Rewriting Rational Expressions - Math 3

1. When $\frac{2x^3 - 3x^2 + x - 5}{x^2 - 2x + 1}$ is written in the form $q(x) + \frac{r(x)}{x^2 - 2x + 1}$, what is the degree of $r(x)$?
   Answer: $1$ When dividing a polynomial by $x^2 - 2x + 1$, the remainder $r(x)$ must have degree less than 2 (the degree of the divisor). Since we're dividing a cubic by a quadratic, we get a linear quotient and a remainder of degree at most 1. Performing the division gives us $r(x) = ax + b$ for some constants $a$ and $b$, so the degree of $r(x)$ is 1. Choice A would only be correct if the remainder were a constant. Choices C and D are impossible since the remainder degree must be less than the divisor degree.
2. For what value of $k$ does $\frac{2x^2 + kx - 6}{x + 3}$ simplify to a linear expression with no remainder?
   Answer: $k = 4$ When you see a rational expression that needs to simplify to a linear expression with no remainder, you're dealing with polynomial division where the numerator must be perfectly divisible by the denominator. For $\frac{2x^2 + kx - 6}{x + 3}$ to have no remainder, the numerator $2x^2 + kx - 6$ must have $(x + 3)$ as a factor. This means $x = -3$ must be a root of the numerator. Substituting $x = -3$ into the numerator: $2(-3)^2 + k(-3) - 6 = 0$ Solving: $2(9) - 3k - 6 = 0$ $18 - 3k - 6 = 0$ $12 - 3k = 0$ $k = 4$ You can verify this by performing polynomial long division with $k = 4$: $\frac{2x^2 + 4x - 6}{x + 3} = 2x - 2$ with remainder 0. Choice A ($k = 12$) comes from incorrectly setting up the equation as $12 - 3k = 0$ but solving $k = 12$ instead of $k = 4$. Choice B ($k = 6$) results from the arithmetic error $18 - 6 = 12$, then $k = 2 \times 6 = 6$. Choice C ($k = 8$) might come from miscalculating $2(-3)^2 = 6$ instead of $18$, leading to $6 - 3k - 6 = 0$, so $k = 0$, but then doubling it. Remember: when a rational expression must simplify with no remainder, use the factor theorem—substitute the root of the denominator into the numerator and set it equal to zero.
3. Which of the following represents the partial fraction decomposition of $\frac{5x + 7}{x^2 + 5x + 6}$?
   Answer: $\frac{-3}{x + 2} + \frac{8}{x + 3}$ First, factor the denominator: $x^2 + 5x + 6 = (x + 2)(x + 3)$. Set up partial fractions: $\frac{5x + 7}{(x + 2)(x + 3)} = \frac{A}{x + 2} + \frac{B}{x + 3}$. Multiply both sides by $(x + 2)(x + 3)$: $5x + 7 = A(x + 3) + B(x + 2)$. Substituting $x = -2$: $5(-2) + 7 = A(1) + B(0)$, so $-3 = A$. Substituting $x = -3$: $5(-3) + 7 = A(0) + B(-1)$, so $-8 = -B$, giving $B = 8$. Therefore, the decomposition is $\frac{-3}{x + 2} + \frac{8}{x + 3}$. Choice A gives coefficients that sum to $5x + 5$, not $5x + 7$. Choice B gives coefficients that sum to $5x + 12$. Choice D has the coefficients reversed.
4. Which form best reveals the horizontal asymptote of $f(x) = \frac{3x^2 + 5x - 2}{x^2 - 4}$?
   Answer: $f(x) = 3 + \frac{5x + 10}{x^2 - 4}$ To find the horizontal asymptote, we need to rewrite the function by dividing: $\frac{3x^2 + 5x - 2}{x^2 - 4} = 3 + \frac{5x + 10}{x^2 - 4}$. As $x \to \infty$, the fraction $\frac{5x + 10}{x^2 - 4} \to 0$, revealing that the horizontal asymptote is $y = 3$. Choice A shows factored form but doesn't reveal the asymptote. Choice C is just a rearrangement that doesn't help. Choice D has the wrong remainder ($5x - 10$ instead of $5x + 10$).
5. The rational expression $\frac{x^2 + 3x + 2}{x^2 - 1}$ is equivalent to which of the following when $x \neq \pm 1$? Give your answer in simplest form.
   Answer: $\frac{x + 2}{x - 1}$ When you see a rational expression that needs simplification, your first instinct should be to factor both the numerator and denominator to look for common factors that can be canceled. Let's factor the given expression $\frac{x^2 + 3x + 2}{x^2 - 1}$. The numerator $x^2 + 3x + 2$ factors as $(x + 1)(x + 2)$ since we need two numbers that multiply to 2 and add to 3. The denominator $x^2 - 1$ is a difference of squares, factoring as $(x + 1)(x - 1)$. This gives us $\frac{(x + 1)(x + 2)}{(x + 1)(x - 1)}$. Since $x \neq -1$, we can cancel the common factor $(x + 1)$, leaving us with $\frac{x + 2}{x - 1}$, which matches answer choice D. Now let's check why the other options are incorrect. Choice A represents a partial fraction decomposition attempt, but $1 + \frac{3x + 3}{x^2 - 1}$ would equal $\frac{x^2 - 1 + 3x + 3}{x^2 - 1} = \frac{x^2 + 3x + 2}{x^2 - 1}$, which is our original expression, not the simplified form. Choice B, $\frac{x + 1}{x - 1}$, would result from incorrectly canceling terms instead of factors. Choice C, $1 + \frac{4x + 3}{x^2 - 1}$, comes from an arithmetic error in partial fraction decomposition. Remember: always factor completely before attempting to simplify rational expressions. Look for common factors in the numerator and denominator that can be canceled, and be careful not to cancel individual terms.
6. Which partial fraction decomposition is correct for $\frac{5x + 7}{(x + 1)(x + 3)}$?
   Answer: $\frac{1}{x + 1} + \frac{4}{x + 3}$ Setting up $\frac{5x + 7}{(x + 1)(x + 3)} = \frac{A}{x + 1} + \frac{B}{x + 3}$, we get $5x + 7 = A(x + 3) + B(x + 1)$. When $x = -1$: $5(-1) + 7 = A(2) + B(0)$, so $2 = 2A$, thus $A = 1$. When $x = -3$: $5(-3) + 7 = A(0) + B(-2)$, so $-8 = -2B$, thus $B = 4$. Choice B reverses the coefficients. Choice C has the wrong sign for A. Choice D uses incorrect coefficient values.
7. Which rewritten form of $\frac{x^4 - 1}{x^2 + 1}$ most clearly shows its behavior for large values of $|x|$?
   Answer: $x^2 - 1$ Since $x^4 - 1 = (x^2)^2 - 1^2 = (x^2 - 1)(x^2 + 1)$, we have $\frac{x^4 - 1}{x^2 + 1} = \frac{(x^2 - 1)(x^2 + 1)}{x^2 + 1} = x^2 - 1$ (provided $x^2 + 1 \neq 0$, which is always true for real $x$). For large $|x|$, this clearly shows the function behaves like $x^2$. Choice A shows factorization but not simplification. Choice B shows the intermediate step. Choice D unnecessarily includes the zero remainder term.
8. Which expression is equivalent to $\frac{x^3 - 8x^2 + 19x - 12}{x^2 - 5x + 6}$ when $x \neq 2$ and $x \neq 3$?
   Answer: $x - 3 + \frac{2}{x^2 - 5x + 6}$ Using polynomial long division: $\frac{x^3 - 8x^2 + 19x - 12}{x^2 - 5x + 6} = (x - 3) + \frac{2}{x^2 - 5x + 6}$. The quotient is $x - 3$ and the remainder is $2$. Choice B incorrectly has $2x$ in the numerator of the remainder term. Choice C results from an error in the division process. Choice D has the wrong quotient and remainder.
9. The rational function $g(x) = \frac{x^3 + x^2 - 6x}{x^2 - 4}$ can be rewritten to reveal a removable discontinuity. Which form best shows this structure?
   Answer: $g(x) = \frac{x(x - 2)(x + 3)}{(x - 2)(x + 2)}$ First, factor completely: $x^3 + x^2 - 6x = x(x^2 + x - 6) = x(x - 2)(x + 3)$ and $x^2 - 4 = (x - 2)(x + 2)$. So $g(x) = \frac{x(x - 2)(x + 3)}{(x - 2)(x + 2)}$. This form clearly shows that $(x - 2)$ appears in both numerator and denominator, indicating a removable discontinuity at $x = 2$. Choice A doesn't factor the numerator completely. Choice C shows the simplified form but not the original structure that reveals the discontinuity. Choice D shows a different rewritten form that doesn't highlight the removable discontinuity.
10. Which expression represents the complete factorization that reveals the structure of $\frac{x^3 - 27}{x^2 + 3x + 9}$?
    Answer: $x - 3$ When you encounter a rational expression where the numerator might factor in a way that cancels with the denominator, look for special factoring patterns first. This is key to simplifying complex-looking fractions. The numerator $x^3 - 27$ is a difference of cubes since $27 = 3^3$. Using the difference of cubes formula $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$, we get: $x^3 - 27 = (x-3)(x^2 + 3x + 9)$ So our expression becomes: $\frac{x^3 - 27}{x^2 + 3x + 9} = \frac{(x-3)(x^2 + 3x + 9)}{x^2 + 3x + 9}$ Since $(x^2 + 3x + 9)$ appears in both numerator and denominator, they cancel out (assuming $x^2 + 3x + 9 \neq 0$), leaving us with $x - 3$. Choice A represents polynomial long division notation but incorrectly suggests there's a remainder when there isn't. Choice B shows the factored form before cancellation but doesn't complete the simplification process. Choice C multiplies by 1, which doesn't change the expression or reveal its simplified structure. Choice D correctly shows the final simplified form after factoring and canceling common factors. Remember: when you see cubes or cube roots in rational expressions, immediately check for sum or difference of cubes patterns. These often lead to dramatic simplifications through cancellation, turning intimidating expressions into simple linear terms.
11. The expression $\frac{x^3 + 27}{x^2 - 3x + 9}$ simplifies to which form?
    Answer: $x + 3$ When you see a rational expression where the numerator and denominator look related, think about factoring. This expression involves a sum of cubes in the numerator and what appears to be a quadratic in the denominator. The numerator $x^3 + 27$ is a sum of cubes since $27 = 3^3$. Using the sum of cubes formula $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$, we get: $x^3 + 27 = x^3 + 3^3 = (x + 3)(x^2 - 3x + 9)$ Now our expression becomes: $\frac{x^3 + 27}{x^2 - 3x + 9} = \frac{(x + 3)(x^2 - 3x + 9)}{x^2 - 3x + 9}$ Since the factor $(x^2 - 3x + 9)$ appears in both numerator and denominator, they cancel out, leaving us with $x + 3$. Choice A ($x^2 + 3x + 9$) represents what you might get if you incorrectly expanded the sum of cubes formula. Choice B ($x - 3$) would result from confusing this with a difference of cubes or making a sign error. Choice C ($\frac{x + 3}{x - 3}$) suggests you incorrectly factored the denominator as $(x - 3)$ instead of recognizing it as part of the sum of cubes pattern. The key strategy here is recognizing sum and difference of cubes patterns. Memorize these formulas: $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$ and $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$. When you see perfect cubes, always check if these special factorizations apply.
12. The expression $\frac{x^3 - 8}{x^2 + 2x + 4}$ can be simplified to which of the following?
    Answer: $x - 2$ Recognize that $x^3 - 8 = x^3 - 2^3$, which is a difference of cubes. Using the formula $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$, we get: $x^3 - 8 = (x - 2)(x^2 + 2x + 4)$. Therefore: $\frac{x^3 - 8}{x^2 + 2x + 4} = \frac{(x - 2)(x^2 + 2x + 4)}{x^2 + 2x + 4} = x - 2$ (provided $x^2 + 2x + 4 \neq 0$). Choice B uses the wrong sign. Choice C represents the other factor in the difference of cubes factorization. Choice D would result from incorrectly attempting to factor the numerator.
13. If $\frac{x^2 - 7x + 12}{x - 3} = x + a + \frac{b}{x - 3}$ for constants $a$ and $b$, what is the value of $ab$?
    Answer: $0$ Notice that $x^2 - 7x + 12 = (x - 3)(x - 4)$, so $\frac{x^2 - 7x + 12}{x - 3} = \frac{(x - 3)(x - 4)}{x - 3} = x - 4$ (for $x \neq 3$). Therefore, $x - 4 = x + a + \frac{b}{x - 3}$, which means $a = -4$ and $b = 0$ (since there's no remainder). Thus, $ab = (-4)(0) = 0$. Choice A comes from multiplying the factors incorrectly. Choice B uses $a$ alone. Choice D ignores the sign.
14. Which expression is equivalent to $\frac{2x^3 - x^2 + 3x - 7}{x^2 + 1}$?
    Answer: $2x - 1 + \frac{x - 6}{x^2 + 1}$ Using polynomial long division: First, $2x^3 ÷ x^2 = 2x$. Multiply: $2x(x^2 + 1) = 2x^3 + 2x$. Subtract: $(2x^3 - x^2 + 3x - 7) - (2x^3 + 2x) = -x^2 + x - 7$. Next, $-x^2 ÷ x^2 = -1$. Multiply: $-1(x^2 + 1) = -x^2 - 1$. Subtract: $(-x^2 + x - 7) - (-x^2 - 1) = x - 6$. Therefore: $\frac{2x^3 - x^2 + 3x - 7}{x^2 + 1} = 2x - 1 + \frac{x - 6}{x^2 + 1}$. Choice B has the wrong sign in the remainder. Choice C has the wrong coefficient for the linear term and wrong remainder. Choice D has wrong coefficient and wrong signs.
15. If $\frac{3x^2 + x - 10}{x - 2} = 3x + 7 + \frac{k}{x - 2}$ for all values of $x$ except $x = 2$, what is the value of $k$?
    Answer: $4$ To find $k$, we multiply both sides by $(x - 2)$: $3x^2 + x - 10 = (3x + 7)(x - 2) + k$. Expanding the right side: $(3x + 7)(x - 2) = 3x^2 - 6x + 7x - 14 = 3x^2 + x - 14$. So we have: $3x^2 + x - 10 = 3x^2 + x - 14 + k$. Comparing coefficients, we get: $-10 = -14 + k$, so $k = 4$. Choice A would result from a sign error when expanding. Choice C would result from incorrectly calculating $(7)(-2)$. Choice D is the constant term from the original numerator, which is a common misconception.

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