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# Division of polynomials (comparing coefficient method) how to find out remainder watch

1. I prefer using the "comparing coefficients" method (I think that's what it's called. It's not actually in my textbook.) because I'm completely hopeless at the other methods.

This is the method I mean. It's just an example. (x^3 + 4x^2 + x -6)/(x - 2) = Ax^2 + Bx + C

And multiply both sides by the divisor and just go through the letters A, B and C to find out the value.

But what do I do when there's a remainder? How would I finish this question?

(x^3 + 3x^2 + 6x + 1)/(x + 1)

I make (x^3 + 3x^2 + 6x + 1) = (Ax^2 + Bx + C)(x + 1)

coefficient of x^3 : 1 = A therefore A = 1
coefficient of x^2 : 3 = A + B therefore B = 2
coefficient of x^1 : 6 = B + C therefore C = 4
coefficient of x^0 : 1 = C therefore C = 1

Obviously my last value of C is wrong. How do I work out the remainder please? Rep on offer. Thanks.
2. Introduce an extra constant, the remainder , by writing
3. (Original post by Kolya)
Introduce an extra constant, the remainder , by writing
Righto! So once I've worked out my A, B and C values what do I do to work out R?
4. (Original post by AnythingButChardonnay)
Righto! So once I've worked out my A, B and C values what do I do to work out R?

What value does the polynomial take when x = -1?
5. (Original post by ghostwalker)
What value does the polynomial take when x = -1?
-3, which is the answer!

So I just have to plug x = -1 into the polynomial every time to find out what the remainder is? If so that's great. Why does it work?
6. (Original post by AnythingButChardonnay)
-3, which is the answer!

So I just have to plug x = -1 into the polynomial every time to find out what the remainder is? If so that's great. Why does it work?

NOOOOOO. You plug in x=-1 because (x+1) is a factor on the RHS, and hence that part will be zero, and you are left with R.

If the factor was (x+a), you'd use x=-a.
7. When you have:

(x+1)(x^2+2x+4), and you multiply them out, by inspection you can see the x^0 coefficient will be 4

hence -3 will be needed to make the x^0 coefficient 1.

(when using your method and working out the value of a,b and c, the higher coefficients always take priority, so c will be 4 rather than 1)

8. (Original post by ghostwalker)
NOOOOOO. You plug in x=-1 because (x+1) is a factor on the RHS, and hence that part will be zero, and you are left with R.

If the factor was (x+a), you'd use x=-a.

So if I was dividing by (2x + 3) I would plug in x = -3/2 ?

(Original post by Robbie10538)
When you have:

(x+1)(x^2+2x+4), and you multiply them out, by inspection you can see the x^0 coefficient will be 4

hence -3 will be needed to make the x^0 coefficient 1.
Ah! That's good too. I would probably use that to check if I had time. It depends which would take longer, and which I'd be more likely to hash up.
9. (Original post by AnythingButChardonnay)

So if I was dividing by (2x + 3) I would plug in x = -3/2 ?
Yes that is correct. The method he showed is the remainder theorem and so works as well. I'd use this to check you have the correct remainder.
10. Brilliant. Thanks so much. Rep to all 3 of you.
11. So if I was dividing by (2x + 3) I would plug in x = -3/2 ?
Yep.
12. This is the method I use and I think it is so much easier to understand than long division.
13. Depends which way is fastest for you.
14. I like this method too, just very easy to see where you've gone wrong, if you have.
15. how about when you are dividing by an x squared, an x and a number
16. (Original post by bruce11)
how about when you are dividing by an x squared, an x and a number
Easiest to have a specific example you're working on, and how far you've got.

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