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Solving a Polynomial Equation

  1. Feb 11, 2010 #1
    1. The problem statement, all variables and given/known data

    I am unsure what to do here. I'm given the equation 5(x+1)3 = -5 , and I was given the equation 5(x+1)3 = 0 beforehand. The latter equation has an 'x' value of -1 only, right. However, for the former equation, because I brought the '-5' across the equal sign I realize the roots are not just -1 anymore.

    3. The attempt at a solution

    5(x+1)3 = -5

    5(x+1)(x+1)(x+1)+5 = 0

    5x3+15x2+15x+5+5 = 0

    5(x3 + 3x2 + 3x + 2) = 0

    But after this I'm lost.

    If I factor:

    5x2(x-5)+5(3x+2) = 0 then I have (x-5) which means x = 5, but I'm lost as to where I go next since there should be three roots. (3x+2) gives me x = -2/3.

    But when I do the 5x2 +5 [should I get this?] I end up with x = sqrt(-1). I don't understand what this is supposed to mean. I figure I must have gone wrong somewhere.
  2. jcsd
  3. Feb 11, 2010 #2
    While I'm not completey sure what happened here, I can tell you that you factored the polynomial incorrectly. Watch what happens when you expand your 'factorization' out: 5x2(x - 5) + 5(3x + 2) = 5x3 - 25x2 + 15x + 10 = 0.

    Try factoring it again.
  4. Feb 11, 2010 #3


    Staff: Mentor

    Or better yet, don't expand the left side. Instead, divide both sides by 5, and then move the term on the RHS over to the LHS.

    Hint: a3 + b3 = (a + b)(a2 -ab + b2)
  5. Feb 11, 2010 #4


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    In the same way that quadratics don't have to cut the x-axis at all, you're finding imaginary roots with that [itex]x=\sqrt{-1}[/itex]. Don't worry about them for now and just come to accept that you'll only be able to find one real root for this cubic polynomial.
  6. Feb 11, 2010 #5


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    The expression above is not factored because it is not the product of two or more expressions. That + sign in the middle keeps the left side from being factored.

    Your original equation can be rewritten as (x + 13 + 1 = 0. Being a sum of cubes, it can be factored as shown in my previous hint. There is one real solution to this equation.
  7. Feb 11, 2010 #6


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    You could also divide the 5 out and then just take the cube root of each side.
  8. Feb 12, 2010 #7


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    This works too, except only in this case when finding the other complex roots is not required.
  9. Feb 12, 2010 #8


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    Putting it at its most general, if you can find one root, as here you can as explained by vela, then you can reduce the degree of the equation to be solved.

    If one root is a, that means that (x - a) is a factor of the polynomial. So you can divide the polynomial by (x - a) and obtain an equation that is satisfied by the other roots.

    In your case the original equation is a cubic, so the division will give you a quadratic which you can certainly solve.
  10. Feb 12, 2010 #9


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    You could get the complex roots of the polynomial if you also write down the complex cube roots of -1 instead of just the real one. I suspect, though, for the OP, Mark44's and epenguin's methods are easier to understand if he or she wants the complex roots.
  11. Feb 12, 2010 #10
    It looks to me like you did the factorisation right in the original post and i agree that you should get: x^3+ 3x^2 + 3x +2 = 0

    Now there are a few different ways to approach this, the simplest being the factor theorem. To use the factor theorem pick some values for x and see if they satisfy the equation, normally +-1 and +-2 are good places to start.

    you can reduce the possibilitiesyou have to try by noting some simple rules. You also can only get fractional answers if the coeff of x^3 (or the highest power) is not 1. (in this example, trying 1/2 as a root would be pointless since you would always end up with 1/8 in the expression). You can't have roots larger than [constant]/[coeff of x^3] which is trickier to show but also holds.

    In this case trying x=-2 gives a solution so factoring gives:
    The quadratic can then be factorised (or in this case not...)

    It's also useful to know there is a formula, similar to the quadratic formula (but more complicated) for solving cubics. It's too cumbersome to use manually, but can be easily programmed into computers to solve cubics very quickly and simply (some graphic calculators have it as a programmed feature). Here's a simple web-based solver http://www.1728.com/cubic.htm?a=1&b=3&c=3&d=2
  12. Feb 15, 2010 #11
    Ok I got factor theorem figured out, but it didn't work for me, and takes forever.

    The sum of cubes way, I'm not sure if it is right the way I did it.

    5(x+1)3 = -5

    [STRIKE]5[/STRIKE](x+1)3 = -[STRIKE]5[/STRIKE]

    (x+1)3 = -1

    (x+1)3 +1 = 0

    Alright so there are only two terms, (x+1) and 1. So (x+1) = a and 1 = b

    a3 + b3 = (a+b)(a2 - ab + b2)

    (x+1)3 + 1[STRIKE]3[/STRIKE] = [(x+1)+(1)][(x+1)2 - 2(x+1)[STRIKE](1)[/STRIKE] + (1)[STRIKE]2[/STRIKE]]

    = (x+2)(x+1)2 - 2x + 2 + 1

    = (x+2)(x+1)2 - 2x + 3

    Was it right to go this far?
  13. Feb 15, 2010 #12


    Staff: Mentor

    You have lost sight of the fact that you're trying to solve an equation. In your 3rd line from the bottom you should have this equation:
    (x + 1)3 + 1 = 0

    In the next line you should have another equation, this time with the left side factored.
    [(x + 1) + 1][(x + 1)2 - (x + 1)(1) + 12] = 0
    Now, simplify the quantities in the brackets and solve the equation.
  14. Feb 15, 2010 #13
    This is actually a very good way to solve it, but it's quite specific to this case/cases where you can get a bracket with (x-a)^3 = c

    By replacing x+1 by z you have:
    Using a roots of unity method (i think that's the name) you can see
    z= ei(pi+2k*pi/3)
    x= ei(pi+2k*pi/3)-1
  15. Feb 15, 2010 #14


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    So far, so good.

    You introduced an incorrect factor of 2 in the middle term of the second factor.

    You dropped for brackets around the second factor. You can't do that without multiplying the (x+2) factor out front into everything inside the brackets first, but that's not generally something you want to do when trying to find the roots.

    You made another mistake when you multiplied the 2 through x+1 term, but the 2 shouldn't have been there in the first place.
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