# Division with polynomials involving the remainder theorem

1. Sep 8, 2006

### Byrgg

Two questions, first, I solved something, but I was just playing around with the numbers, and I didn't really know what I was doing, nor did I really understand it after I was done. The question is as follows:

When x + 2 is divided into f(x), the remainder is 3. Determine the remainder when x + 2 is divided into each of the following:

a) f(x) + 1

b) f(x) + x + 2

I'm only going to use a few of the questions here, just for simplicity. I know how to solve them, because I checked my answers in the back of the book. I just don't really understand why this works out the way it does.

Secondly, something I really have no clue about, at all. The question is as follows:

If f(x) = mx^3 + gx^2 - x + 3 is divided by x + 1, the remainder is 3. If f(x) is divided by x + 2, the remainder is -7. What are the valuse of m and g?

I really don't know what to do, I even wondered if both pieces of information were necessary. I thought that I'd ismplify one, and then rearrenage to solve for either mor g, in terms of the other one, but it didn't work, or maybe I just didn't do it right.

2. Sep 8, 2006

### vsage

For a and b: think of what you are given here. I have no idea what the remainder theorem is so I couldn't help you there, I can tell you that if the remainder of f(x) / (x+2) is 3, then f(x) can be rewritten was k(x)*(x+1) + 3 / (x + 2), where 3 / (x+2) represents the remainder of the function with x+2 and k(x) is the stuff that divided evenly in. Given this way of writing f(x), what is f(x) + x + 2?

f(x) + x + 2 = [k(x) * (x + 2) + 3 / (x + 2)] + x + 2 = (k(x) + 1) * (x + 2) + 3 / (x + 2). Clearly the remainder term is the same. I'll leave a) up to you.

Also, I haven't actually worked out your second issue, but two unknowns *always* requires two givens to completely solve the system of equations you will ultimately get from analyzing the problem. Just my two cents; I'm probably talking outside my scope of knowledge since I didn't understand your title :p

Last edited by a moderator: Sep 8, 2006
3. Sep 9, 2006

### HallsofIvy

If you learn nothing else from this problem, learn this:
When a problem tells you to "use the remainder theorem", for God's sake look up the remainder theorem and make certain you know exactly what it says! vsage did some good work but as he said he didn't know what the remainder theorem was. With the remainder theorem, both problems are trivial!

The remainder theorem just says that if a polynomial P(x) is divided by x-a, the remainder is P(a). That's true because when an nth degree polynomial is divided by a first degree polynomial, the quotient is a n-1 degree polynomial and the remainder is a number:
P(x)/(x-a)= Q(x)+ r/(x-a) or P(x)= Q(x)(x-a)+ r. Setting x= a we get
P(a)= Q(a)(0)+ r so r= P(a).

Since you are told "When x + 2 is divided into f(x), the remainder is 3", since x+2= x-(-2), you know that f(-2)= 3.
a) When f(x)+ 1 is divided by x+2, the remainder is f(-2)+ 1= 3+ 1= 4.
You do (b).

"If f(x) = mx^3 + gx^2 - x + 3 is divided by x + 1, the remainder is 3. If f(x) is divided by x + 2, the remainder is -7. What are the valuse of m and g?"

Since x+1= x-(-1), the remainder theorem tells you that f(-1)= m(-1)3+ g(-1)2- (-1)+ 3= -m+ g+ 4= 3 and, since x+2= x-(-2), that f(-2)= m(-2)3+g(-2)2- (-2)+ 3= -8m+ 4g+ 5= -7. Solve the two equations
m- g= 1 and g- 2m= 6 for m and g.

Last edited by a moderator: Sep 9, 2006
4. Sep 9, 2006

### Byrgg

I already knew what the remainder theorem was, I never said I didn't. As for my first problems, I already solved them, the problem was I didn't understand how my method made any sense, now, thanks to you guys, I understand it. Since x + 2 is being divided into f(x), and you know that remainder to be 3, then f(x) will always be equal to 3 in these cases.

And with the second problem, you are subbing -2 into each x, since you know that f(x) = 3 when x = -2, you can solve the problems. I think I've got it now. Thanks for help with that.

The second also makes more sense now, thanks again.