## Need help solving 9x^3 - 42x^2 - 24x

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### Need help solving 9x^3 - 42x^2 - 24x

9x^3 - 42x^2 - 24x

All I have so far is:

3x(3x^2 - 14x + 8)

oldgoat

Posts: 3
Joined: Fri Jul 20, 2012 7:03 am

oldgoat wrote:9x^3 - 42x^2 - 24x

All I have so far is:

3x(3x^2 - 14x + 8)

To learn how to factor quaddratics, please study this lesson. Thank you!

stapel_eliz

Posts: 1703
Joined: Mon Dec 08, 2008 4:22 pm

### Re: Need help solving 9x^3 - 42x^2 - 24x

Thank you. I'd already reviewed the page before coming here.

What I really need is a solution to the problem I posted.

I don't know if I solved it correctly.
oldgoat

Posts: 3
Joined: Fri Jul 20, 2012 7:03 am

oldgoat wrote:What I really need is a solution to the problem I posted.

I don't know if I solved it correctly.

I'm glad to hear that you've succeeded in doing the factoring. Now please post your fully-factored form, and we'll take a look!

Thank you!

stapel_eliz

Posts: 1703
Joined: Mon Dec 08, 2008 4:22 pm

### Re: Need help solving 9x^3 - 42x^2 - 24x

I did post it in the first reply. I needed to know if I was right or wrong and why. You're of no help.

I posted on yahoo answers and got a detailed explanation within 20 minutes.

Getting the run around was not helpful and I will not recommend this site to others in my math class.
oldgoat

Posts: 3
Joined: Fri Jul 20, 2012 7:03 am

### Re: Need help solving 9x^3 - 42x^2 - 24x

oldgoat wrote:I did post it in the first reply. I needed to know if I was right or wrong and why.

I only see your factoring "so far". Where did you put the work for the insides, like they gave you the link for?
buddy

Posts: 53
Joined: Sun Feb 22, 2009 10:05 pm

### Re: Need help solving 9x^3 - 42x^2 - 24x

When factoring 3x² - 14x + 8, you'll need to find two numbers that multiply to give you ac while adding to give you b. In this case, ac = 3(8) = 24 and b = -14. We'll want to look at negative factors of 24 since we need them to add up to a negative number:

-1 -24, sum = -25 so that's not it
-2 -12, sum = -14 bingo

Now we'll use those two numbers to break up the center term:

3x² - 14x + 8
3x² - 2x - 12x + 8

From here, group the first two terms together and pull out anything you can.

(3x² - 2x) - 12x + 8
x(3x - 2) - 12x + 8

Now do the same thing to the last two terms.

x(3x - 2) - (12x + 8)
x(3x - 2) - 4(3x - 2)

Note that we pulled out -4, so we have to make the last term -2. Now we pull out what we have in the parentheses to get:

x(3x - 2) - 4(3x - 2)
(3x - 2)(x - 4)

You can repeat this process with any trinomial. There are other ways to do it, but this tends to be the most efficient method. Good luck.

MrAlgebra

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