Anybody up for some BASIC algebra?

I've got a take home test for a math108 class and would much rather watch The Adventures of Pete and Pete than struggle through this for the next 3 hours... I've only got about 10 problems left that I don't know how to do, plz help me anonymii.

1) "Simplify the expression. Express the answer so that all exponents are posetive. Whenever an exponent is 0 or negative, we assume that the base is not 0.

-3x^-3 "

dude flip the base from x/1 to 1/x so you can write the exponent -3 as 3 instead.  The -3 at the front just hangs there like a putz until it climbs on top to multiply with the 1.

so... 1/(-3x^3) would be simplified completely?

2) "Factor (Prime if not factorable)

15x^4 +20x^2 +9x^2 -12"

Don't most of these problems usually go in a ^4,^3,^2 etc fashion? Type-o maybe? I dunnoz...

wait, the -3 belongs on the top because it is not part of the base, it is just a thing multiplying to the power expression.

For the next one, you could insert 0x^3 and 0x if you'd like. But wait, is that really two terms with x^2?  Those are like terms:  add those first.

When you have accomplished that, you will have a trinomial that you could pretend is the simpler form Ax^2 + Bx + C if that helps you figure out what you could do next.

Okay, for 1) I got -3/(x^3)

Yeah, I'm not sure if it is supposed to be 20x^3 or not... oh well, I guess I can only assume it is to the ^2

It's probably ^2, because what you get is factorable into a pair of binomials.  Uh, the steps to get further are many - do know about what they call the method of decomposition for factoring a trinomial into a pair of binomial factors?

alright, for 2) I've got (-5x^2+3)(-3x^2-4)... wut now?

That won't do it since if you multiplied that out, you'd get 15x^4 + 11x^2 - 12
You are sort of getting there, just not with that

15x^4 +20x^2 +9x^2 -12 should be
15x^4 +20x^2 -9x^2 -12

Oops...

oh, ha then you got it fine, though it is an unusual configuration of negatives in your factors.  There is a way to rewrite (-5x^2+3)(-3x^2-4) with less negatives and still get the same thing if you multiplied them out.
I got to get going now, tired.

What would the solutions be though? +/-sqrt(.6) rite?

>>12
right, as long as you're ignoring the complex roots

mkay, I watched a couple episodes of Pete & Pete...

3) "Perform the indicated operations and simplify the result. Leave the answer in factored form.

[x-(1/x^2)] / [x-(1/x^3)] "

wtf am i do?

>>14
multiply top and bottom by x^3, to get rid of the 1/x^3

[x^4-x^2]/[x^4-1]

then factor/simplify accordingly

I fail at subtraction

[x^4-x]/[x^4-1]

>>15 so, (-x^2)/-1 ?

>>16 I thought it looked weird, so -x/-1?

Someone try this...

[3x/(x^2-5x-36)] - [x-1/(x^2-16)] + [1/(x^2-13x+36)]

Just factor... I came up with some huge trinomial that I didn't even want to fuck with :'(

>>12
If you don't have an equation with = 0, or any equation for that matter, and you just have an expression, then all you needed to do was factor the expression.  Nothing to actually solve for in that case.

>>18
It won't factor like that because you can't cancel terms in that manner, example (a + b)/(a + c) does not equal b/c.  You have to do some factoring work on the top and on the bottom, then look for matching factors top and bottom that you can then cancel.  The top: x^4 - x has a common factor, so rewrite it, then you will see that one of the factors is a difference of cubes which has a way of factoring like this: a^3 – b^3 = (a – b)(a^2 + ab + b^2).  The bottom: you have a difference of squares which has a way of factoring like this: a^2 - b^2 = (a - b)(a + b).  Now because the first square is actually a power of degree 4, one of the factors will be another difference of squares of degree 2 that you can further factor.

>>19
You are probably meant to compress the expressing into a single fraction form.  First factor all the fraction bottoms.  (Notice that the second fraction has a difference of squares.)  You will then need to make the fractions have a common denominator so that they all have the same factors on the bottom, allowing you to compress the expression into a single fraction.  It is like when you have 1/6 + 1/8 and you have to multiply 1/6 by 4/4 and multiply 1/8 by 3/3 so that both fractions will have 24 on the bottom.  But for you, you have to multiply top and bottom of each fraction by a binomial factor.  Then you simplify the top of the fraction.

>>21


TINY DICK HE KNOWS HIS MATHAMATICKS

Don't call me gay, but I need some mary jay!

Marijuana MUST be legalized.