• Sep 6th 2007, 08:29 PM
echidnajess
Hi, I'm actually in a college precal2 class, but I think this is an alegbra issue...

I'm sure I'm just being an idiot and doing something completely stupid here, as I haven't taken a math class in over a year and have forgotten everything, but this is my problem:

When working on factoring something for a rational function problem, I can't seem to get the quadratic formula to work out properly.

ex. f(x)= -x^2 - x + 6

y= ( 1 +or- sqrt( -1^2 - 4(-1)(6) ) / 2(=1)
solving this gets me y= -3 or 2

which means the factors are (x+3)(x-2), right? But then multiplying those back together gets x^2 + x - 6, which is the same as the start, but with all the signs flipped... am I suppposed to be putting a negative out front, like -(x+3)(x-2)? I'm really confused... does this also mean that if the original function was something like 2x^2 - x +6, I'd need to put a 2 somewhere? I don't get it, and can't remember what I'm supposed to do...

I'm sure this is just a stupid simple mistake, but I'd really like some help, I'm just frustrating myself at the moment.... Thanks!
• Sep 6th 2007, 10:27 PM
CaptainBlack
Quote:

Originally Posted by echidnajess
Hi, I'm actually in a college precal2 class, but I think this is an alegbra issue...

I'm sure I'm just being an idiot and doing something completely stupid here, as I haven't taken a math class in over a year and have forgotten everything, but this is my problem:

When working on factoring something for a rational function problem, I can't seem to get the quadratic formula to work out properly.

ex. f(x)= -x^2 - x + 6

y= ( 1 +or- sqrt( -1^2 - 4(-1)(6) ) / 2(=1)
solving this gets me y= -3 or 2

which means the factors are (x+3)(x-2), right? But then multiplying those back together gets x^2 + x - 6, which is the same as the start, but with all the signs flipped... am I suppposed to be putting a negative out front, like -(x+3)(x-2)? I'm really confused... does this also mean that if the original function was something like 2x^2 - x +6, I'd need to put a 2 somewhere? I don't get it, and can't remember what I'm supposed to do...

I'm sure this is just a stupid simple mistake, but I'd really like some help, I'm just frustrating myself at the moment.... Thanks!

$\displaystyle f(x)= a~x^2 + b~ x + c$

$\displaystyle x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

In your case $\displaystyle a=-1,\ b=-1,\ c=6$, so the roots are:

$\displaystyle x=\frac{1 \pm \sqrt{(-1)^2-4(-1)6}}{2(-1)}= \frac{1 \pm \sqrt{1+24}}{-2}=-3, 2$

Now if we form $\displaystyle (x+3)(x-2)$ the coefficient of $\displaystyle x^2$ is $\displaystyle 1$ so to get the original form back we always have to multiply through by the coefficient of $\displaystyle x^2$. In this case to get:

$\displaystyle f(x)= -x^2 - x + 6=-(x+3)(x-2)$

RonL
• Sep 6th 2007, 11:05 PM
ticbol
Quote:

Originally Posted by echidnajess
Hi, I'm actually in a college precal2 class, but I think this is an alegbra issue...

I'm sure I'm just being an idiot and doing something completely stupid here, as I haven't taken a math class in over a year and have forgotten everything, but this is my problem:

When working on factoring something for a rational function problem, I can't seem to get the quadratic formula to work out properly.

ex. f(x)= -x^2 - x + 6

y= ( 1 +or- sqrt( -1^2 - 4(-1)(6) ) / 2(=1)
solving this gets me y= -3 or 2

which means the factors are (x+3)(x-2), right? But then multiplying those back together gets x^2 + x - 6, which is the same as the start, but with all the signs flipped... am I suppposed to be putting a negative out front, like -(x+3)(x-2)? I'm really confused... does this also mean that if the original function was something like 2x^2 - x +6, I'd need to put a 2 somewhere? I don't get it, and can't remember what I'm supposed to do...

I'm sure this is just a stupid simple mistake, but I'd really like some help, I'm just frustrating myself at the moment.... Thanks!

f(x) = -x^2 -x +6
is the function of x only.
It's graph is a parabola that opens downward.
That's all.

If you want to get the factors of f(x) = -x^2 -x +6, you cannot.
But you can get its zeroes.
Meaning, the values of x when f(x) = 0.

When f(x) = 0, you can now factor that.
So,
0 = -x^2 -x +6
That's when you can use the Quadrtatic Formula to find the x's when f(x) = 0.
That's when you find the factors too.

Now,
0 = -x^2 -x +6 ------------(1)
Transpose all of those in the righthand side of the equation to the lefthand side, and transpose the zero to the righthand side,
x^2 +x -6 = 0 ------------(2)

Whoa, (1) became (2)?
So, (2) = (1) ?

Yes, it is.

Again,
0 = -x^2 -x +6
Divide both sides by -1,
0 = x^2 +x -6

Eh?

2x^2 +6x +34 = 0 -------------(3)
Divide both sides by 2,
x^2 +3x +17 = 0 -------------(4)
Are (3) and (4) equal?
Try graphing them.

------------------------------------------
Edit:

If you want to get the factors of f(x) = -x^2 -x +6, you cannot.

I'd be crucified for that blasphemy.
• Sep 7th 2007, 12:54 AM
CaptainBlack
Quote:

Originally Posted by ticbol
------------------------------------------
Edit:

If you want to get the factors of f(x) = -x^2 -x +6, you cannot.

I'd be crucified for that blasphemy.

f(x) = -x^2 -x +6 = (-x-3)(x-2)

RonL