How do you determine if a polynomial is the difference of two squares?

- anonymous

How do you determine if a polynomial is the difference of two squares?

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- anonymous

waits for amistre

- amistre64

im here lol

- anonymous

a suggestion is try to factor

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## More answers

- amistre64

it got the '-' in it

- amistre64

thats a big clue

- anonymous

yeah big clue

- amistre64

and if you see that the sides can be sqrted thats another

- amistre64

usually; if it has a square in it and has a - they want you to diff it

- anonymous

I don't think you can have a odd exponent as the leading term

- amistre64

you can; but it all depends on how anal a program is lol

- anonymous

oh

- amistre64

you said exponent.... i read coeeff

- amistre64

..... still may apply tho lol

- anonymous

happens to me all the time. im fukin dyslexic or something

- anonymous

how did her question get from facotring to such an analytical question. Interesting text book

- amistre64

\(x^{\sqrt{5}}\)

- anonymous

or maybe th answer is in the introductioin of the cchapter

- amistre64

\[x^{5^{1/2}}\] lol

- anonymous

x^ 1/5 - 0 X^1/5+ 0 ?

- anonymous

I just need a 50 word response for my paper that is all I need

- anonymous

is this a grad course on teaching algebra?

- anonymous

So I am trying to put that all down and in the correct form so that it sounds right

- anonymous

I don't know the definate answer to your question

- anonymous

No it is my algebra/117 my last basic class that I need

- amistre64

uhoh..... if you qoute me youre gonna need alot of ****

- anonymous

Praphrase

- amistre64

amistre64 says yada yada yada; someday it may even be proven correct lol

- anonymous

lol

- anonymous

I am trying to write on my own I just need something to get started with

- anonymous

write P=NP

- anonymous

i'll give you $400,000 if you can prove that

- anonymous

I think im short changing you, too

- amistre64

P=NP when P=0

- anonymous

lol

- anonymous

that's not proving P=NP, that's a case

- amistre64

;)

- anonymous

you can use a case to prove but you can't prove by cases

- amistre64

\[x^y + y^x = xy^{x+y}\]

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