Can I factor a quadratic w/ a neg. x^2?

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Can I factor a quadratic w/ a neg. x^2?

Mathematics
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Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!
you can, but its usually more effort than what its worth
it depends on the quad

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Other answers:

I'm trying to solve a limit question and it has two quadratics undera radical, so I have to simplify. \[2+ x - x^2\]
-x^2+x+2 -(x^2-x-2) -(x-2)(x+1)
thank you -- I still can't figure out my limit, but thanks
what is it?
Lim as X goes to -Sqrt (2\[\lim x \rightarrow -1 \sqrt{2+x-x ^{2 } /x ^{2+4x+3}?}\]
the denominator under the radical is x^2 +4x +3 I couldn't get it to read the right way in the editor. Thanks!
wow that looks crazy
\[\lim_{x \rightarrow -1} \sqrt{\frac{2+x-x^2}{x^2+4x+3}}\]
is that right?
yes! Its confusing to me
\[\lim_{x \rightarrow -1}\sqrt{\frac{-1(x^2-x-2)}{(x+3)(x+1)}}=\lim_{x \rightarrow -1}\sqrt{\frac{-1(x+1)(x-2)}{(x+3)(x+1)}}\]
x+1 's cancel right! :)
\[\lim_{x \rightarrow -1}\sqrt{\frac{-1(x-2)}{x+3}}=\sqrt{\frac{-1(-1-2)}{-1+3}}=\sqrt{\frac{-1(-3)}{2}}=\sqrt{\frac{3}{2}}\]
\[\sqrt{\frac{3}{2}}=\frac{\sqrt{3}}{\sqrt{2}}*\frac{\sqrt2}{\sqrt2}=\frac{\sqrt{6}}{2}\]
yes -- got that great! Thats it! They have the answer as being sqare root of 6 /2. How do they get it in that form? Ah -- Thank you!!!!
lol you didn't see what i typed until you got to the thank part lol right?
Can I give you another tricky one?
yes of course

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