Help!

BabyBlue

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I need help desperately!
I am having problems with-what else? Divining polynomials.
I tried to post this once and somehow all my writing did not get posted properly. Okay here is the original problem:
a^8+a^6+a^4+3a-1 by a+1

So :shock: I add the missing powers: 7, 5,3, 2
Then I get for the begging of the quotient:
a^7
then:-a^6
So I multiply (+) 1 by (-)a^6 and I get -1a^6 correct? Because multiplying in algebra is: opposite=negative and same=positive
So, I try subtracting -1a^6 from +a^6, and they are saying it is wrong. Where did I slip? They are saying the next quotient is 2a^5? I was thinking in my head "what?"

I hope I am not bugging you great guys,
Amy
 

skeeter

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Code:
      a^7-a^6+2a^5
       ---------------------------------------------------------
a+1 | a^8+0a^7+a^6+0a^5+a^4+0a^3+0a^2+3a-1
      a^8+ a^7
      --------------
          -a^7+a^6+0a^5+a^4+0a^3+0a^2+3a-1
          -a^7-a^6
          -----------------
              2a^6+0a^5+a^4+0a^3+0a^2+3a-1
              2a^6+2a^5
             -------------------
                  -2a^5+a^4+0a^3+0a^2+3a-1
can you finish?
 

BabyBlue

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Skeeter,
do you mean that because they are both opposite we add?
Then a+a= 2 and keep the same power?
Okay.. I am trying.
Yuck.. I feel sick. not because of you though
:wink:
 

skeeter

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you do understand that you are subtracting the lower line of terms from the upper ... when you subtract a negative value, what happens?
 

Denis

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Say you are subtracting -5x from +7x:
+7x
- 5x
===
+12x

I find the easiest way "to make sure" is check by going backward:
12x - 5x = 7x
 

BabyBlue

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I think I got it

Okay, so we bring down the a^5
and times the divisor 1 by +2a^5
and we get +2a^5
and we take away that from 0a^5
and we get negative 2a^5?

The answer is (I think)
a^7-a^6-2a^4-2a^3-2a^2-2a+5
remainder=-4
:?
hopefully I am right.
Well got to go!
Love,
Amy
 

tkhunny

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BabyBlue - Emphasis Added said:
I am having problems with-what else? Divining polynomials.
That is an absolutely hilarious typo! It does descibe how some students approach polynomials. I am glad to see that you are trying to be a bit more down-to-earth about it. :lol:
 

tkhunny

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Re: I think I got it

BabyBlue said:
and times the divisor 1 by +2a^5
No, you multiply it.
and we take away that from 0a^5
No, you subtract it.

Were you not ever mercilessly forced to do long division in grammar school? Really, it is EXACTLY the same,
 

stapel

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tkhunny said:
Were you not ever mercilessly forced to do long division in grammar school?
Quite possibly not. Modern progressive (new-new) educationist thought says that there is no need for students to do long numerical division, since calculators can do it for them. Many curricula have had this topic specifically removed, and this is regarded by many educationist authorities, such as the NCTM, as being a good thing. :?

Of course, then we tutors and college teachers have to teach this elementary-school topic to our students, in order to explain so many other topics, lo, these many years later. Oy. :shock:

Eliz.
 

BabyBlue

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People,
I am still here
 

stapel

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BabyBlue said:
People,
I am still here
Okay.... Did you have a question...? :?:

You've been provided with half of the worked solution, and tutors have attempted to figure out how to answer your other (somewhat befuddling) posts. If you are needing help with other details of this process (like how to combine "like" terms, or how to do arithmetic with negatives, or how to multiply polynomials, etc), then please reply with specifics.

Thank you! :D

Eliz.
 

tkhunny

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stapel said:
Quite possibly not. Modern progressive (new-new) educationist thought ...
Well, there I am, out of the loop again.

In any case, I wanted to back off my position just a bit. The numerical long division uses only integers and uses the entire structure of the divisor in each step. The polynomial version is severely generalized, caring only about the first element at each phase and allowing noninteger and negative results. I think a numerical long division would have a problem with the answer 1(½)(-3)4.
 

stapel

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tkhunny said:
The numerical long division uses only integers and uses the entire structure of the divisor in each step.
Roughly, yes. But if you look closely at the raw process, you'll note that the way you often find your first guess (in "A ÷ B"), as to how many times B will go into A, is to look at the first few (leading) digits of A and the first few digits, or rounded value, of B.

For instance, in dividing 295349 by 47, you wouldn't necessarily initially consider all of 295349 or the exact value 47. Instead, you might note that 47 is bigger than 29, so you won't be putting "1" on top of the long-division sign. Adding in another digit, you have 295 and 47. This is, roughly, 300 and 50, so try 6: (47)(6) = 282. This "fits" inside 295, so put "6" on top, put "282" underneath, and subtract.

Lather. Rinse. Repeat. :wink:

Eliz.
 

Mrspi

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stapel said:
Lather. Rinse. Repeat. :wink:

Eliz.
That's wonderful, Eliz!
 

BabyBlue

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I am trying to be patient, but I do not like being laughed at. I am trying to learn. :?
I'll submit the answer soon.
 

morson

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doesn't look like anybody has laughed or made fun of you in this thread ... in fact, not a semblance of mockery/derision can be found in the posts above ...
 

stapel

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Amongst the various partial solutions and explanations said:
...I am glad to see that you are trying to be a bit more down-to-earth about [this]. :lol:
BabyBlue said:
I am trying to be patient, but I do not like being laughed at.
It is to be regretted if you feel that the step-by-step instructions, the explanations, and the encouragement you've been offered are somehow meant to mock you. But I'm afraid there is little we can do to fix that particular situation. Sorry! :oops:

My best wishes to you in your studies! :D

Eliz.
 

BabyBlue

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Sorry if I took offense to easily :p
 

BabyBlue

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my answer

The answer is:
a^7-a^6+2a^5-2a^4+3a^3-3a^2-3a
fraction -1 over a+1
 

stapel

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BabyBlue said:
The answer is:
a^7-a^6+2a^5-2a^4+3a^3-3a^2-3a
fraction -1 over a+1
Do you mean that this was the solution that was provided to you? Or that this is what you've come up with, and you're requesting correction or confirmation?

Do you mean the following?

. . . . .a^7 - a^6 + 2a^5 - 2a^4 + 3a^3 - 3a^2 - 3a - 1/(a + 1)

Or do you mean something else?

Thank you! :D

Eliz.
 
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