How can I solve these harder questions without a tutor?

[MadmanMurray]

I've been struggling to keep up with my college math course since I never finished high school math. I have a test on monday and the teacher gave us a practice test and said if we could get all the questions right we would get 100% in the real test. Unfortunately I could only do 3 of the questions. I'll list the questions I couldn't do at all. Any help at all would be greatly appreciated. I've been studying a fair bit but i can't seem to find info on how to do a lot of these harder questions.

Q1.) "log64 - log128 + log32"
I know that to substract those first 2 logs I'd write 64/128 but I don't know how to add the log32.

Q2.) "X + 1 and X - 2 are factors of X³ + "X² + AX + B. Find A and B and other factor."
I don't know where to start with this one

Q3.) "Solve 2X - 3Y =1 X² - 2XY - 3Y² = -3"
I think this is just a simultaneous equation. I don't know how to do it but I'll figure it out by reading a tutorial

Q4.) "expand (2 + square root of 3)⁵ answer as A + Bsquare root of 3"
I'm really lost here I was out the day we did binomial expansion

Q5.) "Find coefficient of X¹⁰ in expansion (2X - 3)¹⁴"
?

Q6.) "X³ - X² - 5X over X² - 3X + 2"
I think I know how to do this one. If I'm not mistaken I just factorize the denominator then split this fraction into partial fractions.

Q7.) "If roots of PX² - 4X + 1 are equal find P"

Q8.) "(K - 2)X² + 2X - K = 0 prove roots are real"


If anyone can help me figure out the method to solve any of these questions I would be extremely grateful. I have this test on monday so I have until then to figure all this out. I don't really mind failing this test because I'm going to get a math tutor to help me keep up with the class but I'm still going to try my best to pass this test.

 

[cristo]

1. You rightly say that loga-logb=log(a/b). Also note that loga+logb=log(ab).

2. You are given (x+1), (x-2) are factors, i.e. (x+1)(x-2)(x+C)=x^3+x^2+Ax+B. Can you obtain three equations from this which will enable you to solve for A,B and C?

3. Hint: substitute the linear equation into the quadratic.

4. You should read up on the binomial expansion.

5. Again, binomial expansion. Use pascal's triangle to obtain the required coefficient.

6. Your idea sounds correct (supposing the question is to simplify the fraction).

7. Hint: discriminant.

8. Presumably, same hint as 7.

 

[Borek]

Speaking about binomial expansion: this is not something completely different from everything you already know. While there are rules that make these calculations much faster, this is perfectly doable just by multiplying (x+y) several times. Tedious - yes. Time consuming - yes. Impossible - nah.

 

[jacksonpeeble]

I recommend memorizing all of the properties of logarithms, those should help you a lot! If you have any more problems, please don't hesitate to let me know, because I'm enrolled in a course covering the exact same things that you are!

 

[MadmanMurray]

Thanks for the replies but I'm still stuck with a lot of these. I can do 1.) now at least though.

Cristo:
2.) I honestly have no idea what to do here even with these factors. In this equation you gave me "(x+1)(x-2)(x+C)=x^3+x^2+Ax+B" I don't even know where you got the C from.

3.) I haven't learned how to do simultaneous equations fully yet. I know how to substitute simple equations but I've no idea how to substitute equations like these. I know that X = 1 + 3/2Y but trying to substitute that into the quadratic I'm lost.

The rest I'm reading up on right now.

Thanks jackson I need all the help I can get right now. I have a good tutorial on logarithms that the teacher recommended but I don't really have time to read the whole thing right now. I've grasped a fair bit of it though.

 

[cristo]

Thanks for the replies but I'm still stuck with a lot of these. I can do 1.) now at least though.

Good.

Cristo:
2.) I honestly have no idea what to do here even with these factors. In this equation you gave me "(x+1)(x-2)(x+C)=x^3+x^2+Ax+B" I don't even know where you got the C from.

Well, firstly you should recall that a cubic equation has three factors. You are given two: (x+1) and (x-2), so you can write the third as (x+C) (since the coefficient of x^3 is 1). Now, you can expand the left hand side of the equation and group in terms of x.

Now, to complete the question: what can you say about A and B is you have the equation x^2+3x+1=x^2+Ax+B?

3.) I haven't learned how to do simultaneous equations fully yet. I know how to substitute simple equations but I've no idea how to substitute equations like these. I know that X = (1 + 3Y)/2 but trying to substitute that into the quadratic I'm lost.

Ok. Firstly note my amendment to your above comment: the 1 must be divided by 2 also. Then, just literally put the equation for x into the quadratic:

[(1+3Y)/2]^2-2[(1+3Y)/2]-Y^2=-3

and expand and simplify. This will give you a quadratic in Y, which you should be able to solve.