Incidentally in a week or two they will teach you the quadratic formula which takes all of the guess work out of factoring quadratics, and makes much of this moot. But they want to teach you this way first so you can get a feel for what the quadratic formula is doing. Sort of like teaching you to calculate 4*3 as 4+4+4 before they each you the multiplication table.
bolding mine
Finally! After 5 years of intermediate algebra, they will teach him the quadratic formula.
:smack:
Incidentally, if you can factorise a quadratic without recourse to the formula, it is often a lot less work to do so. If there is any doubt then you can check the discriminant b[sup]2[/sup] - 4ac and if this is a perfect square then the quadratic has integer factors.
Well, rational factors, at least. And it’s only “often” less work to do, if by that you mean “for the sorts of problems one is likely to encounter in an algebra textbook”. Across the entire universe of possible problems, it is far more often the case that the quadratic formula is the easiest method, even when other methods can be used. Now add in that the other methods can’t always be used, even in textbook problems. Overall, I would generally advise that if you can’t see the solution within ten or twenty seconds, just go ahead and use the quadratic formula.
Right; for example, consider 4x^2 - 1 = 0. Here, b^2 - 4ac = 4, a perfect square, but the roots are non-integers.
What we can say is that b^2 - 4ac is a perfect square if and only if ax^2 + bx + c can be factored into (integer coefficient) polynomials of degree <= 1. [Proof: being worked on by Chronos in the other thread. :)]
Oh, I see now Your Great Darsh Face was discussing “integer factors”, not “integer roots”. Thus, Your Great Darsh Face was perfectly correct, Chronos’s replacement with “rational” was a miscorrection, and my post was largely pointless. Sorry! :smack: