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anonymous
 one year ago
Someone please explain to me the different methods of factoring a trinomial when the leading coefficient isn't one. Ex: 6x^2+31x30 = 0
anonymous
 one year ago
Someone please explain to me the different methods of factoring a trinomial when the leading coefficient isn't one. Ex: 6x^2+31x30 = 0

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You can use the quadratic formula.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\huge \frac{ b \pm \sqrt{b^24ac} }{2a }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if i were to use the quadratic formula wouldn't I get a fairly large number under the radical though? isn't there some way to avoid that?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0where \(\huge ax^2+bx+c\). No, you wouldn't as long as you know how to solve it.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0earlier I had a question where the number under the radical ended up being something like 1137. In that scenario what route would I take?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Another approach: factor it.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Though it may seem frustrating because it take so much time since it's a trialanderrormethod.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm just not a fan of math.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'll use the quadratic formula to show to you that it's not hard. :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So we have \(6x^2+31x30=0\) where \(a=6\) \(b=31\) and \(c=30\).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0THE QUADRATIC FORMULA: \(\huge \frac{ b \pm \sqrt{b^24ac} }{2a }\) PLUG IN THE VALUES. \(\huge \frac{ 31 \pm \sqrt{31^24(6)(30)} }{2(6) }\) NOW DO THE ALGEBRA. \(\huge x=\)\(\huge \frac{ 31 \pm \sqrt{961+720} }{12 }\) \(\huge x=\)\(\huge \frac{ 31 \pm \sqrt{1681} }{12 }\) \(\huge x=\)\(\huge \frac{ 31 \pm {41} }{12 }\) WE KNOW THAT THIS EQUATION HAS TWO SOLUTIONS. 1.) \(\huge x=\)\(\huge \frac{ 31 + {41} }{12 }\) 2.) \(\huge x=\)\(\huge \frac{ 31  {41} }{12 }\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Are you now okay with this or do you still need more help?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That works for me, thank you very much.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Anytime! Tell me what you got and I'll say if it's right or wrong. :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.06, and 5/6. thank you very much.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You're right! :) You're welcome!
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