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Very next question, no idea what to do...

I am suppose to factor $2\sin^2x + 3\sin x+1$ .

I figure this is pretty simple so I do $(2\sin x)(2\sin x)+3 \sin x+1$ .

For some reason this is incorrect (not sure why) and they give $(2\sin x+1)(\sin x+1)$ which I did verify gives me the original but I don't know how to get there.

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    I like math, I just wish I wasn't so terrible at it. I practice but I really need to learn to practice without getting stuck every other problem and pissed off.2011-06-15

3 Answers 3

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Have you tried multiplying out the book answer to see what you get? Again, the fact that you have $\sin x$ is not important: $2y^2+3y+1=(2y+1)(y+1)$. Your answer is not in a factored form, nor is it correct as $(2\sin x)(2\sin x)=4\sin^2 x$

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Adam:

Looking at: $2\sin^2x + 3\sin x+1$

Note that $\sin^2x$ means nothing more than $(\sin x)^2$, so you can rewrite your expression as $2(\sin x)^2 + 3(\sin x) + 1$. Then, as others have suggested, set $y = \sin x$ --> then substituting into the original expression, we have $2y^2+ 3y + 1$. Once you factor that, remember to replace every $y$ with $\sin x$, and you should have the factored expression that matches your book's answer.

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    Bingo! I think she said something like it. Thanks so much. :-)2013-07-10
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Think of $2sin^2(x) + 3sin(x) + 1$ as a whole function similar to $2x^2 + 3x + 1$ and see if it makes sense now.

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    Thanks I got it, unfortunately I got stuck on the very next problem.2011-06-15