which are solutions of the quadratic, and one of which is the result of equating Q(x) to 0, as if making denominator zero makes the division equal to zero in any way...

They justify this by saying P(x) \ Q(x) = x^2 - 6x + 8 => P(x) = (x^2 - 6x + 8)*Q(x).

EDIT: I think I mispresented the text of the problem. I'm supposed to divide 2 polynomials, then using the solution of the division, find solutions for P(x) = 0. Still, I don't think the path to solving this diverges in any way.

Okay, so now you are saying that the problem is to solve P(x) = 0? Then, yes. you have to include x =-1! You apparently were told that Q(x) divides P(x) in order to give you the x = -1 solution.

-Dan