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Homework Help: Find all roots of x^3 + 3x^2 - 10x + 6

  1. Jun 15, 2008 #1
    find all roots of x^3 + 3x^2 - 10x + 6

    the solution:
    identify the easy root of x=1,
    find the remaining roots from (x-1)(x^2+4x) using quadratic formula.

    The only thing i dont understand here is how to factorize to (x-1)(x^2+4x)... namely the (x^2+4x) part.
     
  2. jcsd
  3. Jun 15, 2008 #2
    factor out a common term in x^2+4x

    and you're pretty much done
     
  4. Jun 15, 2008 #3
    but how did i get to x^2+4x in the first place?... the original equation was x^3 + 3x^2 - 10x + 6.. i merely copied the solution... so i find the easy root of 1.. then what?
     
    Last edited: Jun 15, 2008
  5. Jun 15, 2008 #4
    use synthetic division
     
  6. Jun 15, 2008 #5

    matt grime

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    No, don't use synthetic division (just yet). Pause for a moment and think: is it plausible that x^2+4x is a factor? It isn't. Copying out the answer is never a good idea.
     
  7. Jun 15, 2008 #6
    i just looked up synthetic devision on wikipedia and tried it but it didnt work
     
  8. Jun 15, 2008 #7
    should i take out the common term x first?
     
  9. Jun 15, 2008 #8

    arildno

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    Then you should practice synthetic division once more!

    Further, ponder over matt grime's words:

    WHY should you be suspicious of that particular factorization?

    Hint:
    How could you ascertain whether the factorization is correct or false?
     
  10. Jun 15, 2008 #9
    oh sorry i copied it wrong, it should be (x-1)(x^2+4x-6).. now expanding that get: x^3 + 4x^2 -6x -x^2 -4x + 6 = x^3 + 3x^2 - 10x + 6.. so yeh its right now.. though i still cant get the synthetic devision right (its new to me)

    i tried to learn it now from http://en.wikipedia.org/wiki/Synthetic_division

    though i keep getting 1,2,-12,18

    can someone help show how i would use synthetic devision for the original polynomial ?
     
    Last edited: Jun 15, 2008
  11. Jun 15, 2008 #10

    arildno

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    Okay, we wish to find a second-order polynomial so that:
    [tex](x-1)(ax^{2}+bx+c)=x^{3}+3x^{2}-10x+6[/tex] holds for all x.
    I.e, we must determine a,b and c!!

    Multiplying out the left-hand side, and organizing in powers of x, the lefthandside can be rewritten as:
    [tex]ax^{3}+(b-a)x^{2}+(c-b)x-c= x^{3}+3x^{2}-10x+6[/tex]

    NOw, the coefficients of each power must be equal on the right and left sides, yielding the system of equations:
    a=1
    b-a=3
    c-b=-10
    -c=6

    This yields:
    a=1
    b=4
    c=-6
     
  12. Jun 15, 2008 #11
    thanks arildno, very helpful
     
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