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Polynomial Inequalities

  1. Oct 22, 2004 #1
    We just started these at school, but I have some questions..
    Inequalities on number line (x) graphs..
    Using x = 0 testpoint..
    Let's say I have (x+2)(x-4) > 0, (0 + 2)(0 - 4) would be -8, and -8 !>0, so it would be a disjunction, right? But if it was positive and satisfied the inequality, it would be a conjunction, right?
    OR what if it was < 0 instead, if it satisfied it would be a disjunction and if it didn't satisfy it would be a conjunction, right?
    And let's say I have x(x-2)(x+4) > 0, the teacher taught us this method where we plot -4, 2 and 0 on a number line and place positive signs and negative signes over each one accordingly so that the satisfying inequality would be where the 3 signs ontop of each other would be + or -, according to the inequality...
    IS there an easier way to do this?
    I'm sorry that I don't make much sense, I'm in a hurry...
    Hope someone can make sense of my words and help!
  2. jcsd
  3. Oct 23, 2004 #2
    there are ways ofcourse .... its simple logic most of the times ...

    Anything greater than 0 indicates it is positive and anything less than 0 indicates it is negative ...

    Take your example,
    This suggests to us that (x+2)(x-4) must be positive.

    When is multiplication of two numbers positive?
    we know that,
    (negative)*(negative) = positive
    (negative)*(positive) = negative
    (positive)*(negative) = negative
    (positive)*(positive) = positive

    this suggests to us that,
    either both (x+2) and (x-4) are positive
    or both are negative

    i.e (x+2)>0 and (x-4)>0
    (x+2)<0 and (x-4)<0

    x+2>0 implies x>-2
    x-4>0 implies x>4
    both of these conditions are true only if x>4


    x+2<0 implies x<-2
    x-4<0 implies x<4
    both of these conditions are true only if x<-2

    so u have the solution ....

    I am not saying this is easier than the one proposed by your teacher (infact they are same). This infact is the wordy explanation of the method given by your teacher .... Can u see why??

    -- AI
  4. Oct 23, 2004 #3
    Heh, I still don't understand this method, but was curious if the one I had with the test point is right.
  5. Oct 23, 2004 #4


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    One good way to test such inequalities is to find the points where one of the factors goes to 0 (for (x-2)(x+3)<0, these are 2 and -3) and check a test point between each one.
  6. Oct 24, 2004 #5


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    One key point is that, if a polynomial is positive for x= x0 and negative for x= x1, then the value MUST BE 0 someplace between x0 and x1.

    In particular, (x+2)(x-4)= 0 only at x= -2 and x= 4. Those two points divide the real number line into 3 intervals: x< -2, -2< x< 4, and x> 4. Since (x+2)(x-4) can CHANGE SIGN only at -2 and 4, it must have the SAME SIGN in each interval and it is only necessary to check on point in each.

    Pick ANY number less than -2: -3 will work. (-3+2)(-3-4)= -1(-7)= +7> 0 so EVERY x< -2 makes (x+2)(x-4)>0.
    Pick ANY number between -2 and 4: 0 is a simple choice. (0+2)(0-4)= 2(-4)= -8< 0 so EVERY x between -2 and 4 makes (x+2)(x-4)< 0.
    Pick ANY number larger than 4: 5 is good. (5+2)(5-4)= 7(1)= 7> 0 so EVERY x> 4 makes (x+2)(x-4)> 0.

    (x+ 2)(x- 4)> 0 for all x< -2 and x> 4.
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