Question

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

Answer 1

You can use the quadratic formula.

Answer 2

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

Answer 3

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?

Answer 4

where \(\huge ax^2+bx+c\).

No, you wouldn't as long as you know how to solve it.

Answer 5

Another approach:

Answer 6

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?

Answer 7

Another approach: factor it.

Answer 8

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

Answer 9

I'm just not a fan of math.

Answer 10

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

Answer 11

So we have \(6x^2+31x-30=0\)

where \(a=6\)
\(b=31\) and
\(c=-30\).

Answer 12

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

Answer 13

Now do the math. :)

Answer 14

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

Answer 15

That works for me, thank you very much.

Answer 16

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

Answer 17

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

Answer 18

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