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Mathematics 14 Online
OpenStudy (anonymous):

Helpppp pleaseee >(easy question,easy medal) trying to algebraically simplify this product.. (t^2+1)^3*(t+3)^5 ? do I just re-write as (cubed-root)t+1*5throot(t+3)?

OpenStudy (studygurl14):

I'm thinking what they want you to do is multiply everything out and then combine like terms.

OpenStudy (anonymous):

it seems like multiplying out would give me some crazy polynomial?

OpenStudy (studygurl14):

Yea, I know. Usually they make you do the other way around, so I don't really know what they want...sorry. :(

OpenStudy (wilder.monday):

(t^5 + 1^3)(t^5 + 3^5)

OpenStudy (anonymous):

im terrible about remembering exponential rules...but can I do it that way or does it have to be multiplied completely out?... like.. (t^2+1)(t^2+1)(t^2+1)?... and so on..

OpenStudy (wilder.monday):

Add the exponents of 2 and 3 then distribute the exponent outside of the parentasis to every term

OpenStudy (wilder.monday):

@lahernan , no it doesn't because that's a different rule

OpenStudy (wilder.monday):

Just do it the way @TemporalSorceress did it

OpenStudy (anonymous):

well im not quite sure that's correct?

OpenStudy (wilder.monday):

when you distribute the exponent it simplifies to\[(t ^{5} + 1^{3})(t ^{5} + 3^{5})\]

OpenStudy (studygurl14):

@wilder.monday That's is incorrect. You cannot distribute the exponent like that.

OpenStudy (wilder.monday):

then how would you do it?

OpenStudy (anonymous):

im evaluating it as a limit so .. im just trying to evaluate them as separate equations and then multiply them together...

OpenStudy (wilder.monday):

I believe if I continue with what I did when simplifying it turns out to be \[t ^{15} + 3t ^{5} + 3^{5}\] and 3^5 simplifies further to 243

OpenStudy (anonymous):

I think my instruction were a little confusing, im not trying to solve the polynomial im just trying to evaluate it to a simpler form...

OpenStudy (wilder.monday):

may I see the material given?

OpenStudy (anonymous):

the instructions are "Evaluate the limit and justify each step by indicating the appropriate limit law(s). \[\lim_{t \rightarrow -1}(t ^{2}+1)^3(t+3)^5\]

OpenStudy (wilder.monday):

this is using the multiplication limit law which states the limit as x goes to a of f(x)*g(x) = the limit as x goes to a of f(x) * the limit as x goes to a of g(x)

OpenStudy (anonymous):

correct. which I understand that part, but I wasn't sure if I could just substitute -1 into t and then raise to the third power and do the same for the other side and then just raise to the 5th power if that was an accurate evaluation..?

OpenStudy (anonymous):

u guys r doing great

OpenStudy (wilder.monday):

\[\frac{ (t ^{5} + 1^{5})(t ^{5} + 3^{5}) }{ x - 1 }\]

OpenStudy (wilder.monday):

@Noland__DeWitte thanks :)

OpenStudy (anonymous):

welcome

OpenStudy (anonymous):

I'm confused on how you derived that ?

OpenStudy (anonymous):

its ok if u r confused u will get the hang of it @lahernan

OpenStudy (wilder.monday):

\[\frac{ (t ^{2} + 1)^{3}(t + 3)^{5} }{ x - 1 }\]

OpenStudy (wilder.monday):

give me a second to explain

OpenStudy (wilder.monday):

original question: "Evaluate the limit and justify each step by indicating the appropriate limit law(s). lim t→−1 (t^2 +1)^3(t+3)^5 original is numerator and x - 1 is denominator. which was the last equation I did. when simplifying, the 1^3 stays as 1 and the 3^5 becomes 243. now, plugging in 243 for x in x - 1, the limit comes out to be 242. if you plugged in the 1 for x in x - 1 the limit would be 0, which is not possible.

OpenStudy (wilder.monday):

@lahernan

OpenStudy (anonymous):

what do the squares mean?

OpenStudy (anonymous):

oh nevermind. ok I see what your saying.

OpenStudy (wilder.monday):

it \[\lim_{t \rightarrow -1}\]

OpenStudy (anonymous):

ok. well thank you very much for your help i'm going to print this out and regurgitate it and see if I can't let it soak in. .. :) thank you .

OpenStudy (wilder.monday):

your welcome :) medal maybe?

OpenStudy (anonymous):

sure thing

OpenStudy (wilder.monday):

thanks!!

OpenStudy (anonymous):

http://sketchtoy.com/64186775

OpenStudy (anonymous):

This what you guys get?

OpenStudy (anonymous):

I think so

OpenStudy (anonymous):

The limit is 256.

OpenStudy (anonymous):

thank you

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