# Factoring Trinomials...I think?

• March 9th 2010, 05:46 PM
burgessjkkb
Factoring Trinomials...I think?
Hello all. I've been working on this stuff all day...and I just thought to Google math forums, and ended up here! Well..there are a couple of problems that differ slightly from each other that I'm having problems with. The first one is...

12x²+34x+24

I believe the first step is to find the GCF, which is 6, and divide.

6x (2x+17+4)

-sigh- This is as far as I get though...a little help would be so very much appreciated.(Thinking)
• March 9th 2010, 05:53 PM
pickslides
Quote:

Originally Posted by burgessjkkb

I believe the first step is to find the GCF, which is 6, and divide.

6x (2x+17+4)

Correct, although you now have $6 (2x^2+17x+4)$ not $6x (2x+17+4)$ . The $x$ term is not common to take out.

To factor $2x^2+17x+4$ look for a solution (trial and error) such that $2x^2+17x+4 = (2x+a)(x+b)$ where a and b multiply to give 4.

This must also expand to give $17x$ as the middle term. Could take some time. Good luck!
• March 9th 2010, 06:15 PM
Prove It
First of all, $6 \cdot 17 \neq 34$.

In fact, $6$ does not go into $34$.

The highest common factor is actually $2$.

So $12x^2 + 34x + 24 = 2(6x^2 + 17x + 12)$.

Now, instead of using trial and error, find two numbers that multiply to become $6 \cdot 12 = 72$ and add to become $17$.

I think you'll find that the numbers are $9$ and $8$.

So $2(6x^2 + 17x + 12) = 2(6x^2 + 9x + 8x + 12)$

$= 2[3x(2x + 3) + 4(2x + 3)]$

$= 2(2x + 3)(3x + 4)$.