Polynomial and Rational Inequalities

Scope Label

Core 9758: rational inequalities in which the numerator and denominator are linear or quadratic expressions, together with the sign, restriction, graph, and interval logic needed to solve them. Quadratic inequalities provide essential preparation. Standalone inequalities involving higher-degree polynomials are retained below as Enrichment / transfer practice, not as a stated core requirement.

Role in the Topic

This branch expands the inequality methods in Equations and Inequalities.

The main idea is that an inequality asks where an expression is positive, negative, above another expression, or below another expression. The answer is usually an interval or a union of intervals.

For polynomial and rational inequalities, the most reliable method is sign analysis.

Formulating Inequalities

Some questions do not give the inequality directly. They describe a condition in words, and you must translate it first.

Common language includes:

• “at least” means $\ge$
• “more than” means $>$
• “not more than” means $\le$
• “between” means a double inequality, but the context must state whether the endpoints are included

For example, if a quantity

$$Q=\frac{4x-4}{2x+3}$$

must lie between $1$ and $3$, then the mathematical statement is

$$1\le \frac{4x-4}{2x+3}\le 3.$$

The solving method only begins after this translation is correct.

Why Roots Split the Number Line

For a polynomial expression, the sign can only change when the expression passes through zero.

For a rational expression, the sign can change when:

• the numerator is zero
• the denominator is zero

These values split the number line into intervals. On each interval, the sign of each factor stays fixed.

This is why sign charts work: they replace infinitely many possible $x$-values with a finite set of intervals to test.

Figure reading. First place every zero and undefined value on the number line. These critical values partition the domain into connected intervals on which every continuous factor keeps the same sign. One test input therefore determines the sign throughout each interval. The final inequality symbol tells you which intervals to retain; endpoint inclusion is decided separately.

This argument depends on continuity within each interval. A polynomial factor cannot change sign without becoming zero. A rational expression can also change sign at a denominator zero, because the expression is not continuous—or even defined—there.

Quadratic Inequalities

To solve

$$x^2-4x+1<0,$$

first solve the related equation:

$$x^2-4x+1=0.$$

Using the quadratic formula,

$$x=\frac{4\pm \sqrt{16-4}}{2}=2\pm \sqrt{3}.$$

Since the coefficient of $x^2$ is positive, the parabola opens upward. Therefore the quadratic is negative between its two roots:

$$2-\sqrt{3}<x<2+\sqrt{3}.$$

So

$$x\in (2-\sqrt{3},2+\sqrt{3}).$$

Figure reading. The roots are boundary values where the quadratic equals zero. Because the parabola opens upward, it lies below the $x$-axis between the roots and above it outside them. Strict inequalities exclude the roots; inequalities allowing equality include them.

Enrichment: Higher-Degree Polynomial Inequalities

Scope

This section develops the same sign-chart reasoning beyond the stated linear/quadratic core. It is useful transfer practice, but should not displace revision of prescribed rational and modulus inequalities.

For higher-degree polynomial inequalities:

1. bring everything to one side
2. factorise if possible
3. identify critical values
4. use a sign chart or self-sketch
5. keep intervals with the required sign
6. include endpoints only if equality is allowed

For example, solve

$$(x+1)(x-2)(x-3)\le 0.$$

The critical values are

$$x=-1,x=2,x=3.$$

They split the number line into

$$(-\infty ,-1),(-1,2),(2,3),(3,\infty ).$$

Testing signs gives:

$$-,+,-,+.$$

Since the inequality asks for $\le 0$, keep the negative intervals and include the zeros:

$$x\in (-\infty ,-1]\cup [2,3].$$

The equality sign matters: zeros of the polynomial are included because the expression is allowed to equal zero.

Factor Multiplicity and Sign Changes

The sign does not necessarily alternate at every repeated root. Near a factor $(x-a{)}^m$:

• if $m$ is odd, the factor changes sign as $x$ passes through $a$;
• if $m$ is even, the factor is non-negative on both sides and the sign does not change.

For example,

$$p(x)=(x+1)(x-2{)}^2.$$

At $x=-1$ the odd-multiplicity factor changes sign. At $x=2$ the squared factor becomes zero but does not change sign. A graph crosses the $x$-axis at the first root and touches it at the second.

Figure reading. Both marked values are roots, but their local behaviour differs. At an odd-multiplicity root the curve crosses the axis and the sign changes. At an even-multiplicity root it touches the axis and turns back, so the signs on the two sides agree. Endpoint inclusion still depends on whether equality is allowed.

Rational Inequalities

For a rational expression

$$\frac{f(x)}{g(x)},$$

the denominator must not be zero:

$$g(x)\ne 0.$$

This restriction must be stated before solving and remembered in the final answer.

Core example

Solve

$$\frac{x-2}{x+1}\ge 0.$$

First,

$$x+1\ne 0,$$

so

$$x\ne -1.$$

The critical values are:

• numerator zero: $x=2$
• denominator zero: $x=-1$

These split the number line into

$$(-\infty ,-1),(-1,2),(2,\infty ).$$

The sign pattern is

$$+,-,+.$$

Since the inequality is $\ge 0$, keep the positive intervals and include the numerator zero:

$$x\in (-\infty ,-1)\cup [2,\infty ).$$

The point $x=-1$ is not included because the expression is undefined there.

Figure reading. Both numerator and denominator zeros split the sign chart, but their endpoint roles differ. A numerator zero makes the fraction equal to zero and may be included when equality is allowed. A denominator zero makes the original expression undefined and is always excluded, even if algebraic cancellation later hides it.

Cancellation does not restore a missing domain point

Consider

$$\frac{(x-1)(x+2)}{x-1}.$$

For $x\ne 1$ it simplifies to $x+2$, but the original expression is still undefined at $x=1$. The simplified graph is the line $y=x+2$ with a hole at $(1,3)$. Any inequality involving the original fraction must therefore keep $x=1$ excluded.

Why Cross-Multiplication Is Dangerous

The move

$$\frac{x-2}{x+1}\ge 0\Rightarrow x-2\ge 0$$

is invalid because $x+1$ may be positive or negative.

If $x+1>0$, multiplying preserves the inequality sign.

If $x+1<0$, multiplying reverses the inequality sign.

Since the sign of $x+1$ changes at $x=-1$, careless cross-multiplication loses information.

The safer workflow is:

1. move all terms to one side
2. combine into a single fraction
3. state denominator restrictions
4. identify numerator and denominator zeros
5. use a sign chart

When Multiplying by a Square Is Safe

Sometimes the denominator can be removed by multiplying by a square such as

$$(3x+4{)}^2.$$

It is zero when

$$3x+4=0.$$

So the restriction

$$x\ne -\frac{4}{3}$$

must still be written.

The square removes the sign problem; it does not remove the domain restriction.

More precisely, after first recording $3x+4\ne 0$, the square is strictly positive on the admissible domain. Multiplication then preserves the inequality direction and is reversible there. Saying only that the square is non-negative is not enough, because multiplication by zero would not preserve equivalence.

Rational Comparisons and Graphical Checking

Some inequalities are most naturally treated as comparisons between two expressions.

Solve

$$\frac{12}{x-3}\ge x+1.$$

The restriction is

$$x\ne 3.$$

Move everything to one side:

$$\frac{12}{x-3}-(x+1)\ge 0.$$

Combine into one fraction:

$$\frac{12-(x+1)(x-3)}{x-3}\ge 0.$$

Since

$$12-(x+1)(x-3)=12-(x^2-2x-3)=-x^2+2x+15=-(x-5)(x+3),$$

we solve

$$\frac{-(x-5)(x+3)}{x-3}\ge 0.$$

The critical values are

$$x=-3,x=3,x=5.$$

Here, $x=3$ is excluded because it makes the denominator zero. The values $x=-3$ and $x=5$ may be included because they make the numerator zero and the inequality allows equality.

Testing intervals gives

$$x\le -3\text{or}3<x\le 5.$$

Graphically, the same solution is where the graph of

$$y=\frac{12}{x-3}$$

lies on or above the graph of

$$y=x+1.$$

This graphical view is useful for checking the interval structure, especially near the vertical asymptote $x=3$. The algebraic sign chart is still needed for exact endpoints and the excluded value.

Double Inequalities

For

$$1<\frac{4x-4}{2x+3}<3,$$

split the statement into two inequalities:

$$1<\frac{4x-4}{2x+3}$$

and

$$\frac{4x-4}{2x+3}<3.$$

Solve both carefully, keeping the restriction

$$2x+3\ne 0.$$

For the left inequality,

$$\frac{4x-4}{2x+3}>1⟺\frac{2x-7}{2x+3}>0,$$

so

$$x<-\frac{3}{2}\text{or}x>\frac{7}{2}.$$

For the right inequality,

$$\frac{4x-4}{2x+3}<3⟺\frac{-2x-13}{2x+3}<0,$$

so

$$x<-\frac{13}{2}\text{or}x>-\frac{3}{2}.$$

The original double inequality requires both conditions, so intersect the two sets:

$$x<-\frac{13}{2}\text{or}x>\frac{7}{2}.$$

The forbidden value $x=-\frac{3}{2}$ has remained excluded throughout.

A clear final line should show this intersection explicitly. If the two component solution sets are $S_1$ and $S_2$, then the answer is

$$S_1\cap S_2,$$

not their union.

Substitution Inside Inequalities

Some problems reuse an earlier inequality result after replacing $x$ with another expression such as $e^x$.

For example, if an earlier step gives

$$u>\frac{7}{2},$$

and later

$$u=e^x,$$

then

$$e^x>\frac{7}{2}.$$

Since $e^x$ is positive and increasing,

$$x>\ln \frac{7}{2}.$$

This kind of problem tests whether you understand the range and monotonic behaviour of the substitution, not only the original sign chart.

Two checks are required after a substitution $u=h(x)$:

1. admissibility: retain only $u$-values in the range of $h$ (for $u=e^x$, require $u>0$);
2. back-substitution: solve $h(x)$ in the retained intervals, taking account of whether $h$ is increasing or decreasing.

Exact Algebra Versus Calculator Support

Graphical methods are useful when:

• a polynomial is not easily factorised
• a rational comparison is easier to interpret as one graph above another
• you need to check the shape or approximate roots

However, calculator roots may be decimal approximations. If the question expects exact values, use algebraic methods such as factorisation, the quadratic formula, or exact sign charts wherever possible.

Common Pitfalls

• Forgetting to state denominator restrictions.
• Including a denominator zero in the final answer.
• Cross-multiplying without knowing the sign of the denominator.
• Multiplying by a square but forgetting the excluded zero of the square.
• Treating a graphing-calculator decimal as exact.
• Using “and” when the final answer should be a union of intervals.
• Using “or” when two inequalities in a double inequality must both hold.
• Assuming the sign alternates at a repeated root without checking its multiplicity.
• Cancelling a common factor and then accidentally including a value excluded from the original domain.

Revision Checklist

• Can I explain why roots and undefined points split the number line?
• Can I solve core quadratic/rational sign problems and apply the same reasoning to labelled higher-degree enrichment?
• Can I distinguish numerator zeros from denominator zeros?
• Can I justify endpoint inclusion or exclusion?
• Can I solve rational inequalities without unsafe cross-multiplication?
• Can I handle double inequalities by intersection?
• Can I predict whether the sign changes at a root from its multiplicity?
• Can I retain original domain restrictions after cancellation?
• Can I use graphical calculator results appropriately without losing exactness?

Related

• Equations and Inequalities
• Graphing Techniques