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Homework Help: 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


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    ??? 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


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    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


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    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


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    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)
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