reverse FOIL

[electrolyte23]

hello people.
can somebody explain the reverse FOIL process.
it's the process of factoring equations.
e.g. factoring x<2>-7x+6

thank you

[aches]

I don't use a mathmatical method to do it. If I remember correctly from my high-school math classes there is a formula you can use to figure out the roots. The way I do t is just technically trial and error.

For example let's take a look at your equation:
x<2>-7x+6

I break that down as a = 1, b = -7 and c = 6
then I setup (x - )(x - ) knowing that a = 1, thus the first value is just x

I figure out that both values have negative signs because c is positive and b is negative, thus deducing that the only way b can be negative is if both signs are negative.

I then take the last value 6, and figure out the factors of it, once I do that I check the factors to see which ones add up to the given number.

So we have 3 * 2 = 6, 1*6 = 6. Out of those two I know 1 and 6 will add up to 7 so thus I get
(x-1)(x-6)...

Obviously it's a trial and error type of method, but I've been using it for so long it only takes me a few seconds to figure out the values.

I forget the actual formula, as I'm sure someone else can tell you but I know it was something like b+-sqrt(-4ac -b)/2 or something of that sorts...

Sorry if I wasn't much help :-p

[electrolyte23]

thanks aches.
wasn't searching for a mathematical formula.

just a quick way to get through it!
cheers!

[deren]

hey I am not sure what u r looking for but I did something like that;

x<2
-7x+6<2 => -7x<-4 => 7x>4 => x>4/7
so that; 4/7<x<2

as I said I am not sure what I did but this is the easy way to find the "x"

[vbakre]

The formula is
{-b (+ or -) SQRT (b*b - 4 ac)}/2a
we use + or - because there are 2 roots.

But trial and error works better than using this formula

[yoda_ngen]

I like Aches method. It is a easy and quick way to solve quadratic equations. Let's take another example: x^2-5x+6=0 Here are the steps I follow

1) look at the last term (constant) of the equation. It is 6. Write down the factors of it. 2,3 ; 6;1

2) Look at the middle term. It is 5. the factors determined in the first step should add or subtract such that the answer is 5. Hence, 2+3 is 5 and 6-1=5

3) Now, here is you decide which factor to choose. The sign of last term is positive. So the two factors should be of the same signs. Both positives or both negatives. Since 2+3 are both positives. You choose them.

So the factors are (x-2) (x-3)=0

[olesk]

Is it safe to assume that one doesn't really need the formula for any problems on the GMAT? (It's been my impression that the problems where one could need it are always so simple that it's not really required).

Having endured too many years of math (which oddly enough seems to have little impact on my math scores ) this formula is something I've used so frequently that I don't think I'll ever forget it, and thus I think it's probably the best approach to me.

[CTG1983]

I like Aches method. It is a easy and quick way to solve quadratic equations. Let's take another example: x^2-5x+6=0 Here are the steps I follow

1) look at the last term (constant) of the equation. It is 6. Write down the factors of it. 2,3 ; 6;1

2) Look at the middle term. It is 5. the factors determined in the first step should add or subtract such that the answer is 5. Hence, 2+3 is 5 and 6-1=5

3) Now, here is you decide which factor to choose. The sign of last term is positive. So the two factors should be of the same signs. Both positives or both negatives. Since 2+3 are both positives. You choose them.

So the factors are (x-2) (x-3)=0

This is called "middle factoring"

[yngwie]

Here's how I do it....

Taking electrolyte's example: x^2-7x+6

The co-efficient of the first term is 1 (call this a)
The co-efficient of the middle term is -7 (call this b)
The last term (the constant) is 6 (call this c)

Find two numbers that when :
- multiplied equal the product of first and last terms
- added equal the second term

These two numbers will be among the factors of a*c

So in this case, find two numbers that when
- multiplied, equal 1*6 (product of first and last) = 6
- added, equal -7

Factors of a*c (1*6) are 2,3;1,6

-1 and -6 are the numbers that satisfy these two conditions since
-1*-6 = 6 and -1 + (-6) = -7.

So now we can write the equation as:

x^2-1x-6x+6
Take out the common factor x-1

x(x-1)-6(x-1)
= (x-1)(x-6) are the factors.

==================================
Another example: 3x^2 - 8x + 4 = 0

a=3, b=-8, c=4
Find 2 numbers that when multiplied equal a*c (that is 4*3 = 12)
and when added equal b (-8)

Factors of a*c are 4,3;6,2;12,1
Clearly, -6 and -2 are the two numbers we need
-6*-2 = a*c = 12
-6+(-2)= b = -8

Write the equation as:
3x^2 - 8x + 4 = 0
3x^2 -6x -2x + 4 = 0
3x(x-2) -2(x-2) = 0
(3x-2) (x-2) = 0