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Mathematics 6 Online
OpenStudy (anonymous):

Factor and simplify :cot^4 + 2 cot^2 + 1

OpenStudy (joannablackwelder):

Do you know how to factor quadratics?

OpenStudy (anonymous):

Ax^2 + Bx + c =0

OpenStudy (joannablackwelder):

That is the form, can you factor it into a product of binomials?

OpenStudy (anonymous):

like (cot^4x) + 1(cot^2x) ?

OpenStudy (anonymous):

nvm that's grouping

OpenStudy (joannablackwelder):

No, like x^2 +2x + 1 = (x + 1)(x+1)

OpenStudy (anonymous):

yess

OpenStudy (joannablackwelder):

So, in this problem, cot^2x is like x in my factorization.

OpenStudy (anonymous):

(cot^2 x +1)(cot^2 x +1)

OpenStudy (joannablackwelder):

Awesome!

OpenStudy (joannablackwelder):

But that can still be simplified...

OpenStudy (anonymous):

(cot^2 x)^2

OpenStudy (joannablackwelder):

Don't undo all your hard work! Do you recognize cot^2x + 1?

OpenStudy (anonymous):

yess

OpenStudy (joannablackwelder):

:)

OpenStudy (anonymous):

yeah but the answers to choose from are Tan^4x CSC^2x sec^4x Csc^4x

OpenStudy (anonymous):

and that equals csc^2x

OpenStudy (joannablackwelder):

One of them does, but we had 2.

OpenStudy (anonymous):

so csc^4x

OpenStudy (joannablackwelder):

yup

OpenStudy (anonymous):

yayyyy

OpenStudy (joannablackwelder):

:)

OpenStudy (joannablackwelder):

These kind take practice. Keep at it!

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