Hey , any brainy Sup Forums's here who can explain this shit as simply as possible for dumbass like me ?

hey , any brainy Sup Forums's here who can explain this shit as simply as possible for dumbass like me ?

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khanacademy.org/
youtube.com/watch?v=VrghY1CarA0
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C - 2nd Year Aeronautical Engineer. I should qualify

B sorry mb

Please explain why

If you had a question about a trap thread, Sup Forums is your place. Math, not so much

I fucked up, let me write it and take a picture

ye ye i know , but math is necessary for me and my future , i want to be autistic in US not in my shit country

Ty user

RTFM faggot

Sorry OP,
2nd year pure maths undergrad, I have forgotten most of the little I knew of arithmetic and algebra.

that said, a polynomial `p` is said to be divisible by another `q` if there is some `r` such that r * q = p
> q | p ⇔ ∃r:p=r*q

skipping a whole lotta fucking boring algebra, you just do long division on the things and see if there is a remainder.
if there aint, it's divisible.

this is busting me, got my pal working on it and its not a correctly worded question

I won't do the calculation (because it's mindless and I've got my own mindless calculations for tomorrow's DE test, which I'm about to flunk)

but the algorithm is simple.

Want to find out if a polynomial `p` is divisible by another `q`?

write out the first one, and then the second one. Do long division, which means:

>multiply q by the largest polynomial that doesn't make it go over p;
>subtract that from p
>repeat but with the result from above
> ???
>profit

y tho?

First of all, khanacademy.org/
Second of all, the h(x), p(x), r(x), and s(x) don't mean anything other than that they're different functions.

The question is which of them is divisible by 2x+3.

So A) is h(x) = f(x) + g(x) means h(x) = 2x^3 + 6x^2 + 4x + x^2 +3x + 2
Simplified is h(x) = 2x^3 + 7x^2 + 7x + 2
You have to figure out if you can divide the whole function by 2x+3.
When it says 3g(x) it means 3(x^2 +3x + 2) which you distribute (Multiply each term by 3)
which comes out to 3x^2 + 9x + 6.

The answer is not D because,
s(x) = 3(2x^3 + 6x^2 +4x) + 2(x^2 +3x + 2)
s(x) = 6x^3 + 18x^2 + 12x) + 2x^2 +6x + 4
1 2 3 2 3 4
s(x) = 6x^3 + 20x^2 + 18x + 4
s(x)/2x+3 isn't easily divisible. (That's the hard part to explain. Go to kahnacademy.org and look up dividing polynomials or something)

I cant get it to work...

I mean, if you want to make sure, just multiply out those factors, if it is the same polynomials you began with, than the exercise is wrong.
They often are

The only possible options are B and C
Since if they were a multiple of 2x+3, then their independent coefficient would have to be divisible by 3.

Actually, no. this was me being a fucking moron

Nigga its right there on be (4x+6) its same as 2(2x+3) and there you have it

learn to wolfram.

i found at least one that has that as a root.

youtube.com/watch?v=VrghY1CarA0

I mean if your shit's correct. I didnt check that

>mfw who would win? >600 years of algebra or some wolfram bois
I clearly am wasting my life with this degree
>pic unrelated