A really cool trick to help factoring polynomials...

I don't know about you guys, but I usually have trouble with trinomials such as

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.

Next, find all the products:

Now, find the two that add to 17:

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

The answer turns out to be:

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

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

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.

Next, find all the products:

Now, find the two that add to 17:

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

The answer turns out to be:

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

Thanks for sharing.

-Dan

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

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.

Next, find all the products:

Now, find the two that add to 17:

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

The answer turns out to be:

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 , multiply your a and c values. Then you need to look for two numbers that multiply to give and add to give . Then break up your middle term according to these two numbers you have found, and factorise your quadratic by grouping.

As an example: . Multiply your a and c values to give . 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