Trinomials, a>1

Since trinomials with a > 1 come up when we're finding the roots,  I thought I'd offer an alternative way to factor these types of quadratic equations.  It's definitely something we can talk about in class too.

3x² - 8x + 4

You’re going to start by multiplying your a and c terms.  This product will become your new c term.

x² - 8x + 4(3) =  x² - 8x + 12

Then factor x² - 8x + 12

x² - 8x + 12

 

Using our X-factor, we can see that -6 and -2 multiply to get 12, and add together to get -8.  Since this is a trinomial in which a = 1, we can just insert the factors into their factored form:

(x – 6)(x – 2)

Our only problem is that our -6 and -2 still have that 3 in them (from when we multiplied our a term and our c term.  All we have to do is divide the 3 out of our factor terms:

-6/3  = -2                       -2/3  doesn’t simplify  - but that’s okay

(x – 2) makes sense intuitively.  That’s one piece of our factored form. 

To find the other piece of our factored form, we have to simplify that fraction

(3x – 2)

(x -  2/3)   to simplify the fraction, I simply multiply a 3 through…

(3x – 3(2/3)) = (3x – 2)

So now we have our two factored pieces:  (x – 2)(3x – 2)

If you still have your factoring polynomials guide, this is called the slip-and-slide method for factoring trinomials. 

Here's a video, too.  The method we learned in class is what he refers to as the A.C. method...

http://www.youtube.com/watch?v=yHj1-b9cUzk