[High School- Math] PreCalculus Review

Re: [High School- Math] PreCalculus Review

For number 2, use PEMDAS

1. ()
2. []
3. x or /
4. +
5. -

Re: [High School- Math] PreCalculus Review

1. Simplify:

__-3__
√a + 1

__-3__ √a - 1
√a + 1 √a - 1

Answer: -3√a - 1 or -3√a - 1
.................................1

2. Solve:

4x^4 + 4x^3 - 7x^2 - 8x - 2 = 0

(4x^4) + (4x^3) - (7x^2) - (8x - 2) = 0
(4x^4) + (4x^3) + (-7x^2) + (-8x + 2) = 0

Answer: (4x^4) + (4x^3) + (-7x^2) + (-8x + 2) = 0

Would probably need someone to confirm.. I got some bad habits with this too.
Edit: the lack of () and [] can pretty much turn this wrong.

Re: [High School- Math] PreCalculus Review

Let me add my input, since I finished AP Calc last year.

Problem 1:

By convention, you're not "supposed" to leave a radical in the denominator of a fraction, so by "simplify," it means to get rid of that radical there. So you multiply the numerator and denominator to get rid of it:

......-3(root(a + 1))...
(root(a+1))*(root(a+1))

The denominator then is just a+1, and the numerator becomes -3(√(a+1))

So you're left with

-3(√(a + 1))
.......a+1.......

I think you can simplify that further, but lemme make sure of that before I post anything else. EDIT: Nevermind, I can't seem to simplify that any further.

As for the second problem, you're going to need to factor. That's a quartic function, and from what my AP calc teacher told me, there is no equation to solve quartic functions. Quadratic functions have (-b+-√(b^2-4ac))/2a, cubic functions have something that's several pages long, but mathematicians haven't found a solution to quartic functions yet.

So yeah, you're going to need to factor. But, I was never formally introduced to factoring, so I haven't a clue what you would do there. I guess you could start by factoring out a 4 like this:

4(x^4 + x^3 - (7/4)x^2 - 2x - 0.5) = 0
x^4 + x^3 - (7/4)x^2 - 2x - 0.5 = 0

But I don't know where you'd go from there, or if you can even come to a solution from there.

Also, Meushi, your solution to problem 1 is wrong because you confused some positive and negative signs (edit: or maybe you just assumed something differently. I can't really tell). But the process was right. As for problem two, you can't combine the exponents of two numbers with the same base if they're being added, only if they're being multiplied. So you can't make 4x^4 + 4x^3 into 4x^7.

An example:

10^3 + 10^4 is not equal to 10^7. 10^7 is 10 million. 10^3 + 10^4 is 11 thousand.

Re: [High School- Math] PreCalculus Review

Thanks for the help. I tried to solve as much as I could on the homework, but there's a lot still that I don't get. I talked to my uncle and he said that if it was given as a review, then I should just talk to the teacher about everything I don't get, and she'll probably just help me out and then give me more practice.

Re: [High School- Math] PreCalculus Review

Did the teacher not give you one of the factors for #2?

Re: [High School- Math] PreCalculus Review

the first one depends on whether it is root(a+1) or root(a) + 1. given the notation you used, i assumed the 2nd... but it seems like others have assumed the 1st.

if it's the 2nd, then you'd use [root(a)-1] to factor, but Meushi still did it wrong... you can use FOIL and get a -1 for your denominator, and -3 * [root(a) - 1] for the numerator...

aka: (-3root(a) + 3) / (a - 1)

edit: Relambrien solved it for the root(a+1) denominator... i solved for root(a) + 1. and, from re-reading your initial post... you solved it the same way i did. in which case, your answer is right.

Re: [High School- Math] PreCalculus Review

2. The Rational Zero Theorem states that "If a polynomial has rational roots, then they will be fractions of the form (+/-) (factor of the constant term)/(factor of the leading coefficient)".

Therefore, the possible factors of this polynomial are +/- 2, +/- 1, +/- 1/2, +/- 1/4.

Using the Remainder theorem, which states that for any number [a] that satisfies p(a) = 0, then a is a root of the polynomial function p(x).

Let p(x) = 4x^4 + 4x^3 - 7x^2 - 8x - 2
Substituting +/- 2, +/- 1, +/- 1/2, +/- 1/4 into p(x),
p(2) = 50
p(-2) = 18
p(1) = -9
p(-1) = -1
p(1/2) = -7
p(-1/2) = 0
p(1/4) = 540/7
p(-1/4) = -1/4

From p(-1/2)=0, -1/2 is a root of the function p(x). Therefore, (2x+1) is a factor. Diving 4x^4 + 4x^3 - 7x^2 - 8x - 2 by (2x+1), we have:
p(x) = (2x + 1)(2x^3 + x^2 - 4x - 2)
Factoring:
p(x) = (2x + 1)[ x^2 (2x + 1) - 2(2x + 1) ]
p(x) = (2x + 1)(2x + 1)(x^2 - 2)

The rest is easy:
p(x) = 0
(2x+1)^2 * (x^2 - 2) = 0
x = -1/2 , sqrt(2) , -sqrt(2)

Re: [High School- Math] PreCalculus Review

always rationalize the denominator

makes **** easier later anyway

Re: [High School- Math] PreCalculus Review

Truth. My math teachers couldn't reiterate that enough to me. It's rather simple to do.

Re: [High School- Math] PreCalculus Review

No, she didn't. But I had the class today, thankfully she graded the homework based on the fact that we at least attempted it, so I didn't lose points. She went over all of the problems, then gave us a copy of her work, so we could see how it was done. I think I understand it now, but thanks for the help guys.

Edit: I had the first problem done correctly. I guess I actually do remember a few things from precalc, so that's good.