Solve the rational inequality $\frac{(x - 3)^2(x + 1)}{x - 5} \le 0$ for all real $x$. Pay special attention to the even-multiplicity root at $x = 3$, express the complete solution set in interval notation, and describe the full geometric representation of the solution on a real number line.

Solve the rational inequality $\frac{(x - 3)^2(x + 1)}{x - 5} \le 0$ for all real $x$ . Pay special attention to the even-multiplicity root at $x = 3$ , express the complete solution set in interval notation, and describe the full geometric representation of the solution on a real number line.

Final answer

Interval notation: $[-1, 5) \cup \{3\}$ which simplifies to $[-1, 5)$ ; on the number line, a solid closed circle at $x = -1$ , an open circle at $x = 5$ , and a continuous shaded segment connecting them (with $x = 3$ already included).](streamdown:incomplete-link)

Steps

Identify Critical Numbers and Domain Restrictions

To solve the rational inequality $\frac{(x - 3)^2(x + 1)}{x - 5} \le 0$ , we find the critical numbers, analyze the sign changes across the resulting intervals, and determine the boundary behavior.

Set the numerator and denominator equal to zero to find the critical values:

Numerator roots:
- $x + 1 = 0 \implies x = -1$ (multiplicity $1$ , odd)
- $(x - 3)^2 = 0 \implies x = 3$ (multiplicity $2$ , even)
Denominator root:
- $x - 5 = 0 \implies x = 5$ (multiplicity $1$ , odd)

The denominator cannot equal zero, so $x \ne 5$ .

Analyze the Effect of Multiplicity

The factor $(x - 3)^2$ has an even power. For all $x \ne 3$ , $(x - 3)^2 > 0$ . Therefore, it does not change the sign of the rational expression as $x$ crosses $3$ .
At $x = 3$ , the numerator is $0$ , making the entire fraction equal to $0$ . Since the inequality is non-strict ( $\le 0$ ), $x = 3$ is a valid solution.
For all other values ( $x \ne 3$ ), the sign of the expression depends entirely on $\frac{x + 1}{x - 5}$ .

Test the Intervals

The critical numbers divide the real line into four intervals: $(-\infty, -1)$ , $(-1, 3)$ , $(3, 5)$ , and $(5, \infty)$ .

Interval $(-\infty, -1)$ (e.g., $x = -2$ ): $\frac{(-2 - 3)^2(-2 + 1)}{-2 - 5} = \frac{(+)(-)}{(-)} > 0 \quad (\text{False})$
Interval $(-1, 3)$ (e.g., $x = 0$ ): $\frac{(0 - 3)^2(0 + 1)}{0 - 5} = \frac{(+)(+)}{(-)} < 0 \quad (\text{True})$
Interval $(3, 5)$ (e.g., $x = 4$ ): $\frac{(4 - 3)^2(4 + 1)}{4 - 5} = \frac{(+)(+)}{(-)} < 0 \quad (\text{True})$
Interval $(5, \infty)$ (e.g., $x = 6$ ): $\frac{(6 - 3)^2(6 + 1)}{6 - 5} = \frac{(+)(+)}{(+)} > 0 \quad (\text{False})$

Include Critical Values and Express the Solution

At $x = -1$ : the expression equals $0 \le 0$ (included).
At $x = 3$ : the expression equals $0 \le 0$ (included).
At $x = 5$ : the expression is undefined (excluded).

Combining the valid intervals and points gives: $[-1, 3) \cup \{3\} \cup (3, 5) = [-1, 5)$ ](streamdown:incomplete-link)

Geometric Representation on the Number Line

Place a solid (closed) circle at $x = -1$ to show it is included.
Place an open circle at $x = 5$ to show it is excluded.
Draw a continuous shaded line segment connecting $-1$ to $5$ . The point $x = 3$ lies within this shaded segment, seamlessly included in the solution set.

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student ignores the even multiplicity of (x - 3)² and alternates signs at every boundary point, resulting in the erroneous exclusion of the interval (3, 5).

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
Critical points: $x = -1$ , $x = 3$ , $x = 5 \implies$ Denominator cannot be zero, so $x \ne 5$
Step 2
Sign chart with alternating signs: $x < -1$ : positive $\implies -1 \le x \le 3$ : negative $\implies 3 < x < 5$ : positive $\implies x > 5$ : negative
Correct
Sign of $\frac{(x - 3)^2(x + 1)}{x - 5}$ : positive on $(-\infty, -1)$ , negative on $(-1, 3)$ , zero at $x = 3$ , negative on $(3, 5)$ , and positive on $(5, \infty)$
The factor $(x - 3)^2$ is strictly positive for all $x \ne 3$ , so the expression does not alternate signs when crossing $x = 3$ . Testing $x = 4$ yields $\frac{(1)^2(5)}{-1} = -5 < 0$ , which is negative, and testing $x = 6$ yields $\frac{(9)(7)}{1} = 63 > 0$ , which is positive.
Step 3
Solution set: $[-1, 3] \cup (5, \infty)$
Correct
$[-1, 5)$ ](streamdown:incomplete-link)
Because the sign chart was inverted for $x > 3$ , the resulting intervals selected for $\le 0$ were incorrect. Since the expression is $\le 0$ on $[-1, 3]$ and on $(3, 5)$ , the combined solution is $[-1, 5)$ .](streamdown:incomplete-link)
Step 4
Solid closed circle at -1 connected to a solid circle at 3, and an open circle at 5 shaded rightward indefinitely (with accompanying graph)
Correct
A solid circle at $x = -1$ , an open circle at $x = 5$ , and a single continuous shaded segment from $-1$ to $5$
The graphical representation followed your incorrect solution set instead of shading the interval between $-1$ and $5$ continuously.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly found all the critical points by setting the numerator and denominator factors to zero.

You recognized that the denominator cannot be zero and correctly excluded $x = 5$ from the domain.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Even Multiplicity Sign Alternation

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

You assumed that polynomial and rational expressions always alternate signs across consecutive critical points, overlooking that factors raised to an even power never change sign.

REASONING PATH

Your ACEpath teacher’s suggested approach

Check Multiplicities: Factors raised to an odd power change sign at their zero; factors raised to an even power maintain the same sign on both sides of their zero.
Mark each critical point with its multiplicity. If the exponent is even, keep the sign the same across that boundary.
Test Point Verification: Never rely solely on alternating patterns without evaluating at least one test value or tracking individual factor signs.
Substitute a quick test number from each sub-interval (such as $x = 4$ and $x = 6$ ) to verify whether the fraction is positive or negative.

CORRECTED PATH

ACEpath Teacher's Demonstration

Observe that $(x - 3)^2 \ge 0$ for all real $x$ , and $(x - 3)^2 > 0$ whenever $x \ne 3$ .
Determine the sign of the rational expression purely from $\frac{x + 1}{x - 5}$ for all $x \ne 3$ :
For $x \in (-1, 3)$ , $\frac{(+)}{( - )} < 0$ .
At $x = 3$ , the numerator is $0$ , satisfying $\le 0$ .
For $x \in (3, 5)$ , $\frac{(+)}{( - )} < 0$ .
For $x \in (5, \infty)$ , $\frac{(+)}{(+)} > 0$ .
Combine the valid intervals $[-1, 3) \cup \{3\} \cup (3, 5)$ into $[-1, 5)$ .](streamdown:incomplete-link)

Final answer

Interval notation: $[-1, 5)$ ; Number line: a solid closed circle at $x = -1$ , an open circle at $x = 5$ , and a continuous shaded segment connecting them.](streamdown:incomplete-link)

NEXT STEPS

Final Takeaways from Your Teacher

Practice classifying zeros as having odd multiplicity (sign flips) or even multiplicity (sign stays the same).
Use a factored sign table with individual rows for each factor when solving rational inequalities.
Always plug in a concrete test number in intervals adjacent to even-multiplicity roots to confirm your sign chart.

OverviewACEpath Teacher's Diagnosis

02 The student forgets that the inequality includes non-strict equality (≤ 0) and includes the denominator boundary x = 5 in the solution set.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Partially correct

Step 1
(x - 3)^2 $\ge$ 0 $\text{ for all }$ x $\implies$ $\text{Sign depends on }$ $\frac{x + 1}{x - 5}$ $\le$ 0 $\implies$ $\text{Roots are }$ x = -1,; x = 3,; x = 5
Step 2
$\text{The expression is non-positive on }$ [-1, 5] $\implies$ $\text{Solution set: }$ [-1, 5]
Correct
-1, 5)
Because division by zero is undefined, $x = 5$ cannot be in the domain of the expression. Even for a non-strict inequality ( $\le 0$ ), any value that makes the denominator zero must always use an open boundary parenthesis.
Step 3
$\text{Solid circle at }$ x = -1 $\text{ and solid circle at }$ x = 5 $\text{ connected by a line segment}$
Correct
$\text{Solid circle at }$ x = -1 $\text{ and open circle at }$ x = 5 $\text{ connected by a line segment}$
Carrying over the inclusion of $x = 5$ , you plotted a solid closed dot at $5$ instead of an open circle.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly recognized that the even-powered factor $(x - 3)^2$ is non-negative everywhere and does not change sign.

You correctly identified all the boundary values: $x = -1$ , $x = 3$ , and $x = 5$ .

You correctly found that the rational expression is negative between $-1$ and $5$ .

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Included denominator zero in non-strict inequality

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

You treated all critical boundary points identically under the ' $\le 0$ ' condition, overlooking the fundamental domain restriction that any value making the denominator zero can never be included in the solution set.

REASONING PATH

Your ACEpath teacher’s suggested approach

Check Domain Restrictions First: The denominator of a rational expression cannot equal zero: $q(x) \ne 0$ .
Whenever you encounter a rational expression, immediately write down the domain restriction $x \ne 5$ before testing intervals.
Assign Circle and Bracket Types by Boundary Source: Numerator roots in a non-strict inequality ( $\le$ or $\ge$ ) take solid dots and square brackets; denominator roots always take open circles and round parentheses.
Label denominator boundaries with open circles immediately on your number line so they are never accidentally closed.

CORRECTED PATH

ACEpath Teacher's Demonstration

State the domain restriction: $x - 5 \ne 0 \implies x \ne 5$ .
Combine the tested interval $[-1, 5)$ with the isolated root $x = 3$ , which is already interior to $[-1, 5)$ .](streamdown:incomplete-link)
Graph the solution on the number line using a solid dot at $-1$ and an open dot at $5$ connected by a shaded line segment.

Final answer

Solution set: $[-1, 5)$ ; on the number line, a solid circle at $x = -1$ , an open circle at $x = 5$ , and a continuous shaded segment between them.](streamdown:incomplete-link)

NEXT STEPS

Final Takeaways from Your Teacher

Review the distinction between numerator zeros (where the expression equals 0) and denominator zeros (where the expression is undefined).
Practice marking domain restrictions as open circles on your scratch number line before performing sign analysis.

OverviewACEpath Teacher's Diagnosis

03 The student treats (x - 3)² as strictly positive and cancels it out, mistakenly excluding x = 3 from the interval solution.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
$\text{Since }$ (x - 3)^2 > 0, $\text{ divide both sides by }$ (x - 3)^2 $\implies$ $\frac{x + 1}{x - 5}$ $\le$ 0 $\text{ with }$ x \ne 3
Correct
(x - 3)^2 $\ge$ 0 $\text{ for all }$ x. $\text{ For }$ x = 3, $\text{ the expression equals }$ 0 $\le$ 0 $\text{ (true, so }$ x = 3 $\text{ is included). For }$ x \ne 3, $\frac{x + 1}{x - 5}$ $\le$ 0.
You noted that $(x - 3)^2 > 0$ requires $x \ne 3$ , but because the original inequality is non-strict ( $\le 0$ ), any value that makes the numerator zero is an allowable solution. Substituting $x = 3$ yields $0 \le 0$ , which is true, so $x = 3$ must be included rather than discarded.
Step 2
$\text{Critical points: }$ x = -1,; x = 5 $\implies$ $\text{Expression is negative on }$ -1, 5) $\text{ excluding }$ x = 3
Step 3
$\text{Solution set: } [-1, 3) \cup (3, 5)$ ](streamdown:incomplete-link)
Correct
[-1, 3) $\cup$ $\{$ 3 $\}$ $\cup$ (3, 5) = -1, 5)
Because $x = 3$ was mistakenly removed in Step 1, your interval notation has an unnecessary puncture at $x = 3$ .
Step 4
$\text{Solid closed circle at }$ -1, $\text{ shaded to an open circle at }$ 3, $\text{ and another open circle at }$ 3 $\text{ shaded to an open circle at }$ 5.
Correct
$\text{Solid closed circle at }$ -1, $\text{ open circle at }$ 5, $\text{ and a solid continuous shaded segment connecting }$ -1 $\text{ to }$ 5.
Your number line drawing faithfully represents your written interval, but because $x = 3$ is a solution, the point $x = 3$ should be filled in, resulting in a single continuous segment from $-1$ to $5$ .

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly found the roots of the simplified rational expression at $x = -1$ and $x = 5$ .

You correctly distinguished between the numerator root (closed bracket at $-1$ ) and the denominator restriction (open parenthesis at $5$ ).

You accurately drew a number line that matched your stated interval set.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Lost root from division / exclusion of zero-multiplicity factor

You treated $(x - 3)^2$ as strictly positive and discarded $x = 3$ , overlooking that $x = 3$ directly satisfies the $\le 0$ condition.

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Dividing an inequality by a variable expression that can equal zero can silently eliminate solutions. When solving non-strict inequalities ( $P(x) \le 0$ or $P(x) \ge 0$ ), numerator zeros always produce an equality ( $0 = 0$ ), so they must be checked and retained.

REASONING PATH

Your ACEpath teacher’s suggested approach

Check Zeroes Separately Before Dividing: If you divide an inequality by an expression $A(x) \ge 0$ , you must independently test the cases where $A(x) = 0$ .
Plug the root into the original statement: does $0 \le 0$ hold? If yes, keep that point in the final solution.
Keep Factors Instead of Cancelling: Sign charts and interval testing naturally track even multiplicities without dropping boundary points.
Place $x = 3$ on your sign chart with an even-multiplicity flag (sign does not change across $3$ ), and test the point $x = 3$ directly.

CORRECTED PATH

ACEpath Teacher's Demonstration

Check the excluded point $x = 3$ in the original inequality: $\frac{(3 - 3)^2(3 + 1)}{3 - 5} = \frac{0 \cdot 4}{-2} = 0 \le 0$ , which is true.
Re-unite the isolated solution point with the valid intervals: $[-1, 3) \cup \{3\} \cup (3, 5) = [-1, 5)$ .](streamdown:incomplete-link)
Draw the number line with a solid dot at $-1$ , an open circle at $5$ , and shade continuously across $x = 3$ all the way to $5$ .

Final answer

$[-1, 5)$ ; on the number line, a solid closed circle at $x = -1$ , an open circle at $x = 5$ , and a single continuous shaded segment connecting them.](streamdown:incomplete-link)

NEXT STEPS

Final Takeaways from Your Teacher

Review the rule that non-strict inequalities ( $\le 0$ or $\ge 0$ ) include all numerator roots that do not make the denominator zero.
Practice using a complete critical-number sign chart rather than dividing out squared factors to avoid accidentally omitting boundary values.
Complete the targeted flashcards and practice problems below to master even-multiplicity roots in rational inequalities.

Lock it in

Question

How does the sign of a factor $(x - c)^k$ behave as $x$ crosses $c$ if $k$ is an even integer?

Answer

The sign does not change; $(x - c)^k$ stays non-negative on both sides because any real number squared is non-negative.

Resources

Answer and steps

Final answer

Steps

Typical errors

01 The student ignores the even multiplicity of (x - 3)² and alternates signs at every boundary point, resulting in the erroneous exclusion of the interval (3, 5).

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

02 The student forgets that the inequality includes non-strict equality (≤ 0) and includes the denominator boundary x = 5 in the solution set.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

03 The student treats (x - 3)² as strictly positive and cancels it out, mistakenly excluding x = 3 from the interval solution.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

Lock it in

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