# Review Packet

1. Jul 22, 2008

### epkid08

I have this review packet and I have a couple questions.

1. The problem statement, all variables and given/known data
Solve the fallowing by factoring and making appropriate sign charts.

$$2sin^2(x)>sin(x), 0\leqX\leq2\pi$$

2. Relevant equations

3. The attempt at a solution
Well my main question is, can I divide both sides by sin(x)? In doing so, I got $$2sin(x)\geq1$$, and $$2sinx\leq1$$ because sin(x) can be both positive and negative. After solving I got $$x\leq\pi/6 \leq x$$, which seems pointless.

Okay tex isn't working, the last sentence should say, "x is less than or equal to pi divided by 6 which is less than or equal to x.

1. The problem statement, all variables and given/known data
Use p/q method to factor the polynomial, then solve for P(x)=0.

P(x) = x^3 - 6x^2 + 3x - 10

2. Relevant equations

3. The attempt at a solution
The p/q method doesn't seem to work here because there are two imaginary roots. How can I factor this?

I'll probably end up posting more. Thanks for the help

2. Jul 23, 2008

### yeongil

I think that when you divide by sin(x) you lose a root (or roots). Instead, subtract both sides by sin(x), factor the left side, and go on from there.

01

3. Jul 23, 2008

### HallsofIvy

The problem said "by factoring and making appropriate sign charts."

Look at 2sin2(x)- sin(x)= sin(x)(2sin(x)- 1)> 0. The "appropriate sign charts" refers to the fact that a product is positive as long as the factors have the same sign. What are the signs of sin(x) and 2sin(x)- 1 over various intervals? Where are they the same?

As for the last problem, if two of the roots are complex, then the third must be real: your polynomial can be factored into linear and quadratic factors. I suspect that the "p/q method" is what I would call the "rational root theorem": if a polynomial equation, with integer coefficients, has a rational root, p/q, the p must divide the constant term and q must divide the leading coefficient. Here the leading coefficient is 1 and the constant term is 10. That tells you that there are only 4 possible rational roots. Plug each of them into the equation and see if one satisfies it.

4. Jul 24, 2008

### epkid08

I am sure I checked all of the possible numbers at least twice, I guess I'll check again.

Edit: It is possible that the only real root left is irrational, so there is no way to factor it? (how would I find the root?)

5. Jul 24, 2008

### HallsofIvy

Yes, it is possible that the only real root of a polynomial is irrational. In that case, while it, strictly speaking, can be factored, it would be extremely difficult to find the real root and so factor. But are you sure you have copied the problem correctly?
x3- 6x2+ 3x+ 10 has a simple integer root.