Although Noung's writeup shows a good way for a human to factor a trinomial, trying to program an algorithm using that method would be horribly inefficient. It would have to first factor a and c, and then test every possible combination of factors until it finds the correct one. Luckily, there is another method that is much more efficient:

Start with

ax²+bx+c

as before, but first split the x

term into 2 parts, so that the

product of those two

coefficients equals the product of a and c. An easy way to do this would be to

iterate through the

integers between either ±

sqrt(ac) or ±b, whichever is greater, since this will cover all possible values. If none fits this, its

factorization is

irrational or

imaginary, and

another method will be necessary to factor it. If a number is found, however, the result will look like this:

(ax²+b

_{1}x)+(b

_{2}x+c)

where b

_{1}b

_{2}=ac. Next, factor out everything possible from both sets of

parentheses. The

binomials left inside each set of parentheses will be the same:

px(rx+s)+q(rx+s)

where pr=a, ps=b

_{1}, qr=b

_{2}, and qs=c. Finally, the

distributive property gives you

(px+q)(rx+s)

###### Example:

6x²+5x-4

First, find 2 numbers with a

sum of 5 and a

product of -24. They happen to be 8 and -3. So split the x

term up like this:

(6x²+8x)+(-3x-4)

Then, factor a 2x out of the first

binomial, and a -1 out of the second:

2x(3x+4)-1(3x+4)

Now, use the

distributive property to

factor it:

(2x-1)(3x+4)

Using this method, you don't need to factor any numbers, as you do in the normal method, and the amount of numbers you need to test to find the correct ones is also much smaller.