# A really cool trick to help factoring polynomials...

• Oct 18th 2012, 05:03 PM
Nervous
A really cool trick to help factoring polynomials...
I don't know about you guys, but I usually have trouble with trinomials such as
$\displaystyle 6x^2+17x+5$
Specifically because I have to use combinations of the factors of 5 and of 6, not to mention the order I use the factors in.

Now, here's my little trick. (And all I can say is that no one ever taught me this in school, I don't know about anyone else, but I come from a pretty uneducated county.)

First, take the factors of the first and last digits, that part is a given.

$\displaystyle Factors of 5, \alpha = {1,5}$
$\displaystyle Factors of 6, \beta = {1, 2, 3, 6}$

Next, find all the products:

$\displaystyle Products of 1 and \beta= {1, 2, 3, 6}$
$\displaystyle Products of 5 and \beta= {5, 10, 15, 30}$

Now, find the two that add to 17:
$\displaystyle 2+15=17$

So, now we know the answer involves {1, 5} and {2, 3}:

The answer turns out to be:
$\displaystyle 6x^2+17x+5=0$
$\displaystyle (2x+5)(3x+1)$

So, did anyone else know this, or am I totally awesome?
• Oct 18th 2012, 07:33 PM
topsquark
Re: A really cool trick to help factoring polynomials...
Quote:

Originally Posted by Nervous
I don't know about you guys, but I usually have trouble with trinomials such as
$\displaystyle 6x^2+17x+5$
Specifically because I have to use combinations of the factors of 5 and of 6, not to mention the order I use the factors in.

Now, here's my little trick. (And all I can say is that no one ever taught me this in school, I don't know about anyone else, but I come from a pretty uneducated county.)

First, take the factors of the first and last digits, that part is a given.

$\displaystyle Factors of 5, \alpha = {1,5}$
$\displaystyle Factors of 6, \beta = {1, 2, 3, 6}$

Next, find all the products:

$\displaystyle Products of 1 and \beta= {1, 2, 3, 6}$
$\displaystyle Products of 5 and \beta= {5, 10, 15, 30}$

Now, find the two that add to 17:
$\displaystyle 2+15=17$

So, now we know the answer involves {1, 5} and {2, 3}:

The answer turns out to be:
$\displaystyle 6x^2+17x+5=0$
$\displaystyle (2x+5)(3x+1)$

So, did anyone else know this, or am I totally awesome?

You are awesome! For the record, this is called the "ac" method. Coming from the general form of the quadratic
$\displaystyle ax^2 + bx + c$

Thanks for sharing.

-Dan
• Oct 18th 2012, 08:10 PM
Prove It
Re: A really cool trick to help factoring polynomials...
Quote:

Originally Posted by Nervous
I don't know about you guys, but I usually have trouble with trinomials such as
$\displaystyle 6x^2+17x+5$
Specifically because I have to use combinations of the factors of 5 and of 6, not to mention the order I use the factors in.

Now, here's my little trick. (And all I can say is that no one ever taught me this in school, I don't know about anyone else, but I come from a pretty uneducated county.)

First, take the factors of the first and last digits, that part is a given.

$\displaystyle Factors of 5, \alpha = {1,5}$
$\displaystyle Factors of 6, \beta = {1, 2, 3, 6}$

Next, find all the products:

$\displaystyle Products of 1 and \beta= {1, 2, 3, 6}$
$\displaystyle Products of 5 and \beta= {5, 10, 15, 30}$

Now, find the two that add to 17:
$\displaystyle 2+15=17$

So, now we know the answer involves {1, 5} and {2, 3}:

The answer turns out to be:
$\displaystyle 6x^2+17x+5=0$
$\displaystyle (2x+5)(3x+1)$

So, did anyone else know this, or am I totally awesome?

I knew this, but you are still totally awesome. If you want to write your algorithm more concisely, when you want to factorise \displaystyle \displaystyle \begin{align*} a\,x^2 + b\, x + c \end{align*}, multiply your a and c values. Then you need to look for two numbers that multiply to give \displaystyle \displaystyle \begin{align*} ac \end{align*} and add to give \displaystyle \displaystyle \begin{align*} b \end{align*}. Then break up your middle term according to these two numbers you have found, and factorise your quadratic by grouping.

As an example: \displaystyle \displaystyle \begin{align*} 2x^2 + 3x + 1 \end{align*}. Multiply your a and c values to give \displaystyle \displaystyle \begin{align*} 2 \cdot 1 = 2 \end{align*}. Then you need to look for two numbers that multiply to give 2 and add to give 3. Obviously they are 1 and 2. So breaking up the middle term gives

\displaystyle \displaystyle \begin{align*} 2x^2 + 3x + 1 &= 2x^2 + 1x + 2x + 1 \\ &= x\left( 2x + 1 \right) + 1 \left( 2x + 1 \right) \\ &= \left(2x + 1 \right)\left( x + 1 \right) \end{align*}