anonymous
  • anonymous
More Polynomials ugh... (6x^3+11x^2-4x-4)/3x-2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
I'm understanding them better but I still don't know what to do with the -2 part.
anonymous
  • anonymous
are you doing long division?
anonymous
  • anonymous
Probably. Is that how the answer is normally found?

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anonymous
  • anonymous
sometimes. when you can factor, that way's easier, but it doesn't look like this factors easily
anonymous
  • anonymous
Ok. So how is the long division done?
anonymous
  • anonymous
|dw:1441661061285:dw|
anonymous
  • anonymous
i have it written like that, but how is it done? I'm bad at math so i need it explained step by step.
anonymous
  • anonymous
start off by dividing 6x² by 3x \[\frac{ 6x^2 }{ 3x }=2x^2\] then write 2x above the division sign and multiply it by (3x-2) \[2x^2(3x-2)=6x^3-4x^2\] then subtract |dw:1441661304727:dw|
anonymous
  • anonymous
Oh I see! But do I subtract 6x^3 - 4x^2? Or 11x^2-4x^2?
anonymous
  • anonymous
subtract like terms, so \(6x^3 - 6x^3\) and \(11x^2-(-4x^2)\)
anonymous
  • anonymous
like terms is the exponent right? subtract ^2's with ^2's?
anonymous
  • anonymous
yep
anonymous
  • anonymous
so then i get 7x^2 and divide that by 3x-2 all over again?
anonymous
  • anonymous
no it's 11-(-4), so that's -15 then it's like this once you drop the other terms down |dw:1441661731441:dw|
anonymous
  • anonymous
Now you start over again by dividing \[\frac{ 15x^2 }{ 3x }\]
anonymous
  • anonymous
ok 15 makes more sense lol
anonymous
  • anonymous
yeah, most times the numbers will be nice to work with, so that's usually a good check
anonymous
  • anonymous
I'm writing it all on paper and I think I see the pattern now. It's slowly losing the exponent one power at a time?
anonymous
  • anonymous
exactly
anonymous
  • anonymous
yay! thank you so much!
anonymous
  • anonymous
you're welcome. tag me if you want the answer checked
anonymous
  • anonymous
|dw:1441662259557:dw| um but is it supposed to completely lose the x?
anonymous
  • anonymous
@peachpi
anonymous
  • anonymous
The 10 is supposed to be 10x, so then when you subtract you should have positive 6
anonymous
  • anonymous
so 6x or 6?
anonymous
  • anonymous
6x
anonymous
  • anonymous
ok. i'll post my answer in a bit.
anonymous
  • anonymous
ok.
anonymous
  • anonymous
2x^2+5x+2x?
anonymous
  • anonymous
2x^2+5x+2 There shouldn't be an x on the last 2.
anonymous
  • anonymous
ok
anonymous
  • anonymous
|dw:1441662698472:dw|
anonymous
  • anonymous
yup that's what i've got. thanks!
anonymous
  • anonymous
you're welcome :) are you just supposed to divide this, like do they ask anything else?
dinamix
  • dinamix
the answer must be like ax^2+bx +c cuz is degree Numerator 3 and denumerator degree 1 @xFeathertailx
anonymous
  • anonymous
nope, it doesn't give any instruction other than putting it in descending power of x. just a space to put the answer.
dinamix
  • dinamix
if was mod 0
dinamix
  • dinamix
sorry i forget
anonymous
  • anonymous
ok, that's the answer then xD
dinamix
  • dinamix
yup
dinamix
  • dinamix
good explain @peachpi
anonymous
  • anonymous
one more question. ae you sill here @peachpi ?
anonymous
  • anonymous
or @dinamix
anonymous
  • anonymous
yes
anonymous
  • anonymous
trying to get the pic.
anonymous
  • anonymous
well it's 2x^2+4x+5 x-2|2x^2+x+3 2x^2-4x 4x 3 4x-8 -5
anonymous
  • anonymous
i think soomething went wrong. it's supposed to fit into
1 Attachment
anonymous
  • anonymous
|dw:1441663943278:dw|
anonymous
  • anonymous
It looks like the 2x² you have on top should be 2x. Then you subtracted the x terms wrong
anonymous
  • anonymous
ok. isee that now. but ho w does it fit into the grid i showed you?
anonymous
  • anonymous
|dw:1441664318011:dw| So the solution is \[2x+5+\frac{ 7 }{ x-2 }\]
anonymous
  • anonymous
Oh!!! Because the ^2 shouldn't have been there and the remainder goes above the bar? Now that makes sense!!!
anonymous
  • anonymous
I love this site! So many helpful people!
anonymous
  • anonymous
lol. glad to help

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