When an expression is a product or a quotient, we can determine the
sign of the expression on an interval by looking at the sign of each
factor over the interval.
Examples
For each expression, we determine the values of
x (the
independent variable) which make it positive, negative, zero, or
undefined.

The expression is always positive, and
is positive for any value of x except
x=4, where it equals zero. Therefore,

The expression will be zero when x=1 or x=7. Therefore
product can change sign at x=1 or x=7. We check
the sign of the expression by looking at the sign of each factor
over the intervals established by dividing the number line at
these points. It will be positive when both factors have the
same sign.

()() 



(+)() 



(+)(+) 





1 

7 

x 
Be sure it is clear how the different (+) and () factors
are calculated in the figure above! (If it isn't obvious,
check below, following the answer.)
Therefore,
Ok, how did we find the (+) and () factors? If x<1,
the factor (x+1) must be negative. Thus in the leftmost
section of the number line, the first term is (). Similarly,
if x<1, the factor (x7) is also negative. If
1<x<7, the factor (x+1) is positive, so the
first factor for that section of the number line is (+). But in
this case (x7) is still negative, so the second is ().
And so on.

A fraction is zero when its numerator is zero, so this
expression is zero if r=5/2. The values r=1 and
r=3 make the fraction undefined because the denominator
is zero there. We divide the number line at each of these
points and check the sign of each factor on each interval. This
gives:
Thus,

The expression is zero if y=2. It is not defined if
y is less than 2, because we cannot take the square root
of a negative number. Furthermore, it is not defined when
y=0, because of the zero in the denominator. The radical
is always positive (where defined), so the numerator is
positive. Because the cube of a negative number is negative, we
can visualize the signs on each of the intervals as shown below.
Therefore,
We can often solve polynomial and rational inequalities by starting
out the same way we would with an equation and the using a number line
to find the intervals on which the inequality holds. However,
remember to be careful when using operations that can reverse the
inequality.
Examples
For each expression, we solve the inequality.

The left hand side equals zero for x=0, x=3 and
x=1. To solve the inequality we want to select the
xvalues for which the left hand side is negative. We
check signs over the four intervals created by marking
x=0, 3 and 1.

()()() 



()(+)() 



(+)(+)() 



(+)(+)(+) 





3 

0 

1 

x 
So
And therefore

First, put zero on the right and combine fractions on the left:
We want this quotient to be positive, so the numerator and
denominator must have the same sign. Notice that q
cannot equal 2. Because it is a squared term, the numerator is
never negative, and the denominator is positive if
2q>0, which is q<2. However, the numerator
is zero if q=1, so this must be excluded. Therefore
Practice
Be sure you've gone through each step in the examples above before
doing these. Once you've worked them until you're sure that you
understand them, go on to the next section. There are more problems
of this type in the section test at the end of the section.
Note that you can get new practice problems by clicking the "Refresh"
button at the bottom of the practice set.
precal: 8.2  signs and solving nl ineq
page created: Sun Nov 23 10:35:38 2014
Comments to
mathitc(at)umich(dot)edu
©20032007 Gavin LaRose, Pat Shure /
University of Michigan Math Dept. /
Regents of the University of Michigan