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Mathematics 18 Online
OpenStudy (anonymous):

factor 2x^2+3x+11

OpenStudy (anonymous):

Its not rational, so pretty much put prime

OpenStudy (jhannybean):

You're going to want t complete the square with this one

OpenStudy (jhannybean):

@Kimes are you there?

OpenStudy (anonymous):

yeah sorry, but i got x(2x+3)+11

OpenStudy (anonymous):

Cannot be factored. A plot is attached.

OpenStudy (jhannybean):

The only thing you can do is use the completing the square method to put in a form to solve for x.

OpenStudy (anonymous):

Ah okay I got it, thanks

OpenStudy (jhannybean):

Begin with grouping terms with x in them. \[(2x^2+3x)+11\]\[2\left(x^2+\frac{3}{2}x\right)+11\]\[2\left(x^2+\frac{3}{2}x+\frac{9}{16}\right)+11 - \frac{9}{8}\]\[\boxed{2\left(x+\frac{3}{4}\right)^2+\frac{79}{8}}\]

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