Embarrasing, but I with this.

[Lancelot59]

I've forgotten how to factor quadratics.

Not all of them mind. I've just forgotten how to deal with these sorts:

-x²+4x-3

for example.

I forget how to deal with the negative coefficient in front of a.
I just fell out of doing this stuff and forgot over time.

 

[mgb_phys]

for ax^2 + bx + c = 0;

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

 

[Bohrok]

You could factor out the negative
-(x² - 4x + 3)
and factor what's in the parentheses, or look at the factors of -1 and -3 and find two pairs that add up to 4.

 

[Lancelot59]

Well I remember the quadratic formula. I also remember being taught a manual method of solving these, like seeing what factors of c add to get b. I just forget how to handle negative coefficients on the a term using that method.

EDIT: Your post just slipped in right before I hit post on this. I'll give it a try.

Ok, so do I just ignore the negative that I factored out entirely? I guess I could since the roots would still be the same...

 

[Lancelot59]

Ok, so do I just ignore the negative that I factored out entirely? I guess I could since the roots would still be the same...

 

[diazona]

If the polynomial is set equal to zero, then sure, you just ignore the negative. It's like multiplying both sides of the equation by -1.

If it's not set equal to zero... well, I'm not sure what you'd be trying to do :-/ If you're just trying to factor the polynomial (e.g. if a problem says "factor this polynomial"), you do need to keep the negative sign. You could either keep the sign out in front, or after you've factored the polynomial without the sign, just absorb the sign into one of the factors. So for example:
$$\begin{align*}-x^2 + 3x - 2 \\<br /> -(x^2 -3x + 2) \\<br /> -(x-2)(x-1)\end{align*}$$
That would be the answer if you just leave the sign out in front, or if you prefer to absorb it back in, either
$$(-x+2)(x-1)$$
or
$$(x-2)(-x+1)$$

 

[Lancelot59]

Alright, thanks for the help guys!