# Trinomial factoring help

• Aug 4th 2008, 12:46 PM
blip911
Trinomial factoring help
Hello,

I have been learning how to factor trinomials and I am stuck..

2x^2+x-15 = (2x-5)(x+3)

3a^2-2a-5 = (3a-5)(a+1)

Fc^2+17c+14 = (fc+7)(c+2)

To factor these types of problems, is the only way real to just plug in factors and do a full test with foil?? I mean some of these could have 100 possible factor pairs.. do you have to plug them in and keep going until it works? Am I missing a shortcut?

thanks!
• Aug 4th 2008, 01:00 PM
Jonboy
when you start out learning factoring you normally do it the harder way.

but there's some logic with factoring.

given $\displaystyle ax^2 + bx + c$, if the a is greater than one, you can do what's called the AC method.

For $\displaystyle 2x^2{\color {blue}{+x}}-15$ you first multiply the a and c.

That's -30. Since the linear term has a coefficient of +1, you need to find two numbers that multiply to give you -30 and add up to 1.

That's 6 and -5.

So rewrite this as: $\displaystyle 2x^2 + 6x -5x - 15$

Notice how I keep the -5x near the -15 since they have a common term.

Pull out a greatest common factor between terms: $\displaystyle 2x(x + 3) -5(x + 3)$

Factor by grouping: $\displaystyle \boxed{(2x - 5)(x + 3)}$

I might've went too fast, but we'll explain something more detailed if you ask the question.
• Aug 4th 2008, 01:07 PM
blip911
Pull out a greatest common factor between terms: http://www.mathhelpforum.com/math-he...802f8691-1.gif

will you always get a similar as in the (X+3) in the http://www.mathhelpforum.com/math-he...802f8691-1.gif part? what happens if its different, or will it not be different?

thanks
• Aug 4th 2008, 01:22 PM
Pn0yS0ld13r
You can just use the quadratic formula to factor trinomials. You may find this easier.

For example, factor $\displaystyle 2x^{2}+x-15$.

$\displaystyle x= {\dfrac{-b \pm \sqrt{b^{2}-4ac}}{2a}} = {\dfrac{-(1) \pm \sqrt{(1)^{2}-4(2)(-15)}}{2(2)}}$.

So $\displaystyle x = -3$ or $\displaystyle x=\frac{5}{2}$.

Now take $\displaystyle x = -3$ and add 3 to both sides of the equation to get: $\displaystyle (x+3)=0$.

Now take $\displaystyle x=\frac{5}{2}$ and multiply both sides of the equation by 2 to get $\displaystyle 2x=5$. Then subtract both sides by 5 to get $\displaystyle (2x-5)=0$.

Multiplying the two quantities,
$\displaystyle (2x-5)(x+3)=0$

And you're done. :)
• Aug 4th 2008, 04:15 PM
Jonboy
Hey I never thought of that before! That was spectacular Pn0yS0ld13r
• Aug 4th 2008, 04:17 PM
Jonboy
Quote:

Originally Posted by blip911
Pull out a greatest common factor between terms: http://www.mathhelpforum.com/math-he...802f8691-1.gif