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Math Help - x^4 + 4 - factor

  1. #1
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    x^4 + 4 - factor

    This is a "with the help of your classmates" problem. I am doing self study, so I am going to ask for a little help. According to a certain theorem, if a polynomial like x^4+4 has no real zeros, its factors are irreducible quadratics of the form (x^2 + ax + b)(x^2 + cx + d). We are then asked to apply essentially the same strategy we used with partial fraction decomposition to arrive at a system of equations which we can use to solve for a,b,c and d. That strategy is to take our largely symbolic quadratics, multiply them out, combine like terms, and the equate the coefficient terms (a,b,c and d as they appear naturally in the product's like terms) of the product with the coefficients of the original term that the quadratics are said to be factors of, namely, x^4 + 4. These equations then become our system of equations and should allow us to solve for a,b,c and d.

    Fine. If you multiply the two largely symbolic quadratics out and combine like terms, you get

    x^4 + cx^3 + dx^2 + ax^3 + acx^2 + adx + bx^2 + bcx + bd

    combining like terms, you express them as

    1 = 1 ?? (with fraction decomposition, you usually have a higher degree in the numerator so a higher degree in your symbolic terms, but not so here. This means there's no symbol for the coefficient of x^4.)<edit: this is a wrong or nonsensical comment--the coefficients of the symbolic expansion in fraction decomposition problems are equated to the coefficients of the *numerator* of the original term, which is often of a lower degree for some reason. The issue here is that x^4 provides no help for some reason>

    a + c = 0 (there is no x^3 term in the original term, so its coefficient is zero)

    a + b + c + d= 0 (acx^2 and bx^2 and dx^2 are the same as (a + b + c + d) * x^2 !?!)

    a + b + c + d = 0 (this and the one above are the same so only one need be included I know)

    bd = 4

    And that would be the system, but it is wrong because I looked at the answer (x^2 - 2x + 2)(x^2 +2x + 2). This means that a = -2, and b,c and d = 2. That doesn't hold up in the system above. Also, I don't know what to do with db = 4 in a matrix. That is if my matrix is now
    1 0 1 0 0
    1 1 1 1 0
    1 1 1 1 0
    ?? ?? ?? ?? 4

    how do I make sense of the last entry. <edited the last entry--bd = 4 is the same as the rest, b is a coefficient of 1, as is d, so (b + d) * 1 = 4 is the same as bd = 4, i hope> ack this isn't right. Ok so I still don't understand, how do I use bd = 4 in the matrix.
    Last edited by kjtruitt; February 11th 2014 at 04:49 PM. Reason: bd = 4 , in the matrix, is b + d = 4
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  2. #2
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    Re: x^4 + 4 - factor

    You can also use the web resource
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    Re: x^4 + 4 - factor

    Thanks that looks like a good site, but I wouldn't get any explanation without registering ($$?) and I'm not doing that just now. I would just like to know where I messed up setting up the matrix.
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    Re: x^4 + 4 - factor

    Hello, kjtruitt!

    \text{Factor: }\:x^4+4

    Add 4x^2 and subtract 4x^2.

    . . x^4 {\color{red}\,+\, 4x^2} + 4 {\color{red}\,-\, 4x^2}

    . . =\;(x^4 + 4x^2 + 4) - (4x^2)

    . . =\;(x^2 + 2)^2 - (2x)^2\;\hdots \text{(diff. of squares)}

    . . -\;(x^2 + 2 - 2x)(x^2 + 2 + 2x)

    . . =\;(x^2 - 2x + 2)(x^2 + 2x + 2)
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    Re: x^4 + 4 - factor

    Thanks alot Soroban! That completing the square trick is coming up alot lately for me. This question came at the end of the unit on matrices and systems and we were supposed to apply the steps I outlined in the question. I'm mostly interested in learning where I screwed up applying those steps. I think the idea was to show another application of matrices.
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  6. #6
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    Re: x^4 + 4 - factor

    Those middle two equations are incorrect.

    The x squared coefficient is d+ac+b and you can't write that as a+b+c+d, similarly the x coefficient is ad+bc and that isn't the same as a+b+c+d either.

    To solve the resulting equations, start with a = -c and substitute into the second and third equations.
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  7. #7
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    Re: x^4 + 4 - factor

    Doing it "the hard way":

    x^4 + 4 = (x^2 + ax + b)(x^2 + cx + d)

    = x^4 + (a+c)x^3 + (ac + b + d)x^2 + (ad + bc)x + bd

    \implies a+c = 0, ac + b + d = 0, ad + bc = 0, bd = 4

    Replace c with -a, and d with 4/b. This gives us 2 equations in 2 unknowns:

    -a^2 + b + \frac{4}{b} = 0

    \frac{4a}{b} - ab = 0

    Multiply both equations by b:

    -a^2b + b^2 + 4 = 0
    4a - b^2a = 0.

    The second equation can be written:

    (a)(4 - b^2) = 0.

    This tells us either a = 0, b = -2, or b = 2. If a = 0, then the first equation gives:

    b^2 + 4 = 0, which isn't good, if we're looking for REAL solutions. Let's try b = 2:

    Then the first equation becomes:

    -2a^2 + 4 + 4 = 0, or:

    4 = a^2. Again, we have 2 choices, 2 or -2.

    If we chose b = -2, the first equation gives:

    2a^2 + 8 = 0 which again makes no sense, so b = 2 is the only viable choice.

    It turns out that choosing either a = -2, or a = 2 will work, as both:

    x^2 - 2x + 2,x^2 + 2x + 2 evenly divide x^4 + 4 (I urge you to verify this by long division).

    Another possible approach:

    It is a fact that if an integral polynomial factors over the rationals, it factors over the integers. From bd = 4, we see that:

    b = 1, d = 4
    b = -1, d = -4
    b = 2, d = 2, or
    b = -2, d = -2.

    You could try each of these in turn to see which ones lead to viable solutions for a and c. For example, if d = 4b = 4 or -4, we have:

    ac + b + d = 0 implies ac = 5, or ac = -5. Since a + c = 0, this means we must have ac = -a^2 = -5, since 5 > 0. Since 5 is not a perfect square, a cannot be an integer.

    That leaves b = d = 2 or -2. Again, it is clear that we must have ac = -a^2 = -4 (so b + d = 4 is the only choice possible), so that a = -2, or 2.

    The savings in time here is because we can disregard "integers that don't work".
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    Re: x^4 + 4 - factor

    Doing it the easy way:

    Let y=x2 and factor y2+4: (y=(+/-)2i). Then x=(+/-)sqrt(y)
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    Re: x^4 + 4 - factor

    Actually, that wasn't the easiest way

    -4 = 4eitheta+2kpi, k=0,1,2,3,...

    then take (-4)1/4 to get four unique roots.

    On the road in a rush. It's in elementary section of complex variables in any calculus or algebra book.
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    Re: x^4 + 4 - factor

    Well, to be fair, you have to write out this solution "expanded in full":

    x^4 + 4 = (x - \sqrt{2i})(x + \sqrt{2i})(x - \sqrt{-2i})(x + \sqrt{-2i)

    If we re-arrange this like so:

    x^4 + 4 = (x - \sqrt{2i})(x + \sqrt{-2i})(x - \sqrt{-2i})(x + \sqrt{2i}))

    and multiply the first two factors out in C, we get:

    (x - \sqrt{2i})(x + \sqrt{-2i}) = x^2 + (\sqrt{-2i} - \sqrt{2i})x + \sqrt{-4i^2})

     = x^2 + \sqrt{2}(\sqrt{-i} - \sqrt{i})x + 2.

    If we pick "the usual square roots" (this is tricky, every complex number has 2, and there is no clear method of distinguishing between them):

    \sqrt{i} = \frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2}

    \sqrt{-i} = \frac{-\sqrt{2}}{2} + i\frac{\sqrt{2}}{2}

    we see that:

    \sqrt{-i} - \sqrt{i} = -\sqrt{2}

    so that:

    \sqrt{2}(\sqrt{-i} - \sqrt{i}) = \sqrt{2}(-\sqrt{2}) = -2 so that the first two factors actually give, when multiplied:

    (x - \sqrt{2i})(x + \sqrt{-2i}) = x^2 - 2x + 2.

    It is not clear that from the four complex linear factors we get, we actually get two of them to combine to a REAL polynomial with INTEGER coefficients, although this is indeed the case.

    From a more abstract view, every polynomial splits into linear factors in C, but it is not clear from this alone that we can factor it over some sub-ring of C, such as the real numbers, the rationals or the integers.

    It can be shown (using complex conjugation) that every EVEN degree polynomial does factor into quadratic factors over the reals, but again there is no guarantee that the coefficients will be rational (they usually aren't, unless the discriminant is a perfect square).
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    Re: x^4 + 4 - factor

    Thanks Deveno for the exhaustive reply. I took Bobp's heads up about my mistake (I know better, but I thought I was going to be making a matrix and i didn't know how ac would work in the matrix, so I was moved to wrongly break out acx^2 into (a + c) * x^2 to make it workable for what I know about matrices) and started working with substitutions and got c^2 = 4, meaning c = +/- 2. I think then one is compelled to look into each case for c. Looking at c = 2, and after alot of little equations I got b = d and eventually b = +/- 2. I decided eventually I'd prove that a = -2 and the rest were 2 if I tested each case over all the equations, which sounds alot like what you arrived at. I didn't have to work with two equations in two unknowns in my approach.

    c = -a
    -(a^2) + b + d = 0
    ad - ab = 0
    a(d-b) = 0
    d = b
    db = 4 so ...
    sqrt(4) = +/- 2, so d and b are either -2 or 2
    [a negative quantity or zero] + d + b = 0 : the quantity -(a^2) MUST be negative, so there's no way d and b can be -2
    So d = b = 2.

    a * c = -4 since a = -c and ac + (b + d, or 4) = 0

    -a(a, or c) = 4, sqrt(4) = +/- 2. Can we conclude at this point that either a = 2 and c = -2, or a = -2 and c = 2? I think so.

    I don't see how to determine which is which.
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    Re: x^4 + 4 - factor

    You still have to rule out the possibility that:

    a(d - b) = 0 because a = 0, you can't jump straight-ahead to b = d. But other than that, your solution is fine.

    Choosing a = 2 gives one factor, choosing a = -2 gives the other. It does not matter "which is which" because with polynomials:

    p(x)q(x) = q(x)p(x)
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    Re: x^4 + 4 - factor

    "You still have to rule out the possibility that:

    a(d - b) = 0 because a = 0, you can't jump straight-ahead to b = d."

    True--later it becomes clear since bd = 4 that b+d must be a positive value, and so
    ac + b + d = 0 means that ac must be a non-zero value (also negative) and as long as
    a doesn't equal zero, a(d-b) = 0 means d = b.

    Thanks for all the replies, very helpful.
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    Re: x^4 + 4 - factor

    I thought I would post the follow-up question. Factor x^4 + 6x^2 - 5x + 6.

    The equations are structured similarly because the same theorem still directs that, since there are no real solutions (I proved this by doing synthetic division on +/- 1,2,3,6) the factors are of the form (x^2 + ax + b)(x^2 + cx + d), which still expands to x^4 + cx^3 + ax^3 + acx^2 + bx^2 + dx^2 + adx + bcx + db. The equations are

    c + a = 0

    ac + b + d = 6

    ad + bc = -5

    bd = 6

    I tried substition and got a sixth degree polynomial in d equal to zero. I could have factored that, but I looked for another approach. I solved it by looking at the last equation and reasoning that it had to be 1 and 6, 6 and 1, 2 and 3, or 3 and 2. Then it becomes clear that a/c are going to have to be 1/-1 and from that that b is 6 and d is 1. I don't know what I would have done without that common sense approach.
    Last edited by kjtruitt; February 12th 2014 at 04:45 PM.
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  15. #15
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    Re: x^4 + 4 - factor

    That "common sense approach" is formally codified in something called "Gauss' Lemma" which says that we ought to look for INTEGER coefficients first. Well, integers are "nice" because prime factorization kind of limits the possibilities. In fact, if it DOESN'T work, we're probably looking at some awful formula filled with square and cube roots, which might be easier to solve using a computer, and a numerical algorithm.

    And it gets worse: for a 5th-degree polynomial, there IS NO GENERAL FORMULA, so if the "easy way" doesn't work, it's time for Wolfram|Alpha. In other words, factoring polynomials is only practical for ones of fairly low degree, unless they are a "special case" in which the type of reasoning here works out. In other words, if you can do quadratics, some cubics, and a few quartics, you're on a par with the entire mathematical community (specialists in polynomials know a few more tricks, but these are "advanced topics").
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