General

[quote=nopaycheck][url=http://imageshack.us/photo/my-images/407/81102221.png/]And that is where you are wrong[/url][/quote]
Ty <3

Any ideas on #2 and #3?

for #3 I got -4x^3 + 200x but I can't find the x or the volume.

[quote=InSovietRussia]What I got was:

-2l 3, -1, -8, 12

- l -6, -10, -4
----------------------
3, 5, 2, 16.[/quote]

[url=http://imageshack.us/photo/my-images/407/81102221.png/]And that is where you are wrong[/url]

[quote=nopaycheck]Try again. With the first polynomial you got +/- 1 and 3 from the leading coefficent, and +/- 1,2,3,4,6 and 12 from the constant.
Start with the smaller numbers first and work your way up to the larger factors. It's busy work, but that's the only way to solve polynomials in pre-calc.

EDIT: NVM, me did math wrong. X = -2 is a solution.
Try using synthetic division again with X = -2. You shouldn't get a remainder.[/quote]

What I got was:

-2l 3, -1, -8, 12

- l -6, -10, -4
----------------------
3, 5, 2, 16.
EDIT: NVM I'M AN IDIOT.
Switched the -2 to +2, and it was all K.

Thank you.

*still stuck on #2 and 3 though

im in 8th grade IDK

[quote=InSovietRussia]For #1, I figured out the (x+2) myself. (x+2) = 0, x=-2
3(-2)^3 - (-2)^-8(-2)+12 = 0
and then when I divide it I get remainders...[/quote]
Try again. With the first polynomial you got +/- 1 and 3 from the leading coefficent, and +/- 1,2,3,4,6 and 12 from the constant.
Start with the smaller numbers first and work your way up to the larger factors. It's busy work, but that's the only way to solve polynomials in pre-calc.

EDIT: NVM, me did math wrong. X = -2 is a solution.
Try using synthetic division again with X = -2. You shouldn't get a remainder.

[quote=nopaycheck]Explanining how to do synthetic division in words will be pretty difficult.
[url=http://www.youtube.com/watch?v=bZoMz1Cy1T4]This will give a concise definition. 5min long[/url]
Edit: But if you aren't given an x-value to divide the polynomial from, like #2, you have to take the leading coefficent of the highest power and consider all of its factors, and same must also be done with the constant.
2x^4 + 3x - 17x^2 - 27x-9, meaning you have +/- 1 and 2 for the leading coefficent 2x^4, and +/- 1, 3 and 9 for the constant.
You have to plug in each value to see if the polynomial can be divided without having a remainder. If you try all of them and there is still a remainder, the polynomial cannot be factored.

You're welcome.

@ProBlades: Some schools teach it in pre-calc while others in Alg 2.
Frankly it's more useful to learn in pre-calc since you'll directly use it to find limits in introductory calc, where as if you learn it in Alg 2 you're just left to remember it on your own till you start introductory calc.[/quote]

For #1, I figured out the (x+2) myself. (x+2) = 0, x=-2
3(-2)^3 - (-2)^-8(-2)+12 = 0
and then when I divide it I get remainders...

I'll help you set up the synthetic division for Number [1].

1:

-2|3 -1 -8 12

Note: These are the coefficients of the polynomial. And I should assume you are familiar with synthetic division?

Explanining how to do synthetic division in words will be pretty difficult.
[url=http://www.youtube.com/watch?v=bZoMz1Cy1T4]This will give a concise definition. 5min long[/url]
Edit: But if you aren't given an x-value to divide the polynomial from, like #2, you have to take the leading coefficent of the highest power and consider all of its factors, and same must also be done with the constant.
2x^4 + 3x - 17x^2 - 27x-9, meaning you have +/- 1 and 2 for the leading coefficent 2x^4, and +/- 1, 3 and 9 for the constant.
You have to plug in each value to see if the polynomial can be divided without having a remainder. If you try all of them and there is still a remainder, the polynomial cannot be factored.

You're welcome.

@ProBlades: Some schools teach it in pre-calc while others in Alg 2.
Frankly it's more useful to learn in pre-calc since you'll directly use it to find limits in introductory calc, where as if you learn it in Alg 2 you're just left to remember it on your own for a year.

That's Algebra II, not Pre-calc.