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Algebra 2/trig questions

  1. Mar 5, 2009 #1
    1. The problem statement, all variables and given/known data
    1) how many roots does each polynomial have?
    A) f(x)=x^14 + x^10 -1
    B) g(x)=x + 25.4
    C) P(x)= x^99 + 1
    D) m(x)=x^4 + 5x^7 - x^9

    2) Use synthetic division to find P(2) if P(x)=x^3 - 3x^2 + 10x - 18
    3) Use synthetic division to determine if x-3 is a factor of P(x)=2x^3 -3x^2 - 11x +6
    4) a polynomial of degree 4 has roots -1(multiplicity of 2) and 2i. Also P(-2) = 32. Find the polynomial in standard(multiplied) form.
    5) what is teh remainder when P(x)=x^100 - x^50 + 10 is divided by x+1? explain how you know.
    6) Explain why it is nto possible for a polynomail of degree 3 having integer coefficients to have solutions of i, 2i, and 3i.
    7) List all possible rational roots of the function P(x)=4x^3 - 5x + 9
    8) Find all solutions of P(x)=x^4 +x^3 -3x^2 + 8x + 20

    3. The attempt at a solution
    I will update this as I answer more and more questions:
    1) A) is this 2? I thought i would take [tex]\pm[/tex]1 over [tex]\pm[/tex]1 and that would be 2 solutions.
    B) if what i did was right would this be [tex]\pm[/tex]25.4 over [tex]\pm[/tex]1 which would give me 2 roots?
    C) waiting for a and b
    D) waiting for a and b

    2) Just do synthetic division using 2?
    I got x^2-x+8 R=-2

    3) I would use 3 to divide that equation, and if i got a remainder of 0 then it would be a factor?

    4)not sure how to do this one, all i got so far was (x+1)^2(x-2i)(x+2i)

    5) would I divide by -1?

    6)not sure exactly but i think i have an idea. when u have i, 2i, and 3i, you also have -i,-2i, and -3i. So then you cant have a polynomial of degree 3
    Is that right?

    7)[tex]\pm9[/tex] over [tex]\pm4[/tex] right? then list out all the roots?

    8)list all the possible rational roots, then use synthetic division to get the remainder to 0, and continue doing all that until you get to x^2 right?
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Mar 6, 2009 #2


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    It wasn't until I looked at your answer to (7) that I finally understood this! You are using the "rational root theorem". First you do NOT divide the constant term by the leading coefficient; you look at the factors of each and then divide. Any rational roots must be among those fractions but it does NOT follow that there MUST be that many roots. And, finally, this problem is not asking only about rational roots!

    Once again 25.4 does not satisfy this equation. And it is a linear equation for goodness sake!

    What course is this for? I would be inclined to use "DeCartes' rule of signs" but that seems too advanced if you honestly do not know how to solve linear equations.

    Good! Now what is the answer to the question?

    Yes. Do you understand why? What does "a remainder of 0" tell you?

    Okay, that gives the right roots. But (-2+1)^2(-2- 2i)(-2+ 2i)= 8. What do you have to do to make P(-2)= 32?

    No, not divide by -1. You could, just as said, divide by x+ 1, but it would be easier to find P(-1). If P(x)= (x+1)Q(x)+ r (so that P(x) divided by x+1 has quotient Q(x) and remainder r), what is P(-1)?

    WHY must you "also have -i, -2i, and -3i"? Why is it necessary to say "having integer coefficients"?

    No. First look at the possible factors of 9 and 4 and then look at fractions of those. That will give possible rational roots, not "all the roots".

    That will work as long as there are no more than 2 irrational or complex roots.
    Last edited: Mar 6, 2009
  4. Mar 6, 2009 #3
    sorry, but I have no idea what your talking about for number 1.

    2) would it be p(2)=-2?
    3) got number 3
    4) im guessing i stick in -2 for x and solve? If thats right what would i do after that?
    5) dont know how to find p(-1)
    6) lol I have no idea why, I just know that I do need that.
    7) Im pretty sure your wrong, or you might be doing it a different way because I looked over my notes and i wrote down a question and the answer:
    List all teh possible rational roots of P(x)=3x^3 + 8X^2 -10x + 6
    To solver all factors of constant/all factors of leading coefficent.
  5. Mar 6, 2009 #4


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    Then perhaps I don't understand what you are talking about! Please explain " I thought i would take 1 over 1 and that would be 2 solutions" and " if what i did was right would this be 25.4 over 1 which would give me 2 roots?"


    Solve for what?

    It's hard to believe you are serious. Replace x by -1 and do the arithmetic, of course.

    It's not a laughing matter. You are learning formulas but not learning mathematics. You need to focus more on learning why you do things.

    So your notes say exactly what I did: you look at fractions made from the factors of those coefficients, not just the coefficients themselves. How does that make me wrong?
  6. Mar 6, 2009 #5
    1) a) is it 14?
    4) never mind I read your question wrong, I dont know what I have to do to make p(-2)=32
    5) I got 10 is that right?
    6) Yea I guess im not learning math, blame my teacher (btw Im still in high school, dont know if that makes a difference). I just know that you when you have positive i you cant have negative i. Is that right?
    7) I got 1,3,9,1/2,1/3,3/2,9/2,3/4,9/4. And all of these are [tex]\pm[/tex]
    8)all right so i found that -2(multiplicity of 2) is an answer and i got to x^2-3x+5. Where do i go from there?
  7. Mar 6, 2009 #6
    1) got these
    2) got these
    3) got these
    4) is the answer
    4x^4 + 8x^3 + 20x^2 + 32x + 16
    5) i got the remainder 10
    6)dont have this one
    7) got this one
    8)all right so i found that -2(multiplicity of 2) is an answer and i got to x^2-3x+5. Where do i go from there?
    Last edited by a moderator: Mar 6, 2009
  8. Mar 6, 2009 #7


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    The product you had before was 8 at x= -2 and you wanted 32. Yes, multiplying by 4 is a good idea! Of course, it's easy to check itself, isn't it?

    Yes, but you still have to "explain how you know".

    It isn't really important that the coefficients by integers but that they be real numbers. Any polynomial with roots i, 2i, and 3i must be of the form a(x- i)(x- 2i)(x- 3i) for some real number a (a must be real because it will be the coefficient of x4). Multiply that out. Does it have real coefficients?

    So solve x^2- 3x+ 5= 0. That doesn't factor but you can use the quadratic formula.
  9. Mar 6, 2009 #8
    6) i dont think it does but I still dont understand. Sorry but we never learned the why, just the how.
    8) i got -2(multiplicty of 2) and 3[tex]\pm\sqrt{11}[/tex]/2
  10. Mar 6, 2009 #9


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    This cannot be more true.
    How can it be possible that the questions range in difficulty so much that linear equations and complex numbers / polynomials of large degrees are present?

    Do you understand this:
    [tex]a^2+b^2=a^2-i^2b^2=(a+ib)(a-ib)[/tex] ?
    The difference of 2 squares is the only way you are going to satisfy the fact that all the coefficients of the polynomial are real (integers). So what does this tell you about the polynomial having 3 roots: i, 2i, 3i ?
    Well, as hallsofivy has said, the polynomial can be written as [tex]a(x-i)(x-2i)(x-3i)[/tex] but if you are to ever expand this out and result in all real coefficients (you have reassured me that you probably didn't expand, or didn't know how to with "i dont think it does') you must have the complex conjugate of all the roots. For i, you need -i, for 2i, -2i etc.
    This means the polynomial with those roots must be [tex]a(x-i)(x+i)(x-2i)(x+2i)(x-3i)(x+3i)[/tex] but now this is a polynomial of degree 6, so it thus isn't possible.

    uh, no.

    The quadratic formula: [tex]x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
    try again, and don't change things to make it simpler or neater. Hint: complex roots.
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