## 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+31x-30 = 0

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1. anonymous

You can use the quadratic formula.

2. anonymous

$\huge \frac{ -b \pm \sqrt{b^2-4ac} }{2a }$

3. 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?

4. anonymous

where $$\huge ax^2+bx+c$$. No, you wouldn't as long as you know how to solve it.

5. anonymous

Another approach:

6. 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?

7. anonymous

Another approach: factor it.

8. anonymous

Though it may seem frustrating because it take so much time since it's a trial-and-error-method.

9. anonymous

I'm just not a fan of math.

10. anonymous

I'll use the quadratic formula to show to you that it's not hard. :)

11. anonymous

So we have $$6x^2+31x-30=0$$ where $$a=6$$ $$b=31$$ and $$c=-30$$.

12. 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.) $$\huge x=$$$$\huge \frac{ -31 - {41} }{12 }$$

13. anonymous

Now do the math. :)

14. anonymous

Are you now okay with this or do you still need more help?

15. anonymous

That works for me, thank you very much.

16. anonymous

Anytime! Tell me what you got and I'll say if it's right or wrong. :)

17. anonymous

-6, and 5/6. thank you very much.

18. anonymous

You're right! :) You're welcome!