Smart way to deal with factoring quadratics? For some quadratics such as:
$$10x^2+89x-9$$
I factoring by making the observations $10=(1)(10)=(5)(2)$ and $9=(3)(3)=9(1)$, then I try:
$$(10x ? 1)(x ? 9)$$
$$(10x ? 9)(x ? 1)$$
$$(10 x ? 3)(x ? 3)$$
$$(5x ? 1)(2x ? 10)$$
$$(5x ? 10)(2x ? 1)$$
$$(5x ? 3)(2x ? 3)$$
We got lucky , by our first guess,  we can get;
$$(10x-1)(x+9)$$
This is the way my Algebra $2$ teacher made me switch to because he said it's less magic and more factoring. This method is a bit hard to teach, or is it not in your opinion?
But in many cases I find myself going through a lot of possibilities with this method, sometimes to find that my quadratic cannot be factored.
Of course there is the method what multiplies to $-90$ and adds to $89$...this may take a little less thought then we proceed as:
$$10x^2+90x-x-9$$
And of course we may go straight to the quadratic formula...(if the question is to factor then, using the roots we can find a factored form).
If I'm taking an sat or some other timed test, what method would you suggest. And do you have a better method?
 A: How do you proceed with your method? I was taught your teacher's method 30 years ago but have switched to teaching along the lines of your method because it's nearly foolproof and less guess and check. However the current theory in pedagogy is that different methods are acceptable and even beneficial so making you switch methods isn't ideal. As to which is the fastest, I think they are about the same except with lots of practice you can eventually see the factors very quickly by your teacher's method. I don't think the effort practicing that is worthwhile unless it's for competition math though. 
A: Broadly, unless the factors "jump out at you", factoring is a pretty horrible approach to this. Best reserved for folks who simply cannot "get" any other method, and like most academic (and life things), a group of them does exist.
To factor, just solve it using the quadratic formula. You end up with:
x = root 1 and x = root 2
Not factored, eh? Let's say the roots are x = 3½ and x = 2¾. Using simple algebra, get expressions equal to 0, then multiply by the denominators in the fractions. For example:
x = 3½    move the 3½ to the left side
x - 3½ = 0    multiply each side by 2 (½'s denominator)
2x - 7 = 0

That's the first factor candidate. One step left regarding it and its brother, (4x - 11).
Mutliply the "x" terms in the potential factors. You get 8x in this example. Divide "a" of "ax^2" fame in the quadratic you were given to factor by the result of multiplying them, 8 in this case.
If the result is 1, you are done. You have your factors, (2x - 7) and (4x - 11) in the example.
If it is anything else, 3.24 say, you would simply show it leading the two factors you have:
3.24 * (2x - 7) * (4x - 11)

That's the most proper expression of it. Also, if the quadratic went down, not up (a < 0), this is when you catch that, putting a negative out front.
So, QUICK and easy, and for the SAT your work is not graded, only your answer. If handing it in to a teacher, do the work leaving some space "to the left and above" it and after you KNOW the result, fill the empty space with some maundering about with potential factors and such, don't be careless and write things that couldn't've led to your answer(!!!), be on point, but do it quickly and incompletely... you... didn't have all the room in the world to write so you had to do some of it in your head, eh?
Factoring, for people who can understand other methods and use them, is like the person who spends three minutes deciding what calculational "trick" to use to (cleverly?) solve some problem when 12 seconds was all it would have taken to solve it in the usual, standard way. Awesome if the factors leap out at you, boom, you're done. But a nightmare when you have 20 sets of possible factors, all of which have negative and positive possibilities, and a million combinations. Teachers use them to separate the men from the boys, the A's from the B's. Well, you want the A, right?
Finally, if you have to show every bit of work, well, you have to grunt through it. But doing this you at least know the right answer, without having to check each halfway likely possibility. You're just listing them as if you really evaluated each along the way but NOT spending all that time evaluating. So... faster.
All that said, like anything, there is a huge gain in gradually building a list of common quadratics that you know the factors for, you know, like 4x^2 + 4x + 1 = 0 is (2x + 1)^2 and that kind of thing. Then there's no need for smart ways, kludgy, hard to work ways, or any ways at all. You just write the answer and move on.
