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OpenStudy (anonymous):
Factor completely: 3x^2 + 5x + 1
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OpenStudy (anonymous):
Can it be factored?
OpenStudy (anonymous):
I don't think so because there aren't any like terms?
OpenStudy (anonymous):
This one doesn't factor, as far as I can tell. Using the ac method, there are no factorizations of three that add to five.
mathslover (mathslover):
\[\huge{x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}}\]
\[\huge{x = \frac{-5\pm\sqrt{25-12}}{2(3)}}\]
\[\huge{x=\frac{-5\pm\sqrt{13}}{6}}\]
the solutions are :
\[\Huge{x_1=\frac{-5+\sqrt{13}}{6}}\]
\[\Huge{x_2=\frac{-5-\sqrt{13}}{6}}\]
mathslover (mathslover):
wait
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OpenStudy (anonymous):
Not exactly a factorization yet...
mathslover (mathslover):
wait
OpenStudy (anonymous):
Wait, so it would be prime right?
mathslover (mathslover):
this means that :
\[\Huge{x-(\frac{-5+\sqrt{13}}{6})}\]
\[\Huge{x-(\frac{-5-\sqrt{13}}{6})}\]
mathslover (mathslover):
these are the factors
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OpenStudy (anonymous):
Do y'all need the answer choices?
OpenStudy (anonymous):
That would help.
OpenStudy (anonymous):
It could be shown as\[(x-x_1)(x-x_2)=3x^2+5x+1\]With x_1 and x_2 as above.
OpenStudy (anonymous):
It might be considered prime with respect to the integers or rationals, since the roots are irrational.
OpenStudy (anonymous):
A. (3x + 1)(x + 1)
B. (3x + 5)(x + 1)
C. (3x − 5)(x + 1)
D. Prime
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mathslover (mathslover):
Prime
OpenStudy (anonymous):
D is the only answer of those that makes sense.
OpenStudy (anonymous):
Yup D :).
OpenStudy (anonymous):
Thank you!
OpenStudy (anonymous):
You're welcome!
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