Math 3 Quiz: Rational Expression Operations
11 questions · exam conditions
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Rational Expression OperationsQuestion 1 of 11

If Ax−2+Bx+1=5x−1(x−2)(x+1)\frac{A}{x-2} + \frac{B}{x+1} = \frac{5x-1}{(x-2)(x+1)}, what are the values of constants AA and BB?

A=3,B=2A = 3, B = 2
A=2,B=3A = 2, B = 3
A=1,B=4A = 1, B = 4
A=4,B=1A = 4, B = 1
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Math 3 Quiz

Math 3 Quiz: Rational Expression Operations

Practice Rational Expression Operations in Math 3 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Rational Expression Operations, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 3.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

If Ax−2+Bx+1=5x−1(x−2)(x+1)\frac{A}{x-2} + \frac{B}{x+1} = \frac{5x-1}{(x-2)(x+1)}, what are the values of constants AA and BB?

  1. A=3,B=2A = 3, B = 2 (correct answer)
  2. A=2,B=3A = 2, B = 3
  3. A=1,B=4A = 1, B = 4
  4. A=4,B=1A = 4, B = 1
Explanation: Combine the left side: A(x+1)+B(x−2)(x−2)(x+1)=Ax+A+Bx−2B(x−2)(x+1)=(A+B)x+(A−2B)(x−2)(x+1)\frac{A(x+1) + B(x-2)}{(x-2)(x+1)} = \frac{Ax+A+Bx-2B}{(x-2)(x+1)} = \frac{(A+B)x+(A-2B)}{(x-2)(x+1)}. Comparing with 5x−1(x−2)(x+1)\frac{5x-1}{(x-2)(x+1)}: A+B=5A+B = 5 and A−2B=−1A-2B = -1. Solving: From the second equation, A=2B−1A = 2B-1. Substituting: 2B−1+B=52B-1+B = 5, so 3B=63B = 6, thus B=2B = 2 and A=3A = 3. Choice B reverses the values. Choices C and D don't satisfy both equations.

Question 2

When simplifying x2−9x2+6x+9⋅x2+4x+3x2−2x−3\frac{x^2-9}{x^2+6x+9} \cdot \frac{x^2+4x+3}{x^2-2x-3}, what restriction(s) must be placed on xx for the original expression to be defined?

  1. x≠−3,x≠−1,x≠3x \neq -3, x \neq -1, x \neq 3 (correct answer)
  2. x≠−3,x≠3x \neq -3, x \neq 3
  3. x≠−3,x≠−1x \neq -3, x \neq -1
  4. x≠−3,x≠−1,x≠1,x≠3x \neq -3, x \neq -1, x \neq 1, x \neq 3
Explanation: For the original expression to be defined, all denominators must be non-zero. Factor each denominator: x2+6x+9=(x+3)2x^2+6x+9 = (x+3)^2, so x≠−3x \neq -3. Factor x2−2x−3=(x−3)(x+1)x^2-2x-3 = (x-3)(x+1), so x≠3x \neq 3 and x≠−1x \neq -1. The restrictions are x≠−3,−1,3x \neq -3, -1, 3. Choice B misses x≠−1x \neq -1. Choice C misses x≠3x \neq 3. Choice D incorrectly includes x≠1x \neq 1.

Question 3

What is the domain of the function f(x)=x+1x2−5x+6÷x2−1x−2f(x) = \frac{x+1}{x^2-5x+6} \div \frac{x^2-1}{x-2}?

  1. All real numbers except x=1,2,3x = 1, 2, 3
  2. All real numbers except x=−1,1,2,3x = -1, 1, 2, 3 (correct answer)
  3. All real numbers except x=2,3x = 2, 3
  4. All real numbers except x=1,3x = 1, 3
Explanation: Rewrite as x+1x2−5x+6⋅x−2x2−1\frac{x+1}{x^2-5x+6} \cdot \frac{x-2}{x^2-1}. Factor: x2−5x+6=(x−2)(x−3)x^2-5x+6 = (x-2)(x-3) and x2−1=(x−1)(x+1)x^2-1 = (x-1)(x+1). For the domain, exclude values that make any denominator zero in the original expression: x2−5x+6=0x^2-5x+6 = 0 gives x=2,3x = 2, 3; x2−1=0x^2-1 = 0 gives x=±1x = \pm 1; x−2=0x-2 = 0 gives x=2x = 2. So exclude x=−1,1,2,3x = -1, 1, 2, 3. Choice A misses x=−1x = -1. Choices C and D miss multiple restrictions.

Question 4

When x2−4x+3x2−9⋅x2+6x+9x2−2x+1\frac{x^2-4x+3}{x^2-9} \cdot \frac{x^2+6x+9}{x^2-2x+1} is simplified, which factor cancels completely?

  1. No factors cancel completely
  2. (x+3)(x+3)
  3. (x−1)(x-1)
  4. (x−3)(x-3) (correct answer)
Explanation: When you see rational expressions being multiplied, your goal is to factor everything completely, then identify which factors appear in both a numerator and denominator so they can cancel out. Let's factor each polynomial systematically. For x2−4x+3x^2-4x+3, you need two numbers that multiply to 3 and add to -4: that's -1 and -3, giving us (x−1)(x−3)(x-1)(x-3). For x2−9x^2-9, this is a difference of squares: (x−3)(x+3)(x-3)(x+3). The expression x2+6x+9x^2+6x+9 is a perfect square trinomial: (x+3)2(x+3)^2. Finally, x2−2x+1x^2-2x+1 is also a perfect square: (x−1)2(x-1)^2. Now we can rewrite the multiplication as: (x−1)(x−3)(x−3)(x+3)⋅(x+3)2(x−1)2\frac{(x-1)(x-3)}{(x-3)(x+3)} \cdot \frac{(x+3)^2}{(x-1)^2} Looking at the factors across all numerators and denominators, we see (x−3)(x-3) appears once in a numerator and once in a denominator, so it cancels completely. Choice A is wrong because we clearly have factors that cancel. Choice B is incorrect because (x+3)(x+3) appears twice in numerators but only once in a denominator, so one copy remains. Choice C is wrong because (x−1)(x-1) appears once in a numerator but twice in denominators, leaving one copy in the denominator. Choice D is correct because (x−3)(x-3) cancels entirely. Study tip: When simplifying rational expressions, always factor completely first, then systematically count how many times each factor appears in numerators versus denominators. A factor cancels completely only when it appears the same number of times in both positions.

Question 5

The expression 3xx2−9+2x+3\frac{3x}{x^2-9} + \frac{2}{x+3} simplifies to:

  1. 5x+6x2−9\frac{5x+6}{x^2-9}
  2. 2x−6x2−9\frac{2x-6}{x^2-9}
  3. 5x−6x2−9\frac{5x-6}{x^2-9} (correct answer)
  4. 2x+6x2−9\frac{2x+6}{x^2-9}
Explanation: Factor x2−9=(x−3)(x+3)x^2-9 = (x-3)(x+3). Rewrite with common denominator: 3x(x−3)(x+3)+2(x−3)(x−3)(x+3)=3x+2(x−3)(x−3)(x+3)=3x+2x−6x2−9=5x−6x2−9\frac{3x}{(x-3)(x+3)} + \frac{2(x-3)}{(x-3)(x+3)} = \frac{3x + 2(x-3)}{(x-3)(x+3)} = \frac{3x + 2x - 6}{x^2-9} = \frac{5x-6}{x^2-9}. Choice A adds 6 instead of subtracting. Choice B uses wrong coefficient for xx. Choice D has wrong signs.

Question 6

What is the simplified form of x2−4x+1x2−4x+4x2−1\frac{\frac{x^2-4}{x+1}}{\frac{x^2-4x+4}{x^2-1}}?

  1. x+2x−2\frac{x+2}{x-2}
  2. (x+2)(x−1)x−2\frac{(x+2)(x-1)}{x-2} (correct answer)
  3. x+2x+1\frac{x+2}{x+1}
  4. (x−2)(x+1)x+2\frac{(x-2)(x+1)}{x+2}
Explanation: Rewrite as multiplication: x2−4x+1⋅x2−1x2−4x+4\frac{x^2-4}{x+1} \cdot \frac{x^2-1}{x^2-4x+4}. Factor: (x−2)(x+2)x+1⋅(x−1)(x+1)(x−2)2\frac{(x-2)(x+2)}{x+1} \cdot \frac{(x-1)(x+1)}{(x-2)^2}. Cancel (x−2)(x-2) and (x+1)(x+1): (x+2)(x−1)x−2\frac{(x+2)(x-1)}{x-2}. Choice A incorrectly cancels (x−1)(x-1). Choice C has wrong denominator. Choice D inverts the fraction.

Question 7

The expression 2xx−1+3xx−1−1\frac{\frac{2x}{x-1} + 3}{\frac{x}{x-1} - 1} simplifies to:

  1. 2x+3(x−1)x−(x−1)\frac{2x+3(x-1)}{x-(x-1)}
  2. 5x−31\frac{5x-3}{1}
  3. 5x−35x-3 (correct answer)
  4. 5x−3x−1\frac{5x-3}{x-1}
Explanation: Simplify numerator: 2xx−1+3=2x+3(x−1)x−1=2x+3x−3x−1=5x−3x−1\frac{2x}{x-1} + 3 = \frac{2x + 3(x-1)}{x-1} = \frac{2x + 3x - 3}{x-1} = \frac{5x-3}{x-1}. Simplify denominator: xx−1−1=x−1(x−1)x−1=x−x+1x−1=1x−1\frac{x}{x-1} - 1 = \frac{x - 1(x-1)}{x-1} = \frac{x - x + 1}{x-1} = \frac{1}{x-1}. So the expression becomes: 5x−3x−11x−1=5x−3x−1⋅x−11=5x−3\frac{\frac{5x-3}{x-1}}{\frac{1}{x-1}} = \frac{5x-3}{x-1} \cdot \frac{x-1}{1} = 5x-3. Choice A shows the unreduced form. Choice B adds unnecessary 1 in denominator. Choice D doesn't complete the division.

Question 8

The expression 2x2−8x2−5x+6÷x2−4x2−6x+9\frac{2x^2-8}{x^2-5x+6} \div \frac{x^2-4}{x^2-6x+9} simplifies to which form, and what restrictions apply to the domain?

  1. 2(x−3)x−2\frac{2(x-3)}{x-2}; x≠2,3x \neq 2, 3
  2. 2(x−3)x−2\frac{2(x-3)}{x-2}; x≠2,3,−2x \neq 2, 3, -2 (correct answer)
  3. 2(x+2)(x−3)(x−2)2\frac{2(x+2)(x-3)}{(x-2)^2}; x≠2,3x \neq 2, 3
  4. 2(x+2)(x−3)(x−2)2\frac{2(x+2)(x-3)}{(x-2)^2}; x≠2,3,−2x \neq 2, 3, -2
Explanation: First, factor all expressions: 2x2−8=2(x2−4)=2(x−2)(x+2)2x^2-8 = 2(x^2-4) = 2(x-2)(x+2), x2−5x+6=(x−2)(x−3)x^2-5x+6 = (x-2)(x-3), x2−4=(x−2)(x+2)x^2-4 = (x-2)(x+2), x2−6x+9=(x−3)2x^2-6x+9 = (x-3)^2. The division becomes: 2(x−2)(x+2)(x−2)(x−3)⋅(x−3)2(x−2)(x+2)\frac{2(x-2)(x+2)}{(x-2)(x-3)} \cdot \frac{(x-3)^2}{(x-2)(x+2)}. Simplifying: 2(x−2)(x+2)⋅(x−3)2(x−2)(x−3)⋅(x−2)(x+2)=2(x−3)2(x−2)(x−3)=2(x−3)x−2\frac{2(x-2)(x+2) \cdot (x-3)^2}{(x-2)(x-3) \cdot (x-2)(x+2)} = \frac{2(x-3)^2}{(x-2)(x-3)} = \frac{2(x-3)}{x-2}. For domain restrictions, we exclude values that make any original denominator zero: From x2−5x+6=0x^2-5x+6 = 0: x=2,3x = 2, 3. From x2−6x+9=0x^2-6x+9 = 0: x=3x = 3 (already included). From x2−4=0x^2-4 = 0 (appears in division): x=±2x = \pm 2. So x≠2,3,−2x \neq 2, 3, -2. Choice A misses the restriction x≠−2x \neq -2. Choices C and D show an incompletely simplified form.

Question 9

Given that f(x)=x+1x−3f(x) = \frac{x+1}{x-3} and g(x)=2x−1x+4g(x) = \frac{2x-1}{x+4}, what is the simplified form of f(x)g(x)\frac{f(x)}{g(x)} and what are the domain restrictions?

  1. (x+1)(x+4)(x−3)(2x−1)\frac{(x+1)(x+4)}{(x-3)(2x-1)}; x≠3,−4,12x \neq 3, -4, \frac{1}{2} (correct answer)
  2. (x+1)(x+4)(x−3)(2x−1)\frac{(x+1)(x+4)}{(x-3)(2x-1)}; x≠3,−4x \neq 3, -4
  3. 2x2+7x−42x2−7x+3\frac{2x^2+7x-4}{2x^2-7x+3}; x≠3,−4,12x \neq 3, -4, \frac{1}{2}
  4. 2x2+7x−42x2−7x+3\frac{2x^2+7x-4}{2x^2-7x+3}; x≠3,−4x \neq 3, -4
Explanation: f(x)g(x)=x+1x−32x−1x+4=x+1x−3⋅x+42x−1=(x+1)(x+4)(x−3)(2x−1)\frac{f(x)}{g(x)} = \frac{\frac{x+1}{x-3}}{\frac{2x-1}{x+4}} = \frac{x+1}{x-3} \cdot \frac{x+4}{2x-1} = \frac{(x+1)(x+4)}{(x-3)(2x-1)}. For domain restrictions, we need to exclude values that make any denominator zero in the original functions or the division: From f(x)f(x): x≠3x \neq 3; From g(x)g(x): x≠−4x \neq -4; From the division (where g(x)=0g(x) = 0): 2x−1=02x-1 = 0, so x≠12x \neq \frac{1}{2}. Therefore, x≠3,−4,12x \neq 3, -4, \frac{1}{2}. Choice B misses the restriction from g(x)=0g(x) = 0. Choices C and D show the expanded form but this is not simplified since factored form is preferred when possible.

Question 10

If 2xx−1−x+3x+2=x2+7x+6(x−1)(x+2)\frac{2x}{x-1} - \frac{x+3}{x+2} = \frac{x^2+7x+6}{(x-1)(x+2)}, what is the value of the coefficient of x2x^2 in the numerator on the left side after combining the fractions?

  1. 1 (correct answer)
  2. 2
  3. -1
  4. 0
Explanation: To combine the fractions on the left side, we need a common denominator of (x−1)(x+2)(x-1)(x+2). This gives us: 2x(x+2)−(x+3)(x−1)(x−1)(x+2)\frac{2x(x+2) - (x+3)(x-1)}{(x-1)(x+2)}. Expanding the numerator: 2x(x+2)=2x2+4x2x(x+2) = 2x^2 + 4x and (x+3)(x−1)=x2+3x−x−3=x2+2x−3(x+3)(x-1) = x^2 + 3x - x - 3 = x^2 + 2x - 3. So the numerator becomes: 2x2+4x−(x2+2x−3)=2x2+4x−x2−2x+3=x2+2x+32x^2 + 4x - (x^2 + 2x - 3) = 2x^2 + 4x - x^2 - 2x + 3 = x^2 + 2x + 3. The coefficient of x2x^2 is 1. Choice B (2) comes from not distributing the negative sign. Choice C (-1) comes from incorrectly subtracting 2x2−x22x^2 - x^2. Choice D (0) comes from thinking the x2x^2 terms cancel completely.

Question 11

What is the result when x2+5x+6x2−4⋅x2−x−2x2+4x+3\frac{x^2+5x+6}{x^2-4} \cdot \frac{x^2-x-2}{x^2+4x+3} is multiplied and simplified to lowest terms?

  1. x−2x−1\frac{x-2}{x-1}
  2. x+2x+1\frac{x+2}{x+1}
  3. (x+2)(x−2)(x+1)(x−1)\frac{(x+2)(x-2)}{(x+1)(x-1)}
  4. 11 (correct answer)
Explanation: First, factor each polynomial: x2+5x+6=(x+2)(x+3)x^2+5x+6 = (x+2)(x+3), x2−4=(x−2)(x+2)x^2-4 = (x-2)(x+2), x2−x−2=(x−2)(x+1)x^2-x-2 = (x-2)(x+1), x2+4x+3=(x+1)(x+3)x^2+4x+3 = (x+1)(x+3). The multiplication becomes: (x+2)(x+3)(x−2)(x+2)⋅(x−2)(x+1)(x+1)(x+3)\frac{(x+2)(x+3)}{(x-2)(x+2)} \cdot \frac{(x-2)(x+1)}{(x+1)(x+3)}. Canceling common factors: (x+2)(x+2) cancels, (x+3)(x+3) cancels, (x−2)(x-2) cancels, and (x+1)(x+1) cancels, leaving 11=1\frac{1}{1} = 1. Choice A results from incorrectly canceling factors. Choice B comes from partial cancellation errors. Choice C shows the expression before complete simplification.