The Central Idea
Consider
\[ (x-2)(x+3)>0. \]We are not asking where the product equals a single value. We are asking where the product is positive.
The factors change sign at
\[ x=2 \]and
\[ x=-3. \]These values divide the number line into intervals:
\[ (-\infty,-3),\qquad (-3,2),\qquad (2,\infty). \]On each interval, the sign of every factor remains constant, so the sign of the whole product remains constant.
This is the key strategy:
Find the critical points, divide the number line into intervals, determine the sign on each interval, and keep the intervals that satisfy the inequality.
Why Equation Methods Are Not Enough
For an equation such as
\[ (x-2)(x+3)=0, \]the zero-product property immediately gives
\[ x=2 \]or
\[ x=-3. \]But for
\[ (x-2)(x+3)>0, \]the zeros are only boundary points.
The actual solution consists of intervals where the product has the correct sign.
Polynomial Inequalities
A polynomial inequality compares a polynomial expression with zero or another expression.
Examples include
\[ x^2-5x+6>0, \] \[ x^3-4x\le0, \]and
\[ 2x^2+3x<5. \]A useful first step is usually to move all terms to one side so that the other side is zero.
Put the Inequality in Zero Form
Consider
\[ x^2-3x<4. \]Move everything to one side:
\[ x^2-3x-4<0. \]Factor:
\[ (x-4)(x+1)<0. \]Now the important boundary points are visible:
\[ x=-1,\qquad x=4. \]Critical Points
A critical point for sign analysis is a value where the expression is zero or undefined.
For a polynomial, critical points come from its real zeros.
For
\[ (x-4)(x+1), \]the critical points are
\[ x=-1 \]and
\[ x=4. \]They divide the number line into the intervals
\[ (-\infty,-1), \qquad (-1,4), \qquad (4,\infty). \]Test One Point from Each Interval
To solve
\[ (x-4)(x+1)<0, \]test one convenient value in each interval.
For
\[ x<-1, \]use \(x=-2\):
\[ (-2-4)(-2+1)=(-6)(-1)>0. \]For
\[ -1For
\[ x>4, \]use \(x=5\):
\[ (5-4)(5+1)=(1)(6)>0. \]The expression is negative only between the roots.
Therefore,
\[ \boxed{-1Interval Notation
The solution
\[ -1The endpoints are excluded because the inequality is strict:
\[ (x-4)(x+1)<0. \]At \(x=-1\) or \(x=4\), the product equals zero, not a negative number.
Inclusive Polynomial Inequality
Now solve
\[ (x-4)(x+1)\le0. \]The sign analysis is unchanged, but the zeros now satisfy the inequality because equality is allowed.
Therefore,
\[ -1\le x\le4. \]Interval notation:
\[ [-1,4]. \]Positive Outside the Roots
Consider
\[ (x-2)(x+3)>0. \]The critical points are
\[ -3,\qquad 2. \]Testing the three intervals shows the product is positive on the outside intervals:
\[ x<-3 \]or
\[ x>2. \]So the solution is
\[ (-\infty,-3)\cup(2,\infty). \]Negative Between Simple Roots
For a quadratic with two distinct real roots and a positive leading coefficient, the sign pattern is often
\[ +\quad-\quad+. \]That is, the polynomial is positive outside the roots and negative between them.
This pattern comes from the factor signs, not from a rule that should be memorized without explanation.
Leading Coefficient Matters
Consider
\[ -(x-2)(x+3)>0. \]The negative sign reverses the sign of the entire product.
So the expression is positive where
\[ (x-2)(x+3)<0. \]Therefore,
\[ -3Repeated Roots
Consider
\[ (x-2)^2(x+1)>0. \]The critical points are
\[ x=-1 \]and
\[ x=2. \]The factor
\[ (x-2)^2 \]is nonnegative and has even multiplicity.
As \(x\) passes through \(2\), this factor does not change sign.
So the sign of the full expression does not switch at \(x=2\).
Odd and Even Multiplicity
A zero of odd multiplicity usually causes the sign to change as the graph crosses the axis.
A zero of even multiplicity usually leaves the sign unchanged because the graph touches the axis and turns back.
For example:
\[ (x-1)^1 \]changes sign at \(x=1\), while
\[ (x-1)^2 \]does not.
This connection links algebraic multiplicity with graphical behavior.
Example with a Repeated Root
Solve
\[ (x-2)^2(x+1)\le0. \]The factor
\[ (x-2)^2 \]is always nonnegative.
Therefore, the sign is controlled primarily by
\[ x+1, \]except at \(x=2\), where the entire product is zero.
The product is negative for
\[ x<-1, \]and zero at
\[ x=-1 \]and
\[ x=2. \]Therefore,
\[ \boxed{(-\infty,-1]\cup\{2\}}. \]Higher-Degree Polynomial Inequality
Solve
\[ x(x-2)(x+4)>0. \]The critical points are
\[ -4,\qquad 0,\qquad 2. \]These divide the number line into four intervals:
\[ (-\infty,-4), \] \[ (-4,0), \] \[ (0,2), \]and
\[ (2,\infty). \]Testing one point in each interval gives the sign pattern
\[ -\quad+\quad-\quad+. \]Therefore,
\[ \boxed{(-4,0)\cup(2,\infty)}. \]Sign Charts
A sign chart organizes the sign of each factor across the intervals determined by the critical points.
For
\[ (x-2)(x+3), \]the critical points are \(-3\) and \(2\).
| Interval | \(x+3\) | \(x-2\) | Product |
|---|---|---|---|
| \((-\infty,-3)\) | \(-\) | \(-\) | \(+\) |
| \((-3,2)\) | \(+\) | \(-\) | \(-\) |
| \((2,\infty)\) | \(+\) | \(+\) | \(+\) |
Rational Inequalities
A rational inequality contains a rational expression, often with a variable in the denominator.
For example,
\[ \frac{x-2}{x+1}>0. \]Rational inequalities use the same interval sign logic as polynomial inequalities, but with one important addition:
Denominator zeros are always excluded because the expression is undefined there.
Zeros Versus Undefined Points
Consider
\[ \frac{x-2}{x+1}. \]The numerator is zero at
\[ x=2. \]The denominator is zero at
\[ x=-1. \]These two points play different roles:
- \(x=2\) is a zero of the rational expression and may be included in a non-strict inequality.
- \(x=-1\) is not in the domain and can never be included.
Basic Rational Inequality
Solve
\[ \frac{x-2}{x+1}>0. \]Critical points:
\[ x=-1,\qquad x=2. \]Test the intervals.
For \(x=-2\):
\[ \frac{-4}{-1}>0. \]For \(x=0\):
\[ \frac{-2}{1}<0. \]For \(x=3\):
\[ \frac{1}{4}>0. \]Therefore,
\[ \boxed{(-\infty,-1)\cup(2,\infty)}. \]The point \(-1\) is excluded because the denominator is zero, and \(2\) is excluded because the inequality is strict.
Inclusive Rational Inequality
Solve
\[ \frac{x-2}{x+1}\ge0. \]The positive intervals are the same, but now the zero at \(x=2\) may be included.
The denominator zero at \(x=-1\) remains excluded.
Thus,
\[ \boxed{(-\infty,-1)\cup[2,\infty)}. \]Always Move Everything to One Side
Consider
\[ \frac{1}{x}>2. \]It is dangerous to multiply both sides by \(x\) immediately because the sign of \(x\) is unknown.
Instead, move everything to one side:
\[ \frac{1}{x}-2>0. \]Combine:
\[ \frac{1-2x}{x}>0. \]Now use sign analysis.
Why Blind Cross-Multiplication Can Fail
In an equation, multiplying by a known nonzero denominator can preserve equivalence.
In an inequality, multiplying by a variable denominator is more delicate because its sign may be positive or negative.
If the denominator is negative, the inequality direction should reverse.
If its sign is unknown, direct multiplication can accidentally merge different cases.
Sign analysis avoids that problem.
Worked Rational Example with Unknown-Sign Denominator
Solve
\[ \frac{1}{x}>2. \]Rewrite:
\[ \frac{1-2x}{x}>0. \]Critical points:
\[ x=0 \]and
\[ x=\frac12. \]Test intervals.
For \(x=-1\):
\[ \frac{1-2(-1)}{-1} = \frac{3}{-1}<0. \]For \(x=\frac14\):
\[ \frac{1-\frac12}{\frac14}>0. \]For \(x=1\):
\[ \frac{1-2}{1}<0. \]Therefore,
\[ \boxed{\left(0,\frac12\right)}. \]Rational Inequality with Several Factors
Solve
\[ \frac{(x-3)(x+2)}{x-1}\le0. \]Critical points:
\[ x=-2,\qquad x=1,\qquad x=3. \]The points \(-2\) and \(3\) are zeros.
The point \(1\) is excluded because it makes the denominator zero.
Testing the intervals gives the sign pattern
\[ -\quad+\quad-\quad+. \]Therefore,
\[ \boxed{(-\infty,-2]\cup(1,3]}. \]Canceling Factors Requires Domain Memory
Consider
\[ \frac{(x-2)(x+1)}{x-2}>0. \]The original expression requires
\[ x\ne2. \]For allowed values, it simplifies to
\[ x+1>0. \]So
\[ x>-1. \]But \(x=2\) must still be excluded.
Therefore,
\[ \boxed{(-1,2)\cup(2,\infty)}. \]Cancellation does not erase the original domain restriction.
Holes Matter Even When the Sign Does Not Change
In the previous example,
\[ \frac{(x-2)(x+1)}{x-2}, \]the factor \(x-2\) cancels algebraically, so the sign does not switch at \(x=2\).
But the original rational expression remains undefined there.
This creates a hole in the solution set even though the sign on both sides of \(2\) is the same.
Multiplicity in Rational Inequalities
Multiplicity matters for both numerator and denominator factors.
At a factor of even multiplicity, the sign does not change.
At a factor of odd multiplicity, the sign usually does change.
However, denominator zeros remain excluded regardless of multiplicity.
Example with Repeated Denominator Factor
Consider
\[ \frac{x-1}{(x+2)^2}>0. \]The denominator is positive for every
\[ x\ne-2. \]So the sign is controlled by \(x-1\).
The expression is positive when
\[ x>1. \]The excluded point \(x=-2\) does not affect this positive interval because it lies outside it.
Therefore,
\[ \boxed{(1,\infty)}. \]Example Where a Hole Splits an Interval
Consider
\[ \frac{x+3}{(x-1)^2}\ge0. \]The denominator is positive except at
\[ x=1, \]where the expression is undefined.
The numerator is nonnegative when
\[ x\ge-3. \]So the solution would appear to be
\[ [-3,\infty), \]except \(x=1\) must be removed.
Therefore,
\[ \boxed{[-3,1)\cup(1,\infty)}. \]Graphical Interpretation of Polynomial Inequalities
Solving
\[ f(x)>0 \]means finding where the graph of
\[ y=f(x) \]lies above the \(x\)-axis.
Similarly,
\[ f(x)<0 \]asks where the graph lies below the \(x\)-axis.
The roots are the boundary points where the graph meets the axis.
Graphical Interpretation of Rational Inequalities
For a rational function, sign analysis still asks whether the graph lies above or below the \(x\)-axis.
But denominator zeros may create:
- vertical asymptotes;
- holes;
- or other domain exclusions.
These undefined points must be removed from the solution set even if nearby signs match.
Critical Points Organize the Entire Problem
For polynomial inequalities, critical points are real zeros.
For rational inequalities, critical points include both:
- zeros of the numerator;
- zeros of the denominator.
Once these points are placed in order on the number line, every remaining interval has a constant sign.
Why One Test Point Is Enough
Between two consecutive critical points, none of the factors crosses zero or becomes undefined.
Therefore, the sign of each factor remains constant throughout that interval.
So testing one point tells us the sign of the whole interval.
This is why interval testing is efficient rather than guesswork.
Using Factor Signs Without Numerical Substitution
For
\[ (x-2)(x+5), \]we can reason directly.
If
\[ x<-5, \]both factors are negative, so the product is positive.
If
\[ -5If
\[ x>2, \]both factors are positive, so the product is positive.
With practice, a sign chart can replace repeated arithmetic substitution.
Do Not Solve Factors Separately with the Same Inequality
From
\[ (x-2)(x+3)>0, \]it is incorrect to say
\[ x-2>0 \]and
\[ x+3>0 \]as the only possibility.
A product is also positive when both factors are negative.
That second possibility produces
\[ x<-3. \]Sign analysis automatically captures all sign combinations.
Product Sign Logic
For two nonzero factors:
| First Factor | Second Factor | Product |
|---|---|---|
| Positive | Positive | Positive |
| Positive | Negative | Negative |
| Negative | Positive | Negative |
| Negative | Negative | Positive |
Rational inequalities use the same sign logic because division has the same sign rules as multiplication.
Quotient Sign Logic
For nonzero numerator and denominator:
- same signs produce a positive quotient;
- opposite signs produce a negative quotient.
The difference is that denominator zeros are forbidden rather than ordinary zero boundaries.
Application: Profit Threshold
Suppose profit is modeled by
\[ P(x)=(x-20)(100-x). \]To determine when profit is positive, solve
\[ (x-20)(100-x)>0. \]The critical points are
\[ 20 \]and
\[ 100. \]Testing shows the product is positive between them:
\[ 20Application: Rational Constraint
Suppose a model contains
\[ R(x)=\frac{x-5}{x-10} \]and requires
\[ R(x)\ge0. \]The critical points are
\[ x=5 \]and
\[ x=10. \]The numerator zero may be included because equality is allowed.
The denominator zero must always be excluded.
Sign analysis determines which remaining intervals satisfy the model.
Worked Example: Quadratic Inequality
Solve
\[ x^2-x-6\ge0. \]Factor:
\[ (x-3)(x+2)\ge0. \]Critical points:
\[ -2,\qquad 3. \]The product is nonnegative outside the roots.
Include the roots because equality is allowed:
\[ \boxed{(-\infty,-2]\cup[3,\infty)}. \]Worked Example: Cubic Inequality
Solve
\[ x(x-1)(x+2)<0. \]Critical points:
\[ -2,\qquad 0,\qquad 1. \]Testing the intervals gives
\[ -\quad+\quad-\quad+. \]Therefore,
\[ \boxed{(-\infty,-2)\cup(0,1)}. \]Worked Example: Rational Inequality
Solve
\[ \frac{x+1}{x-4}<0. \]Critical points:
\[ x=-1,\qquad x=4. \]Test intervals:
- for \(x<-1\), numerator and denominator are both negative, so the quotient is positive;
- for \(-1
- for \(x>4\), both are positive, so the quotient is positive.
Therefore,
\[ \boxed{(-1,4)}. \]Worked Example: Rational Inequality with Inclusion
Solve
\[ \frac{x+1}{x-4}\le0. \]The negative interval is
\[ (-1,4). \]The numerator zero at \(x=-1\) may be included.
The denominator zero at \(x=4\) may not.
Therefore,
\[ \boxed{[-1,4)}. \]A Reliable Polynomial-Inequality Workflow
- Move all terms to one side.
- Factor the polynomial when possible.
- Find all real zeros.
- Place the zeros in order on a number line.
- Determine the sign on each interval.
- Select the intervals that satisfy the inequality.
- Include zeros only when equality is allowed.
A Reliable Rational-Inequality Workflow
- Move everything to one side and combine into one rational expression.
- Factor numerator and denominator when useful.
- Find numerator zeros and denominator zeros.
- Mark denominator zeros as permanently excluded.
- Use all critical points to divide the number line.
- Determine the sign on each interval.
- Include numerator zeros only when equality is allowed.
- Never include denominator zeros.
Common Errors
Using the Zero-Product Property as Though the Inequality Were an Equation
The zeros locate boundaries but do not by themselves give the full solution set.
Multiplying a Rational Inequality by an Unknown-Sign Denominator
This may require reversing the inequality in some cases and not others. Sign analysis is safer.
Including Denominator Zeros
An undefined value can never belong to the solution set.
Excluding Numerator Zeros from an Inclusive Inequality
If the inequality uses \(\le\) or \(\ge\), numerator zeros may be included.
Ignoring Multiplicity
Even-multiplicity factors often do not cause a sign change.
Forgetting Original Restrictions After Cancellation
A canceled denominator factor can leave a hole that remains excluded.
Testing Only the Critical Points
The solution usually consists of intervals between critical points, so interval signs must be analyzed.
Practice
1. Quadratic Inequality
Solve:
\[ x^2-9<0. \]Check Answer
\[ (x-3)(x+3)<0. \]The product is negative between the roots:
\[ -32. Inclusive Quadratic Inequality
Solve:
\[ x^2-5x+6\ge0. \]Check Answer
\[ (x-2)(x-3)\ge0. \]The product is nonnegative outside the roots:
\[ x\le2 \]or
\[ x\ge3. \]Therefore,
\[ (-\infty,2]\cup[3,\infty). \]3. Cubic Inequality
Solve:
\[ x(x-4)(x+1)>0. \]Check Answer
Critical points:
\[ -1,\qquad 0,\qquad 4. \]The sign pattern is
\[ -\quad+\quad-\quad+. \]So
\[ (-1,0)\cup(4,\infty). \]4. Basic Rational Inequality
Solve:
\[ \frac{x-3}{x+2}>0. \]Check Answer
Critical points:
\[ -2,\qquad 3. \]The quotient is positive on the outside intervals:
\[ (-\infty,-2)\cup(3,\infty). \]5. Inclusive Rational Inequality
Solve:
\[ \frac{x-3}{x+2}\le0. \]Check Answer
The quotient is negative between the critical points, and the numerator zero may be included:
\[ (-2,3]. \]The denominator zero \(x=-2\) remains excluded.
6. Canceled Factor
Solve:
\[ \frac{(x-1)(x+4)}{x-1}\ge0. \]Check Answer
Original restriction:
\[ x\ne1. \]For allowed values, the expression simplifies to
\[ x+4\ge0. \]So
\[ x\ge-4, \]with \(x=1\) removed.
Therefore,
\[ [-4,1)\cup(1,\infty). \]7. Repeated Root
Solve:
\[ (x-2)^2(x+5)>0. \]Check Answer
The squared factor does not change sign at \(x=2\).
The sign is positive for
\[ x>-5, \]except where the product equals zero.
Because the inequality is strict, exclude both zeros:
\[ (-5,2)\cup(2,\infty). \]Key Ideas
- Polynomial and rational inequalities are usually solved by sign analysis across intervals.
- Move the inequality to one side so the problem becomes a comparison with zero.
- Polynomial zeros and rational zeros or undefined points form the critical points.
- Critical points divide the number line into intervals where the sign is constant.
- One test point is enough to determine the sign on an interval.
- Odd-multiplicity factors usually change sign; even-multiplicity factors usually do not.
- For inclusive inequalities, polynomial zeros and rational numerator zeros may be included.
- Rational denominator zeros are always excluded.
- Cancellation does not erase original domain restrictions.
- Blindly multiplying a rational inequality by an unknown-sign denominator can produce errors.
- Graphically, positive intervals lie above the \(x\)-axis and negative intervals lie below it.
- Sign charts organize the relationship among factors, zeros, undefined points, and solution intervals.
Where This Leads
Polynomial and rational inequalities extend inequality solving from single linear boundaries to solution sets shaped by several critical points.
The same sign-analysis ideas lead naturally to graph-based inequality solving, systems of inequalities, feasible regions, optimization, and later analysis of where functions are positive, negative, increasing, decreasing, or undefined.