# Re: Easiest way of factoring?

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Posted by Jmiah on November 05, 2002 at 11:49:31:

In Reply to: Easiest way of factoring? posted by Lish on November 04, 2002 at 14:33:21:

: I am learning how to factor polynomials of the form ax^2+bx+c. The book explains two different ways to factor using trial factors and grouping. Which is the easiest? Does it matter? And do I use each one for solving a different sort of problem? Hope someone can clear this up for me. Thanks

Okay, this is how I do it. I assume that a b c are coeficients,(1 2 3).

So you have ax²+bx+c. Let's change that to
(1)x²+(3)x+2.
1. Now, to do grouping, you need to find the factors of the last number. Whenever there is only 1 x², the factors of the last term up to the coeficient of the middle term. For example, in x²+3x+2, the factors of 2 are 1 and 2 (1*2=2), and if you add them, it equals 3 (1+2=3).

2. Also, we know that in multiplication, a +*+=+, -*+=-, and -*-=+.

3. Lastly, since the first term is x², we know that the factors will each be x± something. S

Since all terms are positive, we know that the factors have to be x+ something, so from there it is easy.

(x+_)(x+_)
(x+1)(x+2)

You use trial factors for when you have a problem like 6x²+7x+2. Since there is more than 1 of the first term, the above solution won't work. Instead you have to figure out the factors of both the 1st and 2nd terms. This is how I do it.

6x²+7x+2
6x²=(2x*3x) OR (6x*1x)
2=1*2

From here I figure out which works. I already know that both will be (_x+_), so all I have to do is do the trials.

(x+1)(6x+2)=6x²+8x+2Nope

(x+2)(6x+1)=6x²+13x+2 Nope

(2x+2)(3x+1)=6x²+8x+2Nope

(2x+1)(3x+2)=6x²+7x+2 Bingo!

And that's more or less how I do it. Good luck, and remember that practice makes perfect!!

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