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Polynomial and Rational Expressions

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

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Polynomial and Rational Expressions on the SAT

Rational Expressions

A rational expression is a fraction where the numerator and/or denominator are polynomials: P(x)Q(x)where Q(x)≠0\frac{P(x)}{Q(x)} \quad \text{where } Q(x) \neq 0

Simplifying Rational Expressions

Factor and cancel common factors.

x2−4x2+4x+4=(x−2)(x+2)(x+2)2=x−2x+2\frac{x^2 - 4}{x^2 + 4x + 4} = \frac{(x-2)(x+2)}{(x+2)^2} = \frac{x-2}{x+2}

Critical: You can only cancel FACTORS (things being multiplied), never terms (things being added).

WRONG: x+3x+5≠35\frac{x + 3}{x + 5} \neq \frac{3}{5} ← Cannot cancel the xx's!

Operations with Rational Expressions

Multiplication

Factor, cancel, then multiply: x2−1x+3⋅x+3x−1=(x+1)(x−1)x+3⋅x+3x−1=x+1\frac{x^2-1}{x+3} \cdot \frac{x+3}{x-1} = \frac{(x+1)(x-1)}{x+3} \cdot \frac{x+3}{x-1} = x+1

Division

Flip the second fraction and multiply: ab÷cd=ab⋅dc\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \cdot \frac{d}{c}

Addition/Subtraction

Find a common denominator first: 1x+2x+1=x+1x(x+1)+2xx(x+1)=3x+1x(x+1)\frac{1}{x} + \frac{2}{x+1} = \frac{x+1}{x(x+1)} + \frac{2x}{x(x+1)} = \frac{3x+1}{x(x+1)}

Domain Restrictions

A rational expression is undefined when the denominator equals zero.

x+3x2−9=x+3(x+3)(x−3)\frac{x+3}{x^2-9} = \frac{x+3}{(x+3)(x-3)}

Domain restrictions: x≠3x \neq 3 and x≠−3x \neq -3

Even after simplifying to 1x−3\frac{1}{x-3}, the restriction x≠−3x \neq -3 still applies!

Solving Rational Equations

Strategy: Multiply both sides by the LCD to clear fractions.

3x+2x+1=1\frac{3}{x} + \frac{2}{x+1} = 1

LCD = x(x+1)x(x+1): 3(x+1)+2x=x(x+1)3(x+1) + 2x = x(x+1) 3x+3+2x=x2+x3x + 3 + 2x = x^2 + x 5x+3=x2+x5x + 3 = x^2 + x 0=x2−4x−30 = x^2 - 4x - 3

Always check for extraneous solutions! Plug your answers back in to make sure the denominators aren't zero.

SAT Question Types

Type 1: Simplify a Rational Expression

Factor and cancel.

Type 2: Find the Domain

Identify values that make the denominator zero.

Type 3: Add/Subtract Rational Expressions

Find common denominators and combine.

Type 4: Solve a Rational Equation

Clear fractions, solve, and check for extraneous solutions.

Common SAT Mistakes

  1. Canceling terms instead of factors: x+3x+5\frac{x+3}{x+5} cannot be simplified!
  2. Forgetting domain restrictions after simplifying
  3. Not checking for extraneous solutions — solutions that make a denominator zero must be rejected
  4. Incorrect LCD — make sure to include all unique factors
  5. Sign errors when distributing negatives in subtraction of rational expressions

📚 Practice Problems

1Problem 1easy

❓ Question:

Simplify: x2−9x+3\frac{x^2 - 9}{x + 3}

💡 Show Solution

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

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

Answer: x−3x - 3, provided x≠−3x \neq -3

2Problem 2easy

❓ Question:

Simplify: x2−9x+3\frac{x^2 - 9}{x + 3}

💡 Show Solution

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

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

Answer: x−3x - 3, provided x≠−3x \neq -3

3Problem 3easy

❓ Question:

Simplify: x2−9x+3\frac{x^2 - 9}{x + 3}

💡 Show Solution

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

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

Answer: x−3x - 3, provided x≠−3x \neq -3

4Problem 4medium

❓ Question:

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

💡 Show Solution

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

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

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

5Problem 5medium

❓ Question:

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

💡 Show Solution

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

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

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

6Problem 6medium

❓ Question:

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

💡 Show Solution

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

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

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

7Problem 7medium

❓ Question:

For what values of xx is x2+2x−15x2−x−6\frac{x^2 + 2x - 15}{x^2 - x - 6} undefined?

💡 Show Solution

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

x2−x−6=0x^2 - x - 6 = 0

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

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

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

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

8Problem 8medium

❓ Question:

For what values of xx is x2+2x−15x2−x−6\frac{x^2 + 2x - 15}{x^2 - x - 6} undefined?

💡 Show Solution

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

x2−x−6=0x^2 - x - 6 = 0

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

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

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

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

9Problem 9medium

❓ Question:

For what values of xx is x2+2x−15x2−x−6\frac{x^2 + 2x - 15}{x^2 - x - 6} undefined?

💡 Show Solution

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

x2−x−6=0x^2 - x - 6 = 0

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

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

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

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

10Problem 10hard

❓ Question:

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

💡 Show Solution

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

Step 2: Multiply both sides by (x−2)(x-2): 4=x+2(x−2)4 = x + 2(x-2) 4=x+2x−44 = x + 2x - 4 4=3x−44 = 3x - 4 8=3x8 = 3x x=83x = \frac{8}{3}

Step 3: Check: x=83≠2x = \frac{8}{3} \neq 2, so it's valid. ✓

Verify: 483−2=423=6\frac{4}{\frac{8}{3}-2} = \frac{4}{\frac{2}{3}} = 6 and 8323+2=4+2=6\frac{\frac{8}{3}}{\frac{2}{3}} + 2 = 4 + 2 = 6 ✓

Answer: x=83x = \frac{8}{3}

11Problem 11hard

❓ Question:

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

💡 Show Solution

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

Step 2: Multiply both sides by (x−2)(x-2): 4=x+2(x−2)4 = x + 2(x-2) 4=x+2x−44 = x + 2x - 4 4=3x−44 = 3x - 4 8=3x8 = 3x x=83x = \frac{8}{3}

Step 3: Check: x=83≠2x = \frac{8}{3} \neq 2, so it's valid. ✓

Verify: 483−2=423=6\frac{4}{\frac{8}{3}-2} = \frac{4}{\frac{2}{3}} = 6 and 8323+2=4+2=6\frac{\frac{8}{3}}{\frac{2}{3}} + 2 = 4 + 2 = 6 ✓

Answer: x=83x = \frac{8}{3}

12Problem 12hard

❓ Question:

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

💡 Show Solution

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

Step 2: Multiply both sides by (x−2)(x-2): 4=x+2(x−2)4 = x + 2(x-2) 4=x+2x−44 = x + 2x - 4 4=3x−44 = 3x - 4 8=3x8 = 3x x=83x = \frac{8}{3}

Step 3: Check: x=83≠2x = \frac{8}{3} \neq 2, so it's valid. ✓

Verify: 483−2=423=6\frac{4}{\frac{8}{3}-2} = \frac{4}{\frac{2}{3}} = 6 and 8323+2=4+2=6\frac{\frac{8}{3}}{\frac{2}{3}} + 2 = 4 + 2 = 6 ✓

Answer: x=83x = \frac{8}{3}

13Problem 13expert

❓ Question:

Solve: 2x+1+1x−1=4x2−1\frac{2}{x+1} + \frac{1}{x-1} = \frac{4}{x^2-1}

💡 Show Solution

Step 1: Note that x2−1=(x+1)(x−1)x^2 - 1 = (x+1)(x-1), so LCD = (x+1)(x−1)(x+1)(x-1)

Domain restrictions: x≠1x \neq 1 and x≠−1x \neq -1

Step 2: Multiply every term by (x+1)(x−1)(x+1)(x-1): 2(x−1)+1(x+1)=42(x-1) + 1(x+1) = 4

Step 3: Distribute and solve: 2x−2+x+1=42x - 2 + x + 1 = 4 3x−1=43x - 1 = 4 3x=53x = 5 x=53x = \frac{5}{3}

Step 4: Check: x=53≠±1x = \frac{5}{3} \neq \pm 1, so it's valid. ✓

Answer: x=53x = \frac{5}{3}

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

14Problem 14expert

❓ Question:

Solve: 2x+1+1x−1=4x2−1\frac{2}{x+1} + \frac{1}{x-1} = \frac{4}{x^2-1}

💡 Show Solution

Step 1: Note that x2−1=(x+1)(x−1)x^2 - 1 = (x+1)(x-1), so LCD = (x+1)(x−1)(x+1)(x-1)

Domain restrictions: x≠1x \neq 1 and x≠−1x \neq -1

Step 2: Multiply every term by (x+1)(x−1)(x+1)(x-1): 2(x−1)+1(x+1)=42(x-1) + 1(x+1) = 4

Step 3: Distribute and solve: 2x−2+x+1=42x - 2 + x + 1 = 4 3x−1=43x - 1 = 4 3x=53x = 5 x=53x = \frac{5}{3}

Step 4: Check: x=53≠±1x = \frac{5}{3} \neq \pm 1, so it's valid. ✓

Answer: x=53x = \frac{5}{3}

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

15Problem 15expert

❓ Question:

Solve: 2x+1+1x−1=4x2−1\frac{2}{x+1} + \frac{1}{x-1} = \frac{4}{x^2-1}

💡 Show Solution

Step 1: Note that x2−1=(x+1)(x−1)x^2 - 1 = (x+1)(x-1), so LCD = (x+1)(x−1)(x+1)(x-1)

Domain restrictions: x≠1x \neq 1 and x≠−1x \neq -1

Step 2: Multiply every term by (x+1)(x−1)(x+1)(x-1): 2(x−1)+1(x+1)=42(x-1) + 1(x+1) = 4

Step 3: Distribute and solve: 2x−2+x+1=42x - 2 + x + 1 = 4 3x−1=43x - 1 = 4 3x=53x = 5 x=53x = \frac{5}{3}

Step 4: Check: x=53≠±1x = \frac{5}{3} \neq \pm 1, so it's valid. ✓

Answer: x=53x = \frac{5}{3}

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

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❓ Frequently Asked Questions

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.
Is this Polynomial and Rational Expressions study guide free?▾
Yes — all study notes, flashcards, and practice problems for Polynomial and Rational Expressions on Study Mondo are free to access. No account is needed.
What course covers Polynomial and Rational Expressions?▾
Polynomial and Rational Expressions is part of the SAT Prep course on Study Mondo, specifically in the 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.