anonymous
  • anonymous
Someone please explain to me the different methods of factoring a trinomial when the leading coefficient isn't one. Ex: 6x^2+31x-30 = 0
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
You can use the quadratic formula.
anonymous
  • anonymous
\[\huge \frac{ -b \pm \sqrt{b^2-4ac} }{2a }\]
anonymous
  • anonymous
if 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?

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anonymous
  • anonymous
where \(\huge ax^2+bx+c\). No, you wouldn't as long as you know how to solve it.
anonymous
  • anonymous
Another approach:
anonymous
  • anonymous
earlier 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
  • anonymous
Another approach: factor it.
anonymous
  • anonymous
Though it may seem frustrating because it take so much time since it's a trial-and-error-method.
anonymous
  • anonymous
I'm just not a fan of math.
anonymous
  • anonymous
I'll use the quadratic formula to show to you that it's not hard. :)
anonymous
  • anonymous
So we have \(6x^2+31x-30=0\) where \(a=6\) \(b=31\) and \(c=-30\).
anonymous
  • anonymous
THE QUADRATIC FORMULA: \(\huge \frac{ -b \pm \sqrt{b^2-4ac} }{2a }\) PLUG IN THE VALUES. \(\huge \frac{ -31 \pm \sqrt{31^2-4(6)(-30)} }{2(6) }\) 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
  • anonymous
Now do the math. :)
anonymous
  • anonymous
Are you now okay with this or do you still need more help?
anonymous
  • anonymous
That works for me, thank you very much.
anonymous
  • anonymous
Anytime! Tell me what you got and I'll say if it's right or wrong. :)
anonymous
  • anonymous
-6, and 5/6. thank you very much.
anonymous
  • anonymous
You're right! :) You're welcome!

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