Simplifying Rational Expressions - Accuplacer Aaf

Simplify the expression $\frac{ab+3a-2b-6}{b^2+6b+9}$ .
Answer: $\frac{a-2}{b+3}$ Factor the numerator by grouping: $a(b+3)-2(b+3) = (a-2)(b+3)$ . Factor the denominator, which is a perfect square trinomial: $b^2+6b+9 = (b+3)^2$ . The expression becomes $\frac{(a-2)(b+3)}{(b+3)(b+3)}$ . Cancel one common factor of $(b+3)$ . The simplified result is $\frac{a-2}{b+3}$ .
What is the simplified form of the expression $\frac{4x^2 - 100}{2x+10}$ ?
Answer: $2x-10$ First, factor out the greatest common factor from the numerator, which is 4: $4(x^2-25)$ . Then, factor out the GCF from the denominator, which is 2: $2(x+5)$ . The expression is $\frac{4(x^2-25)}{2(x+5)}$ . Factor the difference of squares in the numerator: $x^2-25 = (x-5)(x+5)$ . The expression becomes $\frac{4(x-5)(x+5)}{2(x+5)}$ . Cancel the common factor $(x+5)$ and simplify the constants $\frac{4}{2}=2$ . The result is $2(x-5)$ , which is $2x-10$ .
Which expression is equivalent to $\frac{x^2 - 4y^2}{x^2 - 2xy}$ for all values where it is defined?
Answer: $\frac{x+2y}{x}$ Factor the numerator and the denominator. The numerator $x^2 - 4y^2$ is a difference of squares, factoring to $(x-2y)(x+2y)$ . The denominator $x^2 - 2xy$ has a greatest common factor of $x$ , factoring to $x(x-2y)$ . The expression is $\frac{(x-2y)(x+2y)}{x(x-2y)}$ . Canceling the common factor $(x-2y)$ leaves $\frac{x+2y}{x}$ .
Simplify the expression $\frac{x^4 - 16}{x^2 - x - 2}$ .
Answer: $\frac{(x+2)(x^2+4)}{x+1}$ First, factor the numerator $x^4 - 16$ as a difference of squares: $(x^2-4)(x^2+4)$ . Then factor $(x^2-4)$ further as another difference of squares: $(x-2)(x+2)$ . The fully factored numerator is $(x-2)(x+2)(x^2+4)$ . Next, factor the denominator $x^2 - x - 2$ into $(x-2)(x+1)$ . The expression becomes $\frac{(x-2)(x+2)(x^2+4)}{(x-2)(x+1)}$ . Cancel the common factor $(x-2)$ to get $\frac{(x+2)(x^2+4)}{x+1}$ .
Which of the following is equivalent to $\frac{m^3 - 8}{m^2 + 2m + 4} \div \frac{m - 2}{m + 1}$ ?
Answer: $m + 1$ Recognize that $m^3 - 8 = (m-2)(m^2+2m+4)$ using the difference of cubes formula. The expression becomes $\frac{(m-2)(m^2+2m+4)}{m^2+2m+4} \div \frac{m-2}{m+1} = (m-2) \times \frac{m+1}{m-2} = m+1$ after canceling $(m^2+2m+4)$ and then $(m-2)$ . Choice A stops after the first cancellation. Choice C inverts the division incorrectly. Choice D represents the original divisor unchanged.
For what value(s) of $k$ will the expression $\frac{k^2 - 9}{k^2 + 2k - 15}$ simplify to $\frac{k + 3}{k + 5}$ ?
Answer: All real values except $k = -5$ and $k = 3$ When you encounter rational expressions that need to be simplified, always start by factoring both the numerator and denominator to look for common factors that can be canceled. Let's factor the original expression $\frac{k^2 - 9}{k^2 + 2k - 15}$ . The numerator $k^2 - 9$ is a difference of squares: $(k-3)(k+3)$ . For the denominator $k^2 + 2k - 15$ , you need two numbers that multiply to $-15$ and add to $2$ : those are $5$ and $-3$ , giving us $(k+5)(k-3)$ . So the expression becomes: $\frac{(k-3)(k+3)}{(k+5)(k-3)}$ You can cancel the common factor $(k-3)$ from numerator and denominator, leaving $\frac{k+3}{k+5}$ , which matches the target expression. However, this simplification is only valid when the canceled factor doesn't equal zero. Since we canceled $(k-3)$ , we must exclude $k = 3$ . Additionally, the original denominator cannot equal zero, so $k \neq -5$ and $k \neq 3$ (both values that make $k^2 + 2k - 15 = 0$ ). Choice A incorrectly excludes $k = -3$ , but this value doesn't make any denominator zero. Choice B misses that $k = -5$ must also be excluded since it makes the original denominator zero. Choice D is wrong because the expressions are equivalent for most values of $k$ . The correct answer is C: all real values except $k = -5$ and $k = 3$ . Remember: when simplifying rational expressions, always identify which values make any denominator zero, including factors you cancel out.
Simplify the expression $\frac{5x^4 - 20x^2}{10x^3+20x^2}$ .
Answer: $\frac{x-2}{2}$ First, factor the GCF from the numerator, $5x^2$ , to get $5x^2(x^2-4)$ . Then factor the GCF from the denominator, $10x^2$ , to get $10x^2(x+2)$ . The expression is $\frac{5x^2(x^2-4)}{10x^2(x+2)}$ . Factor the difference of squares in the numerator: $\frac{5x^2(x-2)(x+2)}{10x^2(x+2)}$ . Cancel the common factor $(x+2)$ . Also, simplify the monomial factors: $\frac{5x^2}{10x^2} = \frac{1}{2}$ . The remaining expression is $\frac{1(x-2)}{2}$ , or $\frac{x-2}{2}$ .
If $\frac{3a^2 - 12a}{a^2 - 8a + 16} = \frac{3a}{a - 4}$ for all valid values of $a$ , what restriction must be placed on $a$ ?
Answer: $a \neq 0$ and $a \neq 4$ When working with rational equations, you must identify values that make any denominator equal to zero, as these create undefined expressions. Even when an equation can be simplified, the original restrictions still apply. Let's analyze the given equation by first simplifying the left side. The numerator $3a^2 - 12a$ factors as $3a(a - 4)$ , and the denominator $a^2 - 8a + 16$ is a perfect square trinomial that factors as $(a - 4)^2$ . This gives us: $\frac{3a(a - 4)}{(a - 4)^2} = \frac{3a}{a - 4}$ When $a \neq 4$ , we can cancel one factor of $(a - 4)$ from numerator and denominator, yielding $\frac{3a}{a - 4}$ , which matches the right side of the equation. However, we must consider all values that make any denominator zero in the original equation. From the left side, $a^2 - 8a + 16 = 0$ when $a = 4$ . From the right side (and the simplified left side), $a - 4 = 0$ when $a = 4$ . Additionally, for the equation to be meaningful, we need $3a \neq 0$ , so $a \neq 0$ . Choice A incorrectly includes $a \neq 8$ , which doesn't make any denominator zero. Choice B incorrectly includes $a \neq -4$ , which also doesn't affect the denominators, and misses the restriction $a \neq 0$ . Choice C incorrectly includes $a \neq 2$ and misses $a \neq 4$ . Remember: when simplifying rational expressions, always identify restrictions from the original expression before canceling factors. These restrictions remain even after simplification.
Simplify the expression $\frac{x^3 - 2x^2 + 5x - 10}{x^2+5}$ .
Answer: $x-2$ The numerator $x^3 - 2x^2 + 5x - 10$ can be factored by grouping. Group the first two terms and the last two terms: $x^2(x-2) + 5(x-2)$ . Factor out the common binomial $(x-2)$ to get $(x^2+5)(x-2)$ . The expression is now $\frac{(x^2+5)(x-2)}{x^2+5}$ . The denominator $x^2+5$ is a prime polynomial but is a common factor with the numerator. Canceling $(x^2+5)$ leaves $x-2$ .
Which of the following is equivalent to the expression $\frac{x^2 - 2x - 15}{x^2 - 9}$ for all values of $x$ for which the expression is defined?
Answer: $\frac{x-5}{x-3}$ To simplify the rational expression, factor the numerator and the denominator. The numerator $x^2 - 2x - 15$ factors to $(x-5)(x+3)$ . The denominator $x^2 - 9$ is a difference of squares and factors to $(x-3)(x+3)$ . The expression becomes $\frac{(x-5)(x+3)}{(x-3)(x+3)}$ . The common factor $(x+3)$ can be canceled, leaving $\frac{x-5}{x-3}$ .
Which of the following is equivalent to $\frac{25-x^2}{x^2-2x-15}$ ?
Answer: $\frac{-(x+5)}{x+3}$ Factor the numerator $25-x^2$ as a difference of squares to get $(5-x)(5+x)$ . Factor the denominator $x^2-2x-15$ to get $(x-5)(x+3)$ . The expression is $\frac{(5-x)(5+x)}{(x-5)(x+3)}$ . The terms $(5-x)$ and $(x-5)$ are opposites, so $(5-x) = -1(x-5)$ . Substitute this in: $\frac{-1(x-5)(x+5)}{(x-5)(x+3)}$ . Cancel the common factor $(x-5)$ . The simplified expression is $\frac{-(x+5)}{x+3}$ .
For $x \neq 3$ , what is the simplified form of the expression $\frac{6x-18}{2x-6}$ ?
Answer: $3$ Factor the greatest common factor from the numerator and the denominator. The GCF of the numerator $6x-18$ is 6, so it factors to $6(x-3)$ . The GCF of the denominator $2x-6$ is 2, so it factors to $2(x-3)$ . The expression becomes $\frac{6(x-3)}{2(x-3)}$ . The common factor $(x-3)$ cancels out, leaving $\frac{6}{2}$ , which simplifies to 3.
Which expression is equivalent to $\frac{\frac{x^2 - 4}{x + 1}}{\frac{x^2 - 4x + 4}{x^2 - 1}}$ ?
Answer: $\frac{(x + 2)(x - 1)}{x - 2}$ When you encounter complex fractions like this one, remember that dividing by a fraction is the same as multiplying by its reciprocal. This problem tests your ability to factor polynomials and simplify rational expressions. To solve $\frac{\frac{x^2 - 4}{x + 1}}{\frac{x^2 - 4x + 4}{x^2 - 1}}$ , first rewrite it as a multiplication: $\frac{x^2 - 4}{x + 1} \times \frac{x^2 - 1}{x^2 - 4x + 4}$ Next, factor each polynomial. The numerator $x^2 - 4$ is a difference of squares: $(x + 2)(x - 2)$ . The denominator $x^2 - 4x + 4$ is a perfect square trinomial: $(x - 2)^2$ . The expression $x^2 - 1$ is also a difference of squares: $(x + 1)(x - 1)$ . Substituting the factored forms: $\frac{(x + 2)(x - 2)}{x + 1} \times \frac{(x + 1)(x - 1)}{(x - 2)^2}$ This becomes: $\frac{(x + 2)(x - 2)(x + 1)(x - 1)}{(x + 1)(x - 2)^2}$ Cancel common factors: $(x + 1)$ cancels completely, and one factor of $(x - 2)$ cancels, leaving $\frac{(x + 2)(x - 1)}{x - 2}$ , which is choice C. Choice A is missing the $(x + 2)$ factor. Choice B incorrectly has $(x + 2)$ in the numerator but $(x - 2)$ in the denominator without the $(x - 1)$ term. Choice D fails to cancel the common $(x + 1)$ factor and retains $(x - 2)^2$ in the denominator. Always factor completely before canceling, and double-check that you've simplified all possible common factors.
Simplify the rational expression $\frac{2x^3 - 2x}{x^3 + 4x^2 - 5x}$ .
Answer: $\frac{2(x+1)}{x+5}$ First, factor out the greatest common factor from both the numerator and denominator. Numerator GCF is $2x$ , giving $2x(x^2-1)$ . Denominator GCF is $x$ , giving $x(x^2+4x-5)$ . Then, factor the remaining polynomials. $x^2-1 = (x-1)(x+1)$ and $x^2+4x-5 = (x+5)(x-1)$ . The expression becomes $\frac{2x(x-1)(x+1)}{x(x+5)(x-1)}$ . Cancel the common factors of $x$ and $(x-1)$ . This leaves $\frac{2(x+1)}{x+5}$ .
Simplify the expression $\frac{12x^2 + 10x - 8}{8x^2 - 2}$ .
Answer: $\frac{3x+4}{2x+1}$ First, factor out the GCF from the numerator (2) and the denominator (2). This gives $\frac{2(6x^2+5x-4)}{2(4x^2-1)}$ . The 2s cancel. Now factor the remaining polynomials. The numerator $6x^2+5x-4$ factors to $(3x+4)(2x-1)$ . The denominator $4x^2-1$ is a difference of squares, factoring to $(2x-1)(2x+1)$ . The expression is $\frac{(3x+4)(2x-1)}{(2x-1)(2x+1)}$ . Cancel the common factor $(2x-1)$ to get $\frac{3x+4}{2x+1}$ .
Which expression is equivalent to $\frac{x^2 - 9}{x^2 + 6x + 9} \div \frac{x - 3}{x + 3}$ ?
Answer: $\frac{x + 3}{x - 3}$ First, factor the numerator and denominator: $\frac{x^2 - 9}{x^2 + 6x + 9} = \frac{(x-3)(x+3)}{(x+3)^2}$ . Then convert division to multiplication: $\frac{(x-3)(x+3)}{(x+3)^2} \times \frac{x+3}{x-3} = \frac{(x-3)(x+3)(x+3)}{(x+3)^2(x-3)} = \frac{x+3}{x-3}$ . Choice B incorrectly multiplies instead of canceling factors. Choice C inverts the final result. Choice D fails to properly cancel common factors.
Simplify the expression $\frac{x^3+x^2-4x-4}{x^2-x-2}$ .
Answer: $x+2$ First, factor the numerator $x^3+x^2-4x-4$ by grouping. Group the first two and last two terms: $x^2(x+1) - 4(x+1)$ . This factors to $(x^2-4)(x+1)$ . The factor $(x^2-4)$ is a difference of squares, so it factors further into $(x-2)(x+2)$ . The fully factored numerator is $(x-2)(x+2)(x+1)$ . Next, factor the denominator $x^2-x-2$ into $(x-2)(x+1)$ . The expression is $\frac{(x-2)(x+2)(x+1)}{(x-2)(x+1)}$ . Cancel the common factors $(x-2)$ and $(x+1)$ . The remaining simplified expression is $x+2$ .
Simplify $\frac{x^3+8}{x^2-2x+4}$ .
Answer: $x+2$ The numerator $x^3+8$ is a sum of cubes, $x^3+2^3$ . The sum of cubes formula is $a^3+b^3 = (a+b)(a^2-ab+b^2)$ . Applying this, $x^3+8 = (x+2)(x^2-2x+4)$ . The expression becomes $\frac{(x+2)(x^2-2x+4)}{x^2-2x+4}$ . The factor $(x^2-2x+4)$ cancels out, leaving $x+2$ .
Simplify the rational expression $\frac{3x^2 + 12x}{x^2 + x - 12}$ .
Answer: $\frac{3x}{x-3}$ First, factor the numerator by taking out the greatest common factor, $3x$ , which gives $3x(x+4)$ . Next, factor the denominator, a trinomial, which becomes $(x+4)(x-3)$ . The expression is now $\frac{3x(x+4)}{(x+4)(x-3)}$ . Cancel the common factor $(x+4)$ from the numerator and denominator. The simplified expression is $\frac{3x}{x-3}$ .
What is the result when $\frac{4y^2 - 1}{2y^2 + 5y - 3}$ is simplified, given that $y \neq \frac{1}{2}$ and $y \neq -3$ ?
Answer: $\frac{2y + 1}{y + 3}$ Factor the numerator: $4y^2 - 1 = (2y + 1)(2y - 1)$ . Factor the denominator: $2y^2 + 5y - 3 = (2y - 1)(y + 3)$ . The expression becomes $\frac{(2y+1)(2y-1)}{(2y-1)(y+3)} = \frac{2y+1}{y+3}$ after canceling the common factor $(2y-1)$ . Choice A has the wrong sign in the numerator. Choice C inverts part of the result. Choice D fails to factor and simplify properly.

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