Recent content by mathnovice

  1. M

    How do I Solve for A in Partial Fraction Expansion?

    I found the answer: 1/(x^2+1) - 3/(x^2+4). everybody thanks for the help. I would never been able find the answer without it!
  2. M

    How do I Solve for A in Partial Fraction Expansion?

    how do you guys right fractions and expontents that way? :D would come in handy. Would this step be wrong?: (x^3-3x^2+x-12)/[(x^2+1)(x^2+4)] = [(Ax +B) (x^2+1)]/[(x^2 +1) (x^2+4)] + [(Cx+D)(x^2+4)] /[(x^2 +1) (x^2+4)] and then say (x^3-3x^2+x-12) = (Ax +B) (x^2+1) + (Bx+c)(x^2+4) becose they...
  3. M

    How do I Solve for A in Partial Fraction Expansion?

    Homework Statement divide in partial fractions: x^3 -3x^2+x-12 / x^4+5x^2+4 Homework EquationsThe Attempt at a Solution I factored x^4+5x^2+4 to (x^2 +1) (x^2+4) x^3 -3x^2+x-12/(x^2 +1) (x^2+4) = A (x^2+1)/(x^2 +1) (x^2+4) + B (x^2+4) /(x^2 +1) (x^2+4) they all have the same...
  4. M

    The roots of x^8 - 5x^6 + 7x^4 - 5x^2 +6=0

    Everybody thanks for the replies. I am now improving my knowledge about complex numbers first and will then tackle this problem. But it did motivate me to discover the whole subjet of complex numbers, as I just find out complex numbers are also used in engineering applications and that's my goal...
  5. M

    The roots of x^8 - 5x^6 + 7x^4 - 5x^2 +6=0

    [Thread moved to homework forum by a Mentor] the exercise was to find the roots of x^8 - 5x^6 + 7x^4 - 5x^2 +6=0 I substituded x^2 with y : y^4 - 5y^3 + 7y^2 - 5y +6=0 I factored this by doing the rational roots test and trying those possible roots with the method of horner and got (y-2)...
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