## anonymous one year ago Can someone explain Polynomial Long Division to me?

1. anonymous

I do very well with the division of a polynomial by a monomial, but once I have to divide by a binomial, I get very confused.

2. UsukiDoll

let me find an example somewhere brb

3. UsukiDoll

ah suppose we are given this problem and we need to use long division$\LARGE \frac{x^2-9x-10}{x+1}$

4. UsukiDoll

so we need to rewrite this starting with the denominator outside and the numerator inside|dw:1437820660785:dw|

5. UsukiDoll

so now we have to start finding what variables we need to multiply x+1 with

6. UsukiDoll

so how do we produce that x^2 that we need in the numerator? well we need an x so $\LARGE x(x+1) \rightarrow x^2+x$ after we obtain this result, we need to switch the signs so when we add this to the$\LARGE x^2-9x$ our $\LARGE x^2$ should be gone

7. UsukiDoll

so $\LARGE x^2+x$ becomes $\LARGE -x^2-x$

8. UsukiDoll

|dw:1437820910409:dw|

9. UsukiDoll

we have used all of our variables up, so the only thing remaining is the number

10. UsukiDoll

so based on what we are seeing we have a -10x-10 so there are negatives. If I choose 10 then we will have 10(x+1) = 10x+10 but we're going to have a problem. Our goal is to cancel -10x-10 and not have a remainder if I switch the sign of 10x+10 I will have -10x-10 and that's something we don't want.

11. UsukiDoll

so I need to choose -10 so that -10 is distributed all over the x+1 -10(x+1) = -10x-10 switching those signs gives mean 10x+10 and that's what I need so I can cancel the -10x-10 in the division.

12. UsukiDoll

|dw:1437821086041:dw|

13. UsukiDoll

so our final answer is (x+1)(x-10) and 0 remainders expanding (x+1)(x-10) with FOIL gives back $\LARGE x^2-10x+x-10 \rightarrow x^2-9x-10$

14. UsukiDoll

so we have done this correctly

15. UsukiDoll

I'm going to give another example where we do have a remainder.

16. UsukiDoll

now suppose we are given $\LARGE \frac{2x^3-9x^2+15}{2x-5}$

17. UsukiDoll

whoa oh... we don't have an x... we just put a placeholder like 0x

18. UsukiDoll

|dw:1437821546765:dw|

19. UsukiDoll

our x's in the numerator must increase to decrease and it needs to be present.. in this example the regular x was missing so a placeholder 0x is inserted

20. UsukiDoll

so starting with variables only . I notice that there is a 2x^3 and we have 2x-5 if I multiply $\LARGE x^2(2x-5) =2x^3-5x^2$ switching the signs gives me this $\LARGE -(2x^3-5x^2) => -2x^3+5x^2$

21. UsukiDoll

|dw:1437821703230:dw|

22. UsukiDoll

our new goal is to get rid of the -4x^2 we have already used x^2 so we can get rid of the 2x^3 so now we need an x and a number

23. UsukiDoll

so suppose we choose 2x then $2x(2x-5) => 4x^2$... whoops if I change the sign I'll have a negative -4x^2-4x^2 = -8x^2 and that's a big no no ok we need -2x $\LARGE -2x(2x-5) => -4x^2+10x$ switching signs we have $\LARGE -(-4x^2+10x) =4x^2-10x$

24. UsukiDoll

|dw:1437821921260:dw|

25. UsukiDoll

so now that we have used up all the variables we are down to numbers our new goal is to get rid of -10x+15 if we have a negative we need to multiply 2x-5 with -5 , so that when the signs are switched we have a 10x. $\LARGE -5(2x-5) = -10x+25$ switching signs $\LARGE -(-10x+25) = 10x-25$

26. UsukiDoll

|dw:1437822060614:dw|

27. UsukiDoll

|dw:1437822081374:dw|

28. UsukiDoll

we have a remainder of -10... we need to write our answer a little bit differently

29. UsukiDoll

so we write the $\LARGE x^2-2x-5$ first the remainder $\LARGE-10$ will be on the numerator and $\LARGE 2x-5$ will be on the denominator so the answer will look like $\LARGE x^2-2x-5+\frac{-10}{2x-5}$

30. UsukiDoll

so do my examples make this a bit easier to understand? do you have questions?

31. anonymous

I understand more clearly now in terms of dividing a trinomial by a binomial, but once I get to larger polynomials, I run into trouble.

32. anonymous

Does the process differ at all once you get to the larger polynomials? @UsukiDoll

33. UsukiDoll

hmmmm... the only types of division (for polynomials) is dividing a trinomial with a monomial and a trinomial with a binomial wait brb I have to do something that needs my attention immediately

34. anonymous

Okay, no problem.

35. UsukiDoll

alright I'm back.. so is there any specific example that we can go over together?

36. UsukiDoll

like a polynomial divided by a trinomial? the process should be the same... just takes longer.

37. UsukiDoll

ah .. here's an example So suppose we have a longer polynomial divided by a trinomial like $\LARGE \frac{x^4-3x^3+2x^2-x+1}{x^2-x+1}$

38. UsukiDoll

a big heads up... this is going to take more than one drawing to do .

39. UsukiDoll

|dw:1437823307750:dw|

40. UsukiDoll

so now our goal to get rid of the x^4 so we need x^2 $\LARGE x^2(x^2-x+1) =x^4-x^3+x^2$ switching signs $\LARGE -(x^4-x^3+x^2) = -x^4+x^3-x^2$

41. UsukiDoll

|dw:1437823443863:dw|

42. UsukiDoll

so now our new goal is to get rid of the -2x^3 so we need to multiply -2x with x^2-x+1 $\LARGE -2x(x^2-x+1) = -2x^3+2x^2-2x$ switching signs we have $\LARGE 2x^3-2x^2+2x$

43. UsukiDoll

|dw:1437823607833:dw|

44. anonymous

I see, so it's still the same process for the most part. However, I can see how it all aligns when you're dividing polynomials with a dividend and divisor only 1 term in difference, but some of my class problems are 2, sometimes even 3 terms difference.

45. UsukiDoll

hmm I would do this one at a time though to avoid errors

46. UsukiDoll

anyway we have used up all of our variables and we are left with numbers. since we have a -x^2, we need to multiply -1 with x^2-x+1 and switch signs $\LARGE -1(x^2-x+1) = -x^2+x-1$ switching signs $\LARGE x^2-x+1$

47. UsukiDoll

|dw:1437823911327:dw|

48. UsukiDoll

|dw:1437823935681:dw|

49. UsukiDoll

so our remainder is 2

50. UsukiDoll

$\LARGE x^2-x+1+\frac{2}{x^2-2x-1}$

51. UsukiDoll

sorry got those switched

52. UsukiDoll

$\LARGE x^2-2x-1+\frac{2}{x^2-x+1}$

53. UsukiDoll

the process is the same. It's just that the bigger the polynomial, the more time consuming and tedious it gets

54. alekos

good work, this type of maths is not easy