1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Given f(x) = x^3 + 4x -5, show x-1 is a factor

  1. Apr 27, 2008 #1
    The polynomial f(x) is given by f(x) = x^3 + 4x -5

    show x-1 is factor :

    1+4-5=0 Thus it's a factor.

    express in form (x-1)(x^2+px+q)

    ends up being X^2 + X + 5.

    next question is :

    hence show the equation f(x) = 0 has one real root state it's value.

    i know i have to use b^2 -4ac. If i do that it's -19 which is <0 so it's 1 value. But how do I find the value. And why does all this stuff work.

    My teacher is one that just tells you how to do it. I have no idea why i'm doing it and it makes no sense so it's harder to remember. I would get 1or 2 marks out of 3 for this.

    Firstly i dont know how to find out what the real number is. and i don't know why this works.

    explain plez?! 8D
  2. jcsd
  3. Apr 27, 2008 #2


    User Avatar
    Staff Emeritus
    Science Advisor

    ??? Didn't you start off by showing that f(1)= 0??

    And you are trying x so that f(x)= 0??
  4. Apr 27, 2008 #3
    First a short lesson in discriminants, since you asked why you use them.

    Remember the equation for the roots of any polynomial [tex]f \left(x \right) = a x^{2} + bx + c [/tex] is
    [tex] x = \frac{-b + \sqrt{b^{2} - 4 a c}}{2a}[/tex] and [tex] x = \frac{-b - \sqrt{b^{2} - 4 a c}}{2a}[/tex]

    The term [tex]b^{2} - 4 a c[/tex] is called the "discriminant". If the discriminant is negative, then in you would be taking the square root of a negative number, which you cannot do with the real numbers.

    Now, how many roots must [tex]f \left(x \right) = x^{3} + 4 x - 5[/tex] have?
    Do you know the value of any of the roots already?
  5. Apr 28, 2008 #4


    User Avatar
    Science Advisor
    Homework Helper

    Hi DeanBH! :smile:

    I think you're getting confused … the real number is 1, from the (x - 1) factor that you've already found.

    When the question says "show the equation f(x) = 0 has one real root", it means exactly one real root (and two complex roots), instead of three real roots. That's all! :smile:
  6. Apr 28, 2008 #5
    by working backwards, you can just show that x^3+4x-5 factors into (x^2+5)(x-1)

    not sure if your teacher wants you to use the quadratic formula or not, but working backwards works just fine in some cases.
  7. Apr 29, 2008 #6


    User Avatar
    Staff Emeritus
    Science Advisor

    No, it doesn't.

    (x2+ 5(x-1)= x3- x2+ 5x- 5, NOT x3+ 4x- 5. Did you mean (x2+ x+ 5)(x- 1)?
  8. May 1, 2008 #7
    i don't think it's is a difficult question ? .
  9. May 1, 2008 #8


    User Avatar
    Staff Emeritus
    Science Advisor

    I believe that's already been pointed out several times!
  10. May 1, 2008 #9

    gah my bad, yes i ment (x2+ x+ 5)(x- 1)
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Given f(x) = x^3 + 4x -5, show x-1 is a factor
  1. Factoring 4x^4-x^2-18 (Replies: 6)

  2. X^3 - 4x^2 + 4x +2 = 0 (Replies: 3)