How the fuck do you go from (x^3+6x)-385 to (x-7)(x^2+7x+55) ?? - 4ChanArchives : a 4Chan Archive of /wsr/

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How the fuck do you go from (x^3+6x)-385 to (x-7)(x^2+7x+55) ??

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Anonymous
2016-06-27 23:04:22 Post No. 147620
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Anonymous 2016-06-27 23:04:22 Post No. 147620 [Report]

How the fuck do you go from
(x^3+6x)-385
to
(x-7)(x^2+7x+55) ??

>>

Anonymous 2016-06-28 02:13:28 Post No.147729
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Anonymous 2016-06-28 02:13:28 Post No.147729 [Report]

It looks like an advanced form of factoring.

I haven't done this form of differentials in a long time; the result is right [f'(x) = (3x+2) + 6], but where does the infinitesimal (dx, delta x) show up in this problem?

>>

Anonymous 2016-06-28 02:16:21 Post No.147731
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Anonymous 2016-06-28 02:16:21 Post No.147731 [Report]

>>147729
f'(x) = (3x^2) + 6
I meant

>>

Anonymous 2016-06-28 02:18:06 Post No.147732
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Anonymous 2016-06-28 02:18:06 Post No.147732 [Report]

>>147620
>>147729
What the fuck am I even reading? you need to simplify your expression by taking out the denominator so you just need to divide x^3+6x-385 by x-7

>>

Anonymous 2016-06-28 02:21:34 Post No.147734
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Anonymous 2016-06-28 02:21:34 Post No.147734 [Report]

>>147620
it looks like they factored out the (x-7) to cancle out the one on the bottom.
Thats a bad approach.
instead, just use long division to divide the (x^3+6x-385) by (x-7).

>>

Anonymous 2016-06-28 02:31:52 Post No.147741
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Anonymous 2016-06-28 02:31:52 Post No.147741 [Report]

>>147734
This.

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