What is Polynomial and Rational Expressions?

Factor, simplify, and operate with polynomial and rational expressions.

How can I study Polynomial and Rational Expressions effectively?

Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 15 problems provided, checking solutions as you go. Regular review and active practice are key to retention.

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What course covers Polynomial and Rational Expressions?

Polynomial and Rational Expressions is part of the SAT Prep course on Study Mondo, specifically in the Passport to Advanced Math section. You can explore the full course for more related topics and practice resources.

Are there practice problems for Polynomial and Rational Expressions?

Yes, this page includes 15 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.

Polynomial and Rational Expressions on the SAT

Rational Expressions

❓ Question:

Simplify: $\frac{x^2 - 9}{x + 3}$

Step 1: Factor the numerator (difference of squares): $\frac{(x+3)(x-3)}{x+3}$

Step 2: Cancel the common factor :

❓ Question:

Simplify: $\frac{x^2 - 9}{x + 3}$

❓ Question:

Simplify: $\frac{x^2 - 9}{x + 3}$

❓ Question:

Add: $\frac{2}{x-1} + \frac{3}{x+2}$

❓ Question:

Add: $\frac{2}{x-1} + \frac{3}{x+2}$

❓ Question:

Add: $\frac{2}{x-1} + \frac{3}{x+2}$

❓ Question:

For what values of $x$ is $\frac{x^2 + 2x - 15}{x^2 - x - 6}$ undefined?

❓ Question:

For what values of $x$ is $\frac{x^2 + 2x - 15}{x^2 - x - 6}$ undefined?

❓ Question:

For what values of $x$ is $\frac{x^2 + 2x - 15}{x^2 - x - 6}$ undefined?

❓ Question:

Solve: $\frac{4}{x-2} = \frac{x}{x-2} + 2$

❓ Question:

Solve: $\frac{4}{x-2} = \frac{x}{x-2} + 2$

❓ Question:

Solve: $\frac{4}{x-2} = \frac{x}{x-2} + 2$

❓ Question:

Solve: $\frac{2}{x+1} + \frac{1}{x-1} = \frac{4}{x^2-1}$

❓ Question:

Solve: $\frac{2}{x+1} + \frac{1}{x-1} = \frac{4}{x^2-1}$

❓ Question:

Solve: $\frac{2}{x+1} + \frac{1}{x-1} = \frac{4}{x^2-1}$

Step 1: Factor the numerator (difference of squares): $\frac{(x+3)(x-3)}{x+3}$

Step 2: Cancel the common factor $(x+3)$ : $= x - 3 \quad (x \neq -3)$

Answer: $x - 3$ , provided $x \neq -3$

Step 1: Factor the numerator (difference of squares): $\frac{(x+3)(x-3)}{x+3}$

Step 2: Cancel the common factor $(x+3)$ : $= x - 3 \quad (x \neq -3)$

Answer: $x - 3$ , provided $x \neq -3$

Step 1: Find the LCD: $(x-1)(x+2)$

Step 2: Rewrite each fraction with the LCD: $\frac{2(x+2)}{(x-1)(x+2)} + \frac{3(x-1)}{(x-1)(x+2)}$

Step 3: Add the numerators: $\frac{2(x+2) + 3(x-1)}{(x-1)(x+2)} = \frac{2x+4+3x-3}{(x-1)(x+2)} = \frac{5x+1}{(x-1)(x+2)}$

Answer: $\frac{5x+1}{(x-1)(x+2)}$

Step 1: Find the LCD: $(x-1)(x+2)$

Step 2: Rewrite each fraction with the LCD: $\frac{2(x+2)}{(x-1)(x+2)} + \frac{3(x-1)}{(x-1)(x+2)}$

Step 3: Add the numerators: $\frac{2(x+2) + 3(x-1)}{(x-1)(x+2)} = \frac{2x+4+3x-3}{(x-1)(x+2)} = \frac{5x+1}{(x-1)(x+2)}$

Answer: $\frac{5x+1}{(x-1)(x+2)}$

Step 1: Find the LCD: $(x-1)(x+2)$

Step 2: Rewrite each fraction with the LCD: $\frac{2(x+2)}{(x-1)(x+2)} + \frac{3(x-1)}{(x-1)(x+2)}$

Step 3: Add the numerators: $\frac{2(x+2) + 3(x-1)}{(x-1)(x+2)} = \frac{2x+4+3x-3}{(x-1)(x+2)} = \frac{5x+1}{(x-1)(x+2)}$

Answer: $\frac{5x+1}{(x-1)(x+2)}$

Step 1: The expression is undefined when the denominator = 0.

$x^2 - x - 6 = 0$

Step 2: Factor: $(x-3)(x+2) = 0$

Step 3: Solve: $x = 3 \quad \text{or} \quad x = -2$

Answer: The expression is undefined at $x = 3$ and $x = -2$ .

Note: Even though the full expression simplifies (the numerator factors to $(x+5)(x-3)$ , and $(x-3)$ cancels), $x = 3$ is still a restriction because it was in the original denominator.

Step 1: The expression is undefined when the denominator = 0.

$x^2 - x - 6 = 0$

Step 2: Factor: $(x-3)(x+2) = 0$

Step 3: Solve: $x = 3 \quad \text{or} \quad x = -2$

Answer: The expression is undefined at $x = 3$ and $x = -2$ .

Note: Even though the full expression simplifies (the numerator factors to $(x+5)(x-3)$ , and $(x-3)$ cancels), $x = 3$ is still a restriction because it was in the original denominator.

Step 1: The expression is undefined when the denominator = 0.

$x^2 - x - 6 = 0$

Step 2: Factor: $(x-3)(x+2) = 0$

Step 3: Solve: $x = 3 \quad \text{or} \quad x = -2$

Answer: The expression is undefined at $x = 3$ and $x = -2$ .

Note: Even though the full expression simplifies (the numerator factors to $(x+5)(x-3)$ , and $(x-3)$ cancels), $x = 3$ is still a restriction because it was in the original denominator.

Step 1: Note the domain restriction: $x \neq 2$

Step 2: Multiply both sides by $(x-2)$ : $4 = x + 2(x-2)$ $4 = x + 2x - 4$ $4 = 3x - 4$ $8 = 3x$ $x = \frac{8}{3}$

Step 3: Check: $x = \frac{8}{3} \neq 2$ , so it's valid. ✓

Verify: $\frac{4}{\frac{8}{3}-2} = \frac{4}{\frac{2}{3}} = 6$ and ✓

Answer: $x = \frac{8}{3}$

Step 1: Note the domain restriction: $x \neq 2$

Step 2: Multiply both sides by $(x-2)$ : $4 = x + 2(x-2)$ $4 = x + 2x - 4$ $4 = 3x - 4$ $8 = 3x$ $x = \frac{8}{3}$

Step 3: Check: $x = \frac{8}{3} \neq 2$ , so it's valid. ✓

Verify: $\frac{4}{\frac{8}{3}-2} = \frac{4}{\frac{2}{3}} = 6$ and ✓

Answer: $x = \frac{8}{3}$

Step 1: Note the domain restriction: $x \neq 2$

Step 2: Multiply both sides by $(x-2)$ : $4 = x + 2(x-2)$ $4 = x + 2x - 4$ $4 = 3x - 4$ $8 = 3x$ $x = \frac{8}{3}$

Step 3: Check: $x = \frac{8}{3} \neq 2$ , so it's valid. ✓

Verify: $\frac{4}{\frac{8}{3}-2} = \frac{4}{\frac{2}{3}} = 6$ and ✓

Answer: $x = \frac{8}{3}$

Step 1: Note that $x^2 - 1 = (x+1)(x-1)$ , so LCD = $(x+1)(x-1)$

Domain restrictions: $x \neq 1$ and $x \neq -1$

Step 2: Multiply every term by $(x+1)(x-1)$ : $2(x-1) + 1(x+1) = 4$

Step 3: Distribute and solve: $2x - 2 + x + 1 = 4$ $3x - 1 = 4$

Step 4: Check: $x = \frac{5}{3} \neq \pm 1$ , so it's valid. ✓

Answer: $x = \frac{5}{3}$

SAT Tip: Always factor the denominators first to find the LCD and identify domain restrictions.

Step 1: Note that $x^2 - 1 = (x+1)(x-1)$ , so LCD = $(x+1)(x-1)$

Domain restrictions: $x \neq 1$ and $x \neq -1$

Step 2: Multiply every term by $(x+1)(x-1)$ : $2(x-1) + 1(x+1) = 4$

Step 3: Distribute and solve: $2x - 2 + x + 1 = 4$ $3x - 1 = 4$

Step 4: Check: $x = \frac{5}{3} \neq \pm 1$ , so it's valid. ✓

Answer: $x = \frac{5}{3}$

SAT Tip: Always factor the denominators first to find the LCD and identify domain restrictions.

Step 1: Note that $x^2 - 1 = (x+1)(x-1)$ , so LCD = $(x+1)(x-1)$

Domain restrictions: $x \neq 1$ and $x \neq -1$

Step 2: Multiply every term by $(x+1)(x-1)$ : $2(x-1) + 1(x+1) = 4$

Step 3: Distribute and solve: $2x - 2 + x + 1 = 4$ $3x - 1 = 4$

Step 4: Check: $x = \frac{5}{3} \neq \pm 1$ , so it's valid. ✓

Answer: $x = \frac{5}{3}$

SAT Tip: Always factor the denominators first to find the LCD and identify domain restrictions.

Polynomial and Rational Expressions

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Polynomial and Rational Expressions on the SAT

Rational Expressions

Simplifying Rational Expressions

Operations with Rational Expressions

Multiplication

Division

Addition/Subtraction

Domain Restrictions

Solving Rational Equations

SAT Question Types

Type 1: Simplify a Rational Expression

Type 2: Find the Domain

Type 3: Add/Subtract Rational Expressions

Type 4: Solve a Rational Equation

Common SAT Mistakes

📚 Practice Problems

1Problem 1easy

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2Problem 2easy

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3Problem 3easy

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4Problem 4medium

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5Problem 5medium

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6Problem 6medium

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7Problem 7medium

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8Problem 8medium

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9Problem 9medium

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10Problem 10hard

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11Problem 11hard

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12Problem 12hard

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13Problem 13expert

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14Problem 14expert

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15Problem 15expert

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