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Posted by Floorfiller on Dec-10-2003 02:46:

quick easy math question..

hey,

quick question...i'm trying to help my little brother out with some math homework and they are working on polynomials...and how to factor them. i've been in advance course and i'm totalling spacing the short cut way to factor...

i remember something like you divide the middle number by something and then square that to get the last term or something (sorry hopefully you'll know what i'm talking about ) anyway, now its really bugging me...whats that little trick????

Posted by Dmatrox on Dec-10-2003 02:50:

you mean the opposite of FOILing?

x^2 + 5x + 6

(x+2) (x+3)

like this?

Posted by Floorfiller on Dec-10-2003 02:52:

quote:
Originally posted by Dmatrox
you mean the opposite of FOILing?

x^2 + 5x + 6

(x+2) (x+3)

like this?

yeah, but there is a trick they teach you so that you can always find the zeros

in your case the zeros would be

(x+2) sero=-2

and

(x+3) zero=-3

but there is a trick so that you can find the zeros for really difficult ones...i just wanted to show him the trick...

Posted by Psionic on Dec-10-2003 02:53:

If you have a parabola and want to convert it to working form, you use "completing the square", which is what I think you are talking about.

Let's use this as an example:

y=x^2+4x-7

You bring the 7 over so it's:

y+7=x^2+4x

Next, take half of the coefficient in front of the 'x' and square it (thus, 4 divided by 2 is 2, then square it to get 4 again). This becomes your third term, and you'll need to add it to BOTH sides to equalize the equation. So now it's:

y+7+4=x^2+4x+4
y+11=x^2+4x+4

Now, to complete the square, you "unfoil" the polynomial, so it becomes (x+2)^2. So your final equation is:

y+11=(x+2)^2

To find the vertex of this parabola, you set whatever is with the x or y-coordinate equal to zero:

x+2=0 x=-2
y+11=0 y=-11

Therefore, the vertex (x,y) is (-2,-11).

I hope this helps!

Posted by Floorfiller on Dec-10-2003 03:05:

wait...i'm thinking of something different...you're right, but this is what i mean...

the equation is:

3x^2 + 7x + 2

now i of course can figure out that the factored version is:

(3x + 1)(x + 2)

but isn't there a sure fire way to do this to complicated ones...

Posted by Floorfiller on Dec-10-2003 03:09:

quote:
Originally posted by moncster
Use the quadratic formula to find the zeroes. They are your factors.

i know, but i'm looking for the short cut...i can do the quadratic...once you get into higher levels they teach you the shortcut so you don't have to do that crap everytime...

Posted by moncster on Dec-10-2003 03:10:

quote:
Originally posted by Floorfiller
wait...i'm thinking of something different...you're right, but this is what i mean...

the equation is:

3x^2 + 7x + 2

now i of course can figure out that the factored version is:

(3x + 1)(x + 2)

but isn't there a sure fire way to do this to complicated ones...

The form (ax + b)
in 3x^2 + 7x + 2.
b is one of the following {2 1}/{3 1}
Let's say the equation is
Ax^2 + Bx + C
Then b is one of these combinations
{factors of C}/{factors of A}

Basically, factoring would find you the zeros of the function. BUT remember, not all functions have zeros, therefore not all functions are factorable.

Posted by Psionic on Dec-10-2003 03:39:

Most equations have zeros, but they can also be imaginary or irrational. You would need to use quadratic formula.

Let's say you have the equation Ax^2+Bx+c=0.

Quadratic formula is as follows:

x=[-B+/-sqrt(B^2-4AC)]/2A

So let's say you have 4x^2+2x+9=0 as the equation. This is how you would solve for the zeros:

x=[-2+/-sqrt(2^2-4(4)(9)]/2(4)

x=[-2+/-sqrt(4-144)/8]

x=-2+/-sqrt(-120)/8

x=-2+/-i(sqrt120)/8

x=-2+/-2i*sqrt30/8

x=(-1+i*sqrt30)/4 and (-1-i*sqrt30)/4 are the roots, and they are imaginary.

Posted by moncster on Dec-10-2003 04:03:

Posted by Noisician on Dec-10-2003 04:44:

quote:
Originally posted by Floorfiller

but there is a trick so that you can find the zeros for really difficult ones...i just wanted to show him the trick...

for more difficult ones u can apply Bezout's theorem and Gorner's scheme to have them factored.

Posted by NY1004 on Dec-10-2003 08:22:

quote:
Originally posted by Floorfiller
wait...i'm thinking of something different...you're right, but this is what i mean...

the equation is:

3x^2 + 7x + 2

now i of course can figure out that the factored version is:

(3x + 1)(x + 2)

but isn't there a sure fire way to do this to complicated ones...

OK this is the shortcut I was taught:

-> Take the coefficient of the first term and multiply it to the last constant term so 3x^2 + 7x + 2 = x^2 +7x +6

->Then factor that out: (x+1)(x+6)

->Then put that coefficient (in this case the 3) back in front of the x so we get: (3x+1)(3x+6)

->Then reduce and we get: (3x+1)(x+2)

And vwoila!

Posted by TheFutureIsNear on Dec-10-2003 14:21:

Mm....

If you want to find the zers, on more complicated polynomials, do this

example = 3x^2 + 4x + 6
(a=3)[Coefficient of 3x^2] (b=4)[Coefficient of 4x] (c=6)[Lone #]

then plug it into this formula

The thing below this is square root
_____________
-(b^2) +-(plusminus)\/b^2 - 4(a)(c)

___________________________________
2(a)

Do that, and you find the zeros, if factoring cannot be done