Posted by T.Gracken on November 07, 2002 at 15:05:24:
In Reply to: ? on Quadratic equation example. posted by Tracey on November 07, 2002 at 13:46:50:
: In the example you give Solve 3x^2 + x - 2, when you factor it, I thought you had to find the multiple that would multiply out to equal -2 and add together to get x. I thought this would be (3x+2)(x-1)
: Since -1 * 2 = -2 and 2 - 1 = 1 or in this case x. In the example you have this factored out to be
: (3x - 2)(x + 1) = 0 -2 * 1 = -2 but -2 + 1 = -1 which would be -x. Shouldn't the x be positive? Am I wrong ? If so can you tell me where?
the method you are examining is called the AC method (or at least that's what I call it). But you must consider the coefficient of x2. what you are doing only works if the number multiplied to x2 is 1.
so... the idea is: if you have an expression of the form
Ax2 + Bx + C then we can factor the expression (using integers) IF there are two (integer) factors of A*C whose sum is B (not x but its coefficient).
However, that does not mean the factos of A*C are what is placed in parenthesis when A is not 1. These factors are used to "re-write" the middle term.
So, A=3 and C=-2, thus A*C=-6 and we must find two numbers that multiply to -6 and add to 1 [that is B=1; the coefficient of x, not x]
3 and -2 are numbers that multiply to -6 and add to 1
so the expression 3x2 + x - 2 is rewritten as
3x2 + 3x - 2x - 2
this is done so that the method of grouping can be used.
group the first 2 terms: 3x2 + 3x
this can be factored as 3x(x+1)
now group the second 2 terms: -2x - 2
this can be factored as -2(x+1)
we now have 3x2 + x - 2, rewritten as
3x(x+1)-2(x+1)
the common factor (x+1) can then be factored out (using the distributive property) and we end up with
(x+1)(3x-2)
...or, a more common method is called trial & error (I call it guessing but it is usually faster [with a little practice]).
AND!!! remember you can always check the factored form simply by using the FOIL method to expand it.