Question 23858
A quadratic funtion {{{ax^2+bx+c}}} can be factored as {{{(a)(x-r1)(x-r2)}}}, 
where r1 and r2 are the roots (solutions) to the equation {{{ax^2 + bx + c = 0}}}.

We can find r1 and r2, using the quadratic formula 
{{{(-b +- sqrt( b^2-4*a*c ))/(2*a)}}}.

Here, {{{r1 = (-3 + sqrt( 3^2 - 4*9*(-2))) / (2*9) = (-3 + sqrt(9+72)) / 18 = (-3 + 9) / 18 = 1/3}}}

and

{{{r2 = (-3 - sqrt( 3^2 - 4*9*(-2))) / (2*9) = (-3 - 9) / 18 = -2/3}}}

Therefore, our factors are:  {{{(9) * (x - 1/3) * (x + 2/3)}}}

What I have shown (creating a factored expression by finding the roots of a polynomial equation) is a brute-force method that will always work. However, you can often solve the problem by examination, by posing the question, "What are two numbers whose sum is b/a and product is c/a?" (It is a bit easier to do this if a = 1.) In fact, the usual reason for factoring a quadratic expression is that it gives us a shortcut to finding its roots (which are the negatives of those two numbers). If we can recognize the numbers that create our desired sum and product, we can avoid a somewhat laborious computation with the quadratic formula.

In this problem, 

{{{9x^2 + 3x - 2 = (3x-1)(3x+2) = 9(x-1/3)(x+2/3)}}}