Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Factoring POlynomials and finding Zeros

  1. Mar 15, 2007 #1

    1. 16x^3 - 54 This one, I've broken it to 2(2^3 x^3 - 3^2) but it's still wrong! I don't get how much further it can be broken down!
    2. 3x^4 - 48 This one.. I've gotten far as 3x^2 (x^2 - 4x + 4) ! Still wrong.

    Zeros... dang it I'm kinda embarrassed to even ask...

    1. 9x^3 + 27x^2 = 4x - 12
    I've put x = -3, but I know there's more. I don't know how to get more!

    2. x^5 - 100x = 0
    Dang it, I don't know the zero of this one >< !

  2. jcsd
  3. Mar 15, 2007 #2


    User Avatar
    Homework Helper
    Gold Member

    For the zeros, get x on one side of the equal side and zero on the other. Then solve for x. If you end up with x(Ax+b)=0 or something similar, for this equation to be zero, either x = 0 or Ax+b = 0. For these cases then, what can x be? This should be of help for number 2. For number one you need to factor a third order polynomial.

    For the factoring, in number 1 you have a term containing the difference between two cubes, 8x^3-3^3, this can be factored further. There should be a formula you can follow in your text.In number 2, you have a quadratic,(x^2-4x-4) this can also be factored more.

    Good Luck!

  4. Mar 16, 2007 #3
    3X^4-48 = 3(X^4-16)=3(X^2-4)(X^2+4)=3(X-2)(X+2)(X^2+4)

    X^5-100X=0 -> X(X^4-100)=0 -> X=0 or (X^2-10)(X^2+10)=0 -> X = +- sqrt(10), X=+-i*sqrt(10)
  5. Mar 17, 2007 #4
    Don't forget complex solutions.
  6. Mar 17, 2007 #5
    How far you can factor it depends on what you're factoring over.

    ie. x^2 - 5 cannot be factored over the rational numbers,
    however, it can be factored as [tex](x-\sqrt{5})(x+\sqrt{5})[/tex]
    Similarly, x^2 + 1 cannot be factored over the reals, but it can be factored as [tex](x-i)(x+i)[/tex]

    Note: For x^2 - 3x - 4 = 0, the roots are 4 and -1
    The factors are (x-4) and (x+1)
    And, for 2x^2 - 3x + 1, the roots are 1 and 1/2
    Two of the factors are (x-1) and (x - 1/2) Note: multiplying these doesn't quite give you the original
    2x^2 - 3x + 1. However, if you were to multiply:
    (2)(x-1)(x - 1/2), you'd get that polynomial.
    Almost interestingly, you can distribute that 2 into the 3rd factor and write it as (x-1)(2x-1)

    In short, if you can find the roots of a polynomial, then it can be factored as (leading coefficient)(x-root1)(x-root2)...(x-rootn), n being the degree of the polynomial.
    Last edited: Mar 17, 2007
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook