1. ## 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!

2. 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.

3. Pull out a greatest common factor between terms:

Factor by grouping:

will you always get a similar as in the (X+3) in the part? what happens if its different, or will it not be different?

thanks

4. 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.

5. Hey I never thought of that before! That was spectacular Pn0yS0ld13r

6. Originally Posted by blip911
Pull out a greatest common factor between terms:

Factor by grouping:

will you always get a similar as in the (X+3) in the part? what happens if its different, or will it not be different?

thanks
it might be different sometimes, you'll have to group the terms differently.