Question 901776
Good day

I have something puzzling me for quite a while now

Let't take an example 

In the quadratic equation

1.  72c2 + 24c - 16

The answer is (12c - 4) (6c + 4)

Firstly.. Why 12 and 6? why not something like 9 and 8
Tried the formula with this example numbered as "1."

Secondly.. How do I know from the equation "72c2 + 24c - 16" which bracket is going to get the "+" and which bracket is going to get the "-"

They explained it in school but I just can't remember the principles.

Kind Regards.

Stefan.
<pre>
{{{72c^2 + 24c - 16}}}
The first thing, when factoring this trinomial is to factor out the GCF, which is 8. 
Thus, {{{72c^2 + 24c - 16}}} becomes: {{{8(9c^2 + 3c - 2)}}}, after which, {{{9c^2 + 3c - 2}}} should be factored.
Final factors: {{{highlight_green(8(3c + 2)(3c - 1))}}}

It's confusing when c is the variable in the trinomial, so it's better to 
change {{{72c^2 + 24c - 16}}} to {{{72x^2 + 24x - 16}}}....same thing, just that the variable was changed from c to x.
Without factoring out a GCF, and applying the "ac" method, we would need two factors with a product
of - 1,152 (a * c, or 72 * - 16), and that SUM to "b" (+ 24). These factors are: + 48 and - 24.
We now replace + 24x in the trinomial, {{{72x^2 + 24x - 16}}} with + 48x - 24x. I hope you're following!!

Now, {{{72x^2 + 24x - 16}}} becomes: {{{72x^2 + 48x - 24x - 16}}}. At this point, I'd switch the variable, x back to c,
so we now have: {{{72c^2 + 48c - 24c - 16}}}

The factors are now obtained by grouping the first two binomials, and then the last two binomials, so that results in: 
{{{24c(3c + 2) - 8(3c + 2)}}}, and the final answer: {{{(24c - 8)(3c + 2)}}}, which is the same as your factors: {{{(12c - 4)(6c + 4)}}}.
Factoring these further, results in: {{{highlight_green(8(3c - 1)(3c + 2))}}}, the CORRECT factors.

However, {{{(24c - 8)(3c + 2)}}} and {{{(12c - 4)(6c + 4)}}} are incorrect as the GCF of the original polynomial should be obtained
and factored out first. I only went this far to explain the procedure to determine the correct factors when
leading-coefficient multiples such as 24, 36, 72, and others are part of polynomials that need to be factored.
These and other leading-coefficient multiples pose a problem at times since they have as many as 4 or 5 sets of
factors, and can be pretty tedious to factor correctly.