Factor! Not sure.

**Somatic** · Sep 29th 2009, 08:53 AM

Hello guys, I've stumbled across some problems in which I need to factor.

I'm pretty sure that its in quadratic form, but I'm not too certain on how to factor with numbers that have a power greater than 2.

This is problem number 1:

$18x^4y2 + 21x^5y - 6x^3y^2$

How would I do this even do this? Do I need to factor in trinomials?

Problem number 2:

$3x^3 - 12x^2 - 96x = 0$

I'm not good with these types of problems, don't I usually end up with 2 answers with these type of problems? Can someone give me a description on what does the 0 have to do with this. How would I start this, I would appreciate at least a step on what to do next? Please and thank you in advance.

::EDIT::
I have resolved the second problem.
I'm not sure what to do with this problem, is this a matrix? I don't know how to begin on this problem:

$\{\begin{array}{cc}8x-7y=-9 \\-3x+4y=2 \end{array}$

**masters** · Sep 29th 2009, 11:47 AM

Originally Posted by Somatic

Hello guys, I've stumbled across some problems in which I need to factor.

I'm pretty sure that its in quadratic form, but I'm not too certain on how to factor with numbers that have a power greater than 2.

This is problem number 1:

$18x^4y^2 + 21x^5y - 6x^3y^2$

How would I do this even do this? Do I need to factor in trinomials?

Problem number 2:

$3x^3 - 12x^2 - 96x = 0$

I'm not good with these types of problems, don't I usually end up with 2 answers with these type of problems? Can someone give me a description on what does the 0 have to do with this. How would I start this, I would appreciate at least a step on what to do next? Please and thank you in advance.

::EDIT::
I have resolved the second problem.
I'm not sure what to do with this problem, is this a matrix? I don't know how to begin on this problem:

$\{\begin{array}{cc}8x-7y=-9 \\-3x+4y=2 \end{array}$

Hi Somatic,

You need to factor out a common monomial factor from the group.

$18x^4y^2+21x^5y-6x^3y^2=3x^3y(6xy+7x^2-2y)$

The pair of linear equations you have there might represent a system in which you might want to find the intersection. What were your instructions?

**Somatic** · Sep 29th 2009, 11:55 AM

Thank you masters.

As for the other one, my instructions were to "solve for the given variables:"
So I think I just have to solve them. Do I just solve each equation like a regular equation or what?

**masters** · Sep 29th 2009, 12:24 PM

Originally Posted by Somatic

Thank you masters.

As for the other one, my instructions were to "solve for the given variables:"
So I think I just have to solve them. Do I just solve each equation like a regular equation or what?

No, you have to solve them as a system. Each equation has 2 variables, so you need both equations to solve for the common (x, y) that satisfies both equations.

Use a method called 'elimination'. We'll eliminate one of the variables by manipulating one or both of the equations so that when we add the equations together, that variable will add to zero.

(1) $8x-7y=-9$

(2) $-3x + 4y = 2$

Multiply (1) by 3 and multiply (2) by 8. This should eliminate the x variable.

(3) $24x-21y=-27$

(4) $-24x+32y=16$

Add (1) + (2)

(5) $11y=-11$

(6) $\boxed{{\color{red}y=-1}}$

Substitute this value for y into (1) and solve for x.

**Somatic** · Sep 29th 2009, 12:55 PM

Once again thank you very much, masters.

I'll remember to use the 'elimination' method. I solved it.

I substituted the y into (1) which made it:
$8x-7(-1)=-9$

From there it was easy. After solving this I got:
$x = -2$

So the solution to the system is:
$\{\begin{array}{cc}x = -2\\y = -1 \end{array}$

Thanks again, I appreciate it.

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