/sci/ - Science & Math

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Are you fucking retarded? Just divide the whole equation by the coefficient for X^2.

quadratic formula

(-b(+/-)sqrt(b^2-4ac))/2a

there you go

I think I know your problem. Firstly ignore the coefficient of x^2. Look at the constant, find two number which product equals that. Then simply (2x+1)(x+2). Just do loads of examples. What'd you get in C1 and C2 brah?

http://www.youtube.com/watch?v=Dz_szBawT90

Just see what two numbers when multiplied are the constant? Fucking easy bro. What'd you get in C1 and C2?

you're going to have problems in C4 if you don't get it down, try khanacademy.org

>>2614406
http://www.khanacademy.org/video/ca-algebra-i--factoring-quadratics?playlist=California%20Standards%
20Test%3A%20Algebra%20I

Consider the quadratic y=3x^3+12x+9

Of course, in this case it would be easiest to divide it all by 3, but I'm too lazy to think of an example for which there isn't a common factor greater than one, so deal with it.

So set it equal to 0.

0=3x^2+12x+9

Now take the coefficient on the x^2 term and multiply it by the constant and temporarily ignore the fact that there is a leading coefficient.

0= 3x^2+12x+27

So we know that 3 and 9 multiply 27 (since we just did it) and add up to twelve. Now recognize that this is essentially what happens to the b term in the quadratic. You get it by adding two numbers when you expand something like (x^2+3)*(x^2+1). Note that when you have a leading coefficient that one of the two numbers you're summing will be multiplied by that leading coefficient, which is why this works. Anyway, using that knowledge we can rewrite the quadratic:

0=3x^2+9x+3x+9

Now we factor by grouping, by which I mean break it up into two pieces and look for a common x+c term that can be pulled out of both:

3x^2+9x+3x+9
(3x^2+9x)+(3x+9)
Both contain (x+3)
3x(x+3)+3(x+3)
Pull out (x+3)
(x+3)(3x+3)
And there are your factors.

Do I see crippled formulas?