the maffs

[Anonymous No.3370842]

Hi. I'm an idiot.

I'm taking a math class this semester. I missed class friday, and now have to do a whole slew of problems like this one:

12x^2 + 5x - 2 = 0

Can someone explain to me or hand me a helpful link to explain how I solve problems like this? I tried solving it like a normal equation, but that didn't work. (For this example, I ended up at x=-1/4x for instance, which is obviously wrong).

I really don't know what to do here, so anything you guys can offer would be appreciated.

[Community College !!Z5OU4C4Yj5W No.3370860]

(4x-1)(3x+2)=0

x=1/4, -2/3

[Anonymous No.3370868]

learn to factorise!

[Anonymous No.3370876]

http://www.youtube.com/watch?v=N30tN9158Kc&feature=relmfu

[Anonymous No.3370885]

Just apply http://en.wikipedia.org/wiki/Quadratic_formula#Quadratic_formula .

[Anonymous No.3371410]

Thanks for the replies. >>3370876 was especially helpful.

>>3370868
I'm trying, I am. Like I said, I'm an idiot. If I wasn't, I wouldn't be taking a class I don't get any credit for over the summer.

Anyway, I have one last thing to ask.

I'm nearing the end of my work for the night and I'm getting some really, really crazy shit. I don't know if I know how to do it or not, the stuff is just so daunting to look at I "feel" stuck. The one I'm looking at right now is:

(x^2 + x-56)(11x^2 - 8x - 3) = 0

I then FOIL it out to:

11x^4 - 8x^3 - 3x^2 + 11x^3 - 8x^2 - 3x - 616x + 448 + 168

And combine like terms to end up with:

11x^4 + 3x^3 - 11x^2 - 619x + 616 = 0

And I have no idea how to even start factoring something this complex.

[Anonymous No.3371421]

>>3371410
The Factor Theorem.

It's literally guess and check at this point.

Take the ending value that has no variable, and put it over the leading coefficient of the highest x term.
Then, use a calculator or your head to factorize them, one over the other.

Then, sub in all the values you got right there, plus or minus, until you get one that makes the equation = 0.

Then, do your long division/synthetic division to make it a cubic polynomial.

Then do that again until you get a quadratic.

Then, just factor or use the quadratic formula.

[Anonymous No.3371433]

not sure if trolling.

[Anonymous No.3371450]

all you need to do is factorize it. once it's factorized you can use the null factor law for anything in the form.
(x-a)(x-b)(x-c)(x-d)=0
then x=a,b,c,d

i'm assuming you either did extra derp math in high school or you are yet to finish school.

[Rebirth !NuJBxbKoek No.3371459]

>>3370842

Utilize the MP method.

MP stands for Master Product.

Step 1: You have your equation in the ax^2 + bx + c = 0 format. Multiply a and c. So, your product is ac.
Step 2: Find two factors of ac whereby when you add/subtract them, you obtain your bx.
Step 3: Rewrite your original equation in the format of ax^2 + ax + cx + c, whereby ax + cx = bx.
Step 4: Factor by grouping.

Let's do this with your question.

12x^2 + 5x - 2 = 0

So, 12 * -2 = -24. Factors of -24? Let's list them.

6, 4; 12, 2; 3, 8; 24, 1.

Okay, so let's analyze these.
6+4 = 10, and 6-4 = 2. So, it can't be 6, 4. Crossed out.
12 + 2 = 14, 12 - 2 = 10. Crossed out of the realm of possibilities.
8 + 3 = 11, 8 - 3 = 5. Oh, now we're getting somewhere. So, 8*(-3) = -24 and 8 + (-3) = 5. Perfect!

Rewritten, we see:
12x^2 - 3x + 8x - 2 = 0
3x(4x - 1) + 2(4x - 1) = 0
(3x + 2)(4x - 1) = 0

Solve for x. x = 1/4; x = -2/3

Hope I helped.

[Anonymous No.3371457]

>>3371421

[Anonymous No.3371468]

>>3371410
>(x^2 + x-56)(11x^2 - 8x - 3) = 0
>11x^4 + 3x^3 - 11x^2 - 619x + 616 = 0
DON'T EXPAND IT OUT! they are already quadratics in the brackets so you simply factorize them both as opposed to solving a quartic.

[Rebirth !NuJBxbKoek No.3371469]

>>3371421
>>3371433
>>3371450

The fuck are you guys doing? Why so complex? The MP method is fine for problems like these.

[Anonymous No.3371490]

>>3371459
> Step 3: Rewrite your original equation in the format of ax^2 + ax + cx + c, whereby ax + cx = bx.

wtf is this shit. a + c = b for every quadratic? no.

[Anonymous No.3371507]

>>3371468
THANK YOU!

It was just so long I couldn't "see it," even though I was looking right at it.

Thanks, guys!

[Anonymous No.3371515]

>>3371410
easymode

[Rebirth !NuJBxbKoek No.3371529]

>>3371490

Not for every quadratic, no. For these types, though? Yes.

Some quadratics can easily be solved utilizing the MP method or the basic factoring method whereby a=1 in the ax^2 + bx + c = 0 format. Others need the quadratic formula.

[Rebirth !NuJBxbKoek No.3371537]

>>3371490

Here: http://www.austincc.edu/jthom/MasterProductMethod.html

[Anonymous No.3371561]

>>3371529
i think i just realised it was a matter of you using horrid notation. in line 3 you should have written a' or something. the equation has been multiplied by a factor amirong?