Offering Math Help

[Jessica]

Hey, guys! I'd love to help you guys with your math questions. I'm a sophomore in college, majoring in mathematics, and I can try to help you up to questions in Differential Equations. If I can't, you will get my sincerest apologies, and I'll try to find something/someone who can help you.

[Maggie Finn]

Are you any good with Geometry?

[Jessica]

Yes.... I am very sorry it took me so long to reply. Do you still need help?

[Sky Rene Birdcreek]

I dunno if you're still doing this, but I sure hope you are! We're doing a lesson in my Algebra 2 Class called "Find Rational Zeros" which was fine, Because they gave us an X and we worked from there. Then they took away the nice answer consequently leaving me totally lost. D; So any help at all would be very much appreciated!

[Jessica]

Still helping! I'll be here for a while. Anyway, finding rational zeroes is not a difficult business at all, once you've gotten some practice. Say you have a basic function like:

 

ax^2+bx+c=0

 

All you are doing when you're finding rational zeroes is finding the values of x where the function will equal zero. You could plug and chug, but with longer polynomials, that can get to be rather impossible, or at the least tedious.

 

For the smaller polynomials, though, (like the one above) you have a variety of tools. One of which is the quadratic formula, which works for functions like the one above. If you're having trouble remembering it, sing the following to the tune of "Pop Goes the Weasel!"

 

X equals opposite b

Plus or minus the square root

Of b-squared minus 4-a-c,

All over 2-a!

 

I learned this trick as a junior in high school, and four years later, that tune still runs through my head every time I use it!

 

For something longer, this link will be helpful:

http://dlc.k12.ar.us/Resources/Mathematics/Algebra_II/FindingRation...

 

Let me know how it goes.

[Sky Rene Birdcreek]

We did actually learn that tune in class. And my teacher showed us videos of youtube with the Quadratic Rap! (: 

 

So if the question was f(x)=x^3-5x^2-22x+56 could I still do the quadratic formula? Or something different? Our teacher had the class do synthetic substitution, but one answer (x-2 or something) would be given. Now it's not. Which has thrown me off. I even went to the math book website. :/ Math is not my strong suit. 

[Jessica]

Excellent! So that will help you with polynomials that are quadratics. Quadratics, by the way, have a highest power of 2. You cannot use the quadratic formula for a polynomial like the one you asked about or longer.

 

Synthetic substitution is certainly one way to evaluate the polynomial, but you're right; it does help when you're given one of the zeroes. In the link that I showed you above, though, it shows you a method to find possible values for zeroes. When you would find the possible values, you would just use synthetic substitution for each of the possibilities. This can get long and tedious, but it's certainly better than trial and error.

 

So if all that is throwing you off is the fact that you're not given one of the zeroes, take a look at the link that I gave you above. Read over that, and if something confuses you, just let me know. :)

[Sky Rene Birdcreek]

Oh my god, I fee like a total spaz. I thought the number outside the synthetic division HAD to be 1 or -1! I didn't know I could move up numbers. This will be very helpful!

 

Thank you so much! 

[Jessica]

You are very welcome!

[Dave Sunny-O]

Are you any good with factoring or rational expressions?

[Jessica]

Sure. What's up?

[Dave Sunny-O]

The basic factoring with FOIL  is pretty simple and I understand the concept of that but then more types of factoring come into the chapter and now I'm pretty confused. One example is:

 

6xy - 5xz -6wy + 5wz

 

Since there's four variables, the answer would have one of each, I can't ever get the numbers to meld right though. Could you help me with finding what numbers to pair up with what variables in this situation? Thanks.

[Jessica]

Well, I can sure try.

 

Have you though of pairing it this way?

 

6xy - 6wy + 5wz - 5xz

 

Pull out a 6y from the first pair and a -5z from the second...

 

6y(x - w) - 5z(x - w)

 

You'll notice that you've got the same variables in the parentheses. So the 6y and the -5z (don't forget the negative sign) go in their own parentheses, and the (x - w) is its own parentheses.

 

(x - w)(6y - 5z)