long division polynomial
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The Highlander
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There seems to be a bit missing from your picture. Where it says: "(Simplify your answers. Do not factor. U" there is, clearly, a missing (closing) bracket, therefore it would appear that the "U" is the beginning of another word or phrase and we need to see all of what is written there in the original problem. Please post another picture that shows your original question in its entirety.
However, that said, this is "how" (the way) you "solve" it:-
[math]\frac{31}{7} = \frac{28+3}{7} = \frac{28}{7} + \frac{3}{7} = 4+\frac{3}{7} \left(since \frac{28}{7} = 4\right)[/math]
Do you see the parallel?
EXCEPT, of course, you are not faced with a simple division like: \(\displaystyle \frac{28}{7} = 4\), so you need to use Polynomial Long Division (just as your topic heading implies).
Clearly, (since you chose that topic heading) this is an area of Mathematics that you are currently studying or are required to know about.
If you are really as "stuck" as you suggest and have not made any attempts at solving this yourself (because you know little or nothing about how to handle Polynomial Long Division) then I suggest you go here and read up about it.
Then come back and show us what attempt(s) you have made to reach an answer; it doesn't matter how "good" or "bad" your attempt(s) are, please just post something that you have done! (Along with the complete picture of the original problem, as mentioned above.)
We will then confirm your solution as correct or offer advice on how to rectify your approach but we don't just supply the answers to problems like this, we need to see some input from you first.
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Cubist
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Code:
3x² + 8x - 1
--------------------------------
2x²+3 ) 6x⁴ + 16x³ + 7x²
-( 6x⁴ + 9x² )
-----------------
16x³ - 2x²
-( 16x³ + 24x )
-----------------
-2x² - 24x
-( -2x² - 3)
---------------
- 24x + 3
...[math]3x^2 + 8x - 1 + \dfrac{-24x+3}{2x^2 + 3}[/math]
BigBeachBanana
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I think you have the form you asked for without factoring.Then I got stuck on, "simplify your answers, do not factor". Can this be simplified without factoringCode:3x² + 8x - 1 -------------------------------- 2x²+3 ) 6x⁴ + 16x³ + 7x² -( 6x⁴ + 9x² ) ----------------- 16x³ - 2x² -( 16x³ + 24x ) ----------------- -2x² - 24x -( -2x² - 3) --------------- - 24x + 3
[math]3x^2 + 8x - 1 + \dfrac{-24x+3}{2x^2 + 3}[/math]
I guess you could also write in quotient and remainder form, which is a bit nicer.
[imath](3x^2 + 8x - 1)(2x^2 + 3) +(-24x+3)[/imath]
One of the things I do not like about the way polynomial multiplication is taught is that it is not clear why it works.
[math]\dfrac{6x^4 + 16x^3 + 7x^2}{2x^2 + 3} = \dfrac{3x^2(2x^2) + 16x^3 + 7x^2}{2x^2 + 3} = \dfrac{3x^2(2x^2 + 3 - 3) + 16x^3 + 7x^2}{2x^2 + 3} =\\ \dfrac{3x^2(2x^2 + 3) - 9x^2 + 16x^3 + 7x^2}{2x^2 + 3} = 3x^2 + \dfrac{16x^3 - 2x^2}{2x^2 + 3} =\\ 3x^2 + \dfrac{8x(2x^2) - 2x^2}{2x^2 + 3} = 3x^2 + \dfrac{8x(2x^2 + 3 - 3) - 2x^2}{2x^2 + 3} = \\ 3x^2 + \dfrac{8x(2x^2 + 3) - 24x - 2x^2}{2x + 3} = 3x^2 + 8x + \dfrac{- 2x^2 - 24x}{2x^2 + 3} =\\ 3x^2 + 8x + \dfrac{(-1)(2x^2) - 24x}{2x^2 + 3} = 3x^2 + 8x + \dfrac{(-1)(2x^2 + 3 - 3) - 24x}{2x^2 + 3} = \\ 3x^2 + 8x + \dfrac{(-1)(2x^2 + 3) + 3 - 24x}{2x^2 + 3} = 3x^2 + 8x - 1 + \dfrac{3 - 24x}{2x^2 + 3}. [/math]
Long division is just a way to do the preceding with far fewer steps.
Do the long division yourself as you were taught and see how it skips many of the preceding steps.
[math]\dfrac{6x^4 + 16x^3 + 7x^2}{2x^2 + 3} = \dfrac{3x^2(2x^2) + 16x^3 + 7x^2}{2x^2 + 3} = \dfrac{3x^2(2x^2 + 3 - 3) + 16x^3 + 7x^2}{2x^2 + 3} =\\ \dfrac{3x^2(2x^2 + 3) - 9x^2 + 16x^3 + 7x^2}{2x^2 + 3} = 3x^2 + \dfrac{16x^3 - 2x^2}{2x^2 + 3} =\\ 3x^2 + \dfrac{8x(2x^2) - 2x^2}{2x^2 + 3} = 3x^2 + \dfrac{8x(2x^2 + 3 - 3) - 2x^2}{2x^2 + 3} = \\ 3x^2 + \dfrac{8x(2x^2 + 3) - 24x - 2x^2}{2x + 3} = 3x^2 + 8x + \dfrac{- 2x^2 - 24x}{2x^2 + 3} =\\ 3x^2 + 8x + \dfrac{(-1)(2x^2) - 24x}{2x^2 + 3} = 3x^2 + 8x + \dfrac{(-1)(2x^2 + 3 - 3) - 24x}{2x^2 + 3} = \\ 3x^2 + 8x + \dfrac{(-1)(2x^2 + 3) + 3 - 24x}{2x^2 + 3} = 3x^2 + 8x - 1 + \dfrac{3 - 24x}{2x^2 + 3}. [/math]
Long division is just a way to do the preceding with far fewer steps.
Do the long division yourself as you were taught and see how it skips many of the preceding steps.
The Highlander
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Are you guys providing worked solutions "just for the sake of completeness" because the OP has never been seen since the day after his/her original post? (ie: two months ago. ?)
If so, why are you still "addressing" him/her at the end of each of your posts? ?
If so, why are you still "addressing" him/her at the end of each of your posts? ?
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Maybe because I did not look at the original posting date?
BigBeachBanana
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Posting Worked Solutions - Comments Please
Hi, So they are asking to us optimize cost in terms of the surface area of the tank. The equation for the surface of the tank is the surface of a sphere (2 hemispheres put together) and the surface area of the cylinder (ignoring the circle top and bottom); SA=4Pir^2+2Pirh Now to make the...
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Cubist
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My fault, sorry for causing confusion. I fancied a bit of a change today so I trawled through some old unanswered posts. I think that we're ok to post an answer if they're over a month old. In the future I'll put a prominent comment like "Answering an old post", if I do this again
The Highlander
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Yes, I was vaguely aware that there was a "rule of thumb" that it was 'fine' to post a worked solution after a sensible period (a month?) had elapsed and that might well provide other members searching the forum with useful information/exemplars.
It was not my intention to criticise the provision of the solutions as they all made very interesting reading (esp. Jeff's ?).
It just seemed ironic to me that they appeared to be "addressed" to the OP (eg: Jeff's exhortation to: "Do the long division yourself as you were taught...") despite the likelihood the s/he was never, ever going to come back again (after two months' absence). ?