```Question 12560
Well, going from:{{{(3x-4)(2x+1)}}} to {{{6x^2-5x-5}}} is a whole lot easier than going the other way.

Here are some things you might try on factoring {{{6x^2 - 5x - 4}}} for example:

1)  Look at the first term {{{6x^2}}} and ask...how can I factor {{{6x^2}}}?
There aren't that many ways to factor it.
{{{(6x)(x) = 6x^2}}}
{{{(2x)(3x) = 6x^2}}}
{{{(-6x)(-x) = 6x^2}}}
{{{(-2x)(-3x) = 6x^2}}}

Now look at the last term {{{-4}}} and go through the same process.
{{{(-1)(4) = -4)}}}
{{{(1)(-4) = -4}}}
{{{(2)(-2) = -4}}}
{{{(-2)(2) = -4}}} This is really the same as the one above.

Now, keeping in mind that you want -5x as the middle term, you can try a pair of factors from the first group with a pair of factors from the second group.

For example: Try (6x - 1)(x + 4) and you see, almost right away that, while you get the first and last terms of your quadratic, you do not get -5x as the middle term. The same with (6x + 1)(x - 4) and (6x + 4)(x - 1).
Now try (3x + 1)(2x - 4) again, you don't get the middle term of -5x, so try (3x - 4)(2x + 1) Now you get a middle term of -5x from 3x - 8x = -5x.

So, these are the required factors.

It's a little harder when, as in this case, the coefficient of the x^2 term is greater than 1. Indeed, in some cases, you may not be able to factor the quadratic at all. But, you can always solve a quadratic equation by using the quadratic formula: {{{x = (-b +- sqrt(b^2 - 4ac))/2a}}}

I hope this is some help.  I have one math book on Intermediate Algebra that devotes some 36 pages to the topic of factoring polynomials.
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