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Rational Expression Operations - Math 3

  1. 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?
    Answer: (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} 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.
  2. 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?
    Answer: A=3,B=2A = 3, B = 2 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.
  3. 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?
    Answer: (x−3)(x-3) 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.
  4. The expression 3xx2−9+2x+3\frac{3x}{x^2-9} + \frac{2}{x+3} simplifies to:
    Answer: 5x−6x2−9\frac{5x-6}{x^2-9} 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.
  5. 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}?
    Answer: All real numbers except x=−1,1,2,3x = -1, 1, 2, 3 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.
  6. 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?
    Answer: x≠−3,x≠−1,x≠3x \neq -3, x \neq -1, x \neq 3 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.
  7. The expression 2xx−1+3xx−1−1\frac{\frac{2x}{x-1} + 3}{\frac{x}{x-1} - 1} simplifies to:
    Answer: 5x−35x-3 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.
  8. 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?
    Answer: 11 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.
  9. 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}}?
    Answer: (x+2)(x−1)x−2\frac{(x+2)(x-1)}{x-2} 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.
  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?
    Answer: 1 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.
  11. 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?
    Answer: 2(x−3)x−2\frac{2(x-3)}{x-2}; x≠2,3,−2x \neq 2, 3, -2 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.

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