[__sarah__]

anybody know how to factor this?

3x^2 + 3x - 2 =0

thanks

[labman]

Make a table of all possible factors of 3X^2 and -2.

3X,X;-3X,-X
-2,1;2,-1

I can't find any combination of factors that will yield +3X. Not all expressions have intragel factors.

[CroCivic91]

... however, factorization IS possible.

All right, as english is not my prime language I can't talk well in english about maths, so try to follow me on this.

First of all, to see that the factorization IS possible, lets look at the graph of the function. It's minimal value is for x=-(1/2) and f(x):=3x^2 + 3x - 2
=> f( -(1/2) ) < 0. As a coefficient next to the "biggest" power is positive (3*x^2) that means that function grows when going to the left and right from the point x=-(1/2). Which means it intersects with the X at two points.

Now if we took a standard formula for solving equations of form:
a*x^2 + b*x + c = 0
which (formula) looks like:
x1 = ( -b + square root( b^2 - 4*a*c ) ) / 2*a
x2 = ( -b - square root( b^2 - 4*a*c ) ) / 2*a

and applied it to our polynom f(x)
we would get:
x1 = ( -3 + square root( 33 ) ) / 6
and
x2 = ( -3 - square root( 33 ) ) / 6

Now that means that f(x) = ( x - x1 ) * ( x - x2 ) * 3 ( 3 comes because of the coefficient next to the "biggest" power (3*x^2) )

End result:
3*x^2 + 3x - 2 = ( x + ( 3 - sqrt( 33 ) ) / 6 )*( x + ( 3 + sqrt( 33 ) ) / 6 )*3
= ( 6x + 3 + sqrt( 33 ) ) * ( 6x + 3 - sqrt( 33 ) ) / 12 = 0

As I said, not nice at all, but its right.

Kresho

[labman]

Kresho's answer is correct, but not easy to do. Most math teachers would accept the answer of no integral solution.

[CroCivic91]

That's right.

All problems of that sort have a solution in a set of complex numbers (numbers of type x + I*y) (which is often called set "C").
If we want a solution in a set of real numbers (numbers with no imaginary part) (which is often called set "R"), we cannot guarantee we will always have a solution.
If we want a solution in a set of integers (numbers like 1,2,3... or -1,-2,-3... ) (which is often called set "Z") we will have even less "solvable" problems.

Also, notice that set Z is a subset of set R and R is a subset of set C.

Well, that's just a little info, maybe not really useful.

Kresho