Simplifying Rational Expressions - Math 3

What is the completely simplified form of $\frac{\frac{x^2-1}{x+2}}{\frac{x^2+2x+1}{x^2+3x+2}}$ ?
Answer: $x-1$ with restrictions $x \neq -1, -2$ First, rewrite the complex fraction as division: $\frac{x^2-1}{x+2} \div \frac{x^2+2x+1}{x^2+3x+2}$ . Factor everything: $x^2-1 = (x-1)(x+1)$ , $x^2+2x+1 = (x+1)^2$ , $x^2+3x+2 = (x+1)(x+2)$ . The expression becomes: $\frac{(x-1)(x+1)}{x+2} \cdot \frac{(x+1)(x+2)}{(x+1)^2} = \frac{(x-1)(x+1)(x+1)(x+2)}{(x+2)(x+1)^2} = \frac{(x-1)(x+1)^2(x+2)}{(x+2)(x+1)^2} = x-1$ . Domain restrictions come from all original denominators: $x+2 \neq 0$ , $(x+1)^2 \neq 0$ , and $(x+1)(x+2) \neq 0$ , giving $x \neq -1, -2$ .
Consider the expression $\frac{a^3 + b^3}{a^2 - b^2} \div \frac{a^2 + ab + b^2}{a + b}$ . After simplifying completely, what restrictions must be placed on the variables?
Answer: $a \neq b$ and $a \neq -b$ and $a^2 + ab + b^2 \neq 0$ When working with rational expressions involving division, you need to identify all values that make any denominator zero, both in the original expression and during the simplification process. Let's simplify this step by step. First, rewrite the division as multiplication: $\frac{a^3 + b^3}{a^2 - b^2} \times \frac{a + b}{a^2 + ab + b^2}$ Now factor what you can. The numerator $a^3 + b^3$ factors as $(a + b)(a^2 - ab + b^2)$ using the sum of cubes formula, and the denominator $a^2 - b^2$ factors as $(a + b)(a - b)$ using difference of squares. This gives us: $\frac{(a + b)(a^2 - ab + b^2)}{(a + b)(a - b)} \times \frac{a + b}{a^2 + ab + b^2} = \frac{(a + b)^2(a^2 - ab + b^2)}{(a - b)(a^2 + ab + b^2)}$ The restrictions come from setting each factor in any denominator equal to zero: $a - b = 0$ gives $a \neq b$ , $a + b = 0$ gives $a \neq -b$ , and $a^2 + ab + b^2 = 0$ gives $a^2 + ab + b^2 \neq 0$ . Answer choice A misses the restriction $a^2 + ab + b^2 \neq 0$ and incorrectly adds $a \neq 0$ and $b \neq 0$ . Answer choice B omits the crucial restriction $a^2 + ab + b^2 \neq 0$ . Answer choice C only identifies one restriction, missing $a \neq b$ and $a^2 + ab + b^2 \neq 0$ . Answer choice D correctly identifies all three necessary restrictions. Remember: when simplifying rational expressions, always consider restrictions from the original denominators before canceling common factors.
For what value(s) of $k$ will the rational expression $\frac{x^2 + kx - 12}{x^2 - 9}$ simplify to have a removable discontinuity rather than a vertical asymptote at $x = 3$ ?
Answer: $k = 1$ only When you encounter rational expressions with potential discontinuities, the key distinction is between removable discontinuities (holes) and vertical asymptotes. A removable discontinuity occurs when both the numerator and denominator have the same factor that can be canceled out. First, let's factor the denominator: $x^2 - 9 = (x-3)(x+3)$ . Since we want a removable discontinuity at $x = 3$ , the numerator must also have $(x-3)$ as a factor, allowing it to cancel with the denominator. For $x^2 + kx - 12$ to have $(x-3)$ as a factor, substituting $x = 3$ must equal zero: $3^2 + k(3) - 12 = 0$ $9 + 3k - 12 = 0$ $3k - 3 = 0$ $k = 1$ Let's verify: when $k = 1$ , we have $x^2 + x - 12 = (x-3)(x+4)$ . The expression becomes: $\frac{(x-3)(x+4)}{(x-3)(x+3)} = \frac{x+4}{x+3}$ (for $x \neq 3$ ) This creates a hole at $x = 3$ , not a vertical asymptote. Answer A suggests both $k = 1$ and $k = -1$ work, but $k = -1$ gives us $x^2 - x - 12 = (x-4)(x+3)$ , which doesn't contain the factor $(x-3)$ . Answer B incorrectly claims only $k = -1$ works. Answer D is wrong because we've shown $k = 1$ does create a removable discontinuity. Strategy tip: For removable discontinuities, always check if the troublesome factor appears in both numerator and denominator by substituting the x-value that makes the denominator zero into the numerator.
What is the result when $\frac{a^3-b^3}{a^2+ab+b^2} \div \frac{a^2-b^2}{a^2+2ab+b^2}$ is simplified, and what restrictions apply?
Answer: $(a+b)$ with restrictions $a \neq b, -b$ and $a^2+ab+b^2 \neq 0$ Use difference of cubes and other factoring formulas: $a^3-b^3 = (a-b)(a^2+ab+b^2)$ , $a^2-b^2 = (a-b)(a+b)$ , $a^2+2ab+b^2 = (a+b)^2$ . The division becomes: $\frac{(a-b)(a^2+ab+b^2)}{a^2+ab+b^2} \cdot \frac{(a+b)^2}{(a-b)(a+b)} = \frac{(a-b)(a^2+ab+b^2)(a+b)^2}{(a^2+ab+b^2)(a-b)(a+b)} = \frac{(a+b)^2}{(a+b)} = a+b$ . Domain restrictions: The original expression is undefined when $a^2+ab+b^2 = 0$ (first denominator), when $a^2-b^2 = 0$ (second numerator becomes second denominator after division), or when $a^2+2ab+b^2 = 0$ (second denominator). From $a^2-b^2 = 0$ , we get $a = \pm b$ . From $a^2+2ab+b^2 = (a+b)^2 = 0$ , we get $a = -b$ . The condition $a^2+ab+b^2 = 0$ has no real solutions when $a,b$ are real (discriminant is $b^2-4b^2 = -3b^2 < 0$ for $b \neq 0$ ). So restrictions are $a \neq b, -b$ , but we should include the $a^2+ab+b^2 \neq 0$ condition for completeness.
Consider $\frac{2x^2 + x - 3}{x^2 - 1} \cdot \frac{x^2 + 3x + 2}{2x^2 + 5x + 3}$ . After complete simplification, what is the numerator?
Answer: $x + 2$ When you encounter rational expressions being multiplied, your goal is to factor everything completely, then cancel common factors between numerators and denominators. Let's factor each polynomial step by step: - $2x^2 + x - 3 = (2x + 3)(x - 1)$ - $x^2 - 1 = (x + 1)(x - 1)$ - $x^2 + 3x + 2 = (x + 1)(x + 2)$ - $2x^2 + 5x + 3 = (2x + 3)(x + 1)$ Substituting these factored forms: $\frac{(2x + 3)(x - 1)}{(x + 1)(x - 1)} \cdot \frac{(x + 1)(x + 2)}{(2x + 3)(x + 1)}$ Now you can cancel common factors across the entire expression: - The factor $(2x + 3)$ appears once in the numerator and once in the denominator - The factor $(x - 1)$ appears once in the numerator and once in the denominator - The factor $(x + 1)$ appears twice in the numerator and twice in the denominator After canceling all common factors, only $(x + 2)$ remains in the numerator, making D the correct answer. Let's examine the wrong answers: A) $1$ would result if everything canceled completely, but $(x + 2)$ has no matching factor to cancel with. B) $(x + 1)$ appears equally in numerator and denominator, so it cancels out rather than remaining. C) $(2x + 3)$ also cancels completely since it appears once in both numerator and denominator. Study tip: Always factor completely before attempting to cancel. Look for the same factors appearing in both numerator and denominator across the entire multiplication—they don't have to be in the same fraction to cancel.
Which rational expression, when simplified, has the same form as $\frac{x-1}{x+1}$ but requires different domain restrictions?
Answer: $\frac{(x-1)^2}{(x+1)(x-1)}$ Choice C simplifies to $\frac{x-1}{x+1}$ after canceling $(x-1)$ , but the original expression is undefined at both $x = -1$ and $x = 1$ , while $\frac{x-1}{x+1}$ is only undefined at $x = -1$ . Choice A simplifies to $\frac{x+1}{x+1} = 1$ . Choice B simplifies to $\frac{(x-1)^2}{(x+1)^2}$ . Choice D simplifies to $\frac{x-1}{x+1}$ but has the same domain restrictions.
A student attempts to simplify $\frac{x^2 + 5x + 6}{x^2 + 4x + 4} \cdot \frac{x^2 - 4}{x^2 + 7x + 12}$ and claims the answer is $\frac{x-2}{x+2}$ . What error did the student most likely make?
Answer: Failed to recognize that $x^2 + 4x + 4 = (x+2)^2$ and incorrectly factored it as $(x+2)(x-2)$ Correct factorization: $\frac{(x+2)(x+3)}{(x+2)^2} \cdot \frac{(x+2)(x-2)}{(x+3)(x+4)} = \frac{x-2}{x+4}$ . To get $\frac{x-2}{x+2}$ , the student must have treated $(x+2)^2$ as $(x+2)(x-2)$ . Choice B describes a correct factorization. Choice C would produce a completely different form. Choice D wouldn't affect the simplified expression's form.
What is the simplified form of $\frac{x^3 - 8}{x^2 - 4} \div \frac{x^2 + 2x + 4}{x + 2}$ and what values must be excluded from the domain?
Answer: $x - 2$ ; exclude $x = 2, -2$ First, factor: $x^3 - 8 = (x-2)(x^2+2x+4)$ and $x^2 - 4 = (x-2)(x+2)$ . The division becomes: $\frac{(x-2)(x^2+2x+4)}{(x-2)(x+2)} \cdot \frac{x+2}{x^2+2x+4} = \frac{(x-2)(x^2+2x+4)(x+2)}{(x-2)(x+2)(x^2+2x+4)} = x-2$ . Domain restrictions come from denominators in the original expression: $x^2-4=0$ gives $x = \pm 2$ , and $x^2+2x+4=0$ has no real solutions. So exclude $x = 2, -2$ .
When working with the expression $\frac{3x^2 - 27}{x^3 - 3x^2 - 9x + 27}$ , a student factors the numerator as $3(x-3)(x+3)$ and the denominator as $(x-3)(x^2-9)$ . What error will this lead to in the simplification?
Answer: The student will get $\frac{3}{x-3}$ instead of the correct $\frac{3}{x+3}$ The correct factorization of the denominator is $x^3-3x^2-9x+27 = (x-3)(x^2-9) = (x-3)(x+3)(x-3) = (x-3)^2(x+3)$ . The correct simplification is $\frac{3(x-3)(x+3)}{(x-3)^2(x+3)} = \frac{3}{x-3}$ after canceling one factor of $(x-3)$ and the factor $(x+3)$ . However, if a student incorrectly treats the denominator as having only one factor of $(x-3)$ , they would cancel all common factors and get $\frac{3}{x+3}$ instead of the correct $\frac{3}{x-3}$ .
Consider the rational expression $\frac{2x^2 - 18}{x^2 - 6x + 9}$ . After simplifying completely, which statement correctly describes both the simplified form and a key characteristic of this expression?
Answer: Simplifies to $\frac{2(x+3)}{x-3}$ with a vertical asymptote at $x = 3$ Factor: numerator = $2(x^2-9) = 2(x+3)(x-3)$ , denominator = $(x-3)^2$ . Simplifying gives $\frac{2(x+3)(x-3)}{(x-3)^2} = \frac{2(x+3)}{x-3}$ . Since one factor of $(x-3)$ remains in the denominator, there's a vertical asymptote at $x = 3$ . Choice B incorrectly identifies this as removable. Choice C incorrectly cancels both factors. Choice D doesn't factor the numerator completely.
A rational function has the form $f(x) = \frac{P(x)}{Q(x)}$ where $P(x) = x^3 - x^2 - 6x$ and $Q(x) = x^2 - 9$ . After simplification, what type of discontinuity occurs at $x = 3$ ?
Answer: Removable discontinuity because the factor $(x-3)$ appears in both numerator and denominator Factor $P(x) = x(x^2-x-6) = x(x-3)(x+2)$ and $Q(x) = (x-3)(x+3)$ . The factor $(x-3)$ appears in both, so it cancels, creating a removable discontinuity (hole) at $x = 3$ . The simplified form is $\frac{x(x+2)}{x+3}$ . Choice A would be true if $(x-3)$ only appeared in the denominator. Choice C describes a different type of discontinuity. Choice D is incorrect since $x = 3$ is indeed a zero of the original denominator.
When simplifying the rational expression $\frac{x^3 - 8}{x^2 + 2x - 8}$ , what is the simplified form and what values must be excluded from the domain?
Answer: $\frac{x^2 + 2x + 4}{x + 4}$ ; $x \neq -4, x \neq 2$ Factor the numerator as a difference of cubes: $x^3 - 8 = (x-2)(x^2+2x+4)$ . Factor the denominator: $x^2 + 2x - 8 = (x+4)(x-2)$ . After canceling $(x-2)$ , we get $\frac{x^2+2x+4}{x+4}$ . Domain restrictions come from the original denominator: $x \neq 2$ and $x \neq -4$ . Choice B incorrectly identifies domain restrictions. Choice C has the wrong denominator after simplification. Choice D uses the wrong difference of cubes factorization.

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