Rewriting Rational Expressions - Algebra 1

1. Use polynomial division to express $\frac{2x^3+3x^2-5x+1}{x-2}$ in the form $q(x)+\frac{r(x)}{x-2}$, where $\deg(r)<\deg(x-2)$.
   Answer: $2x^2+7x+9+\frac{19}{x-2}$ This question tests your understanding of polynomial division—rewriting a rational expression like $\frac{2x^3 + 3x^2 - 5x + 1}{x - 2}$ in the form quotient + remainder/divisor, just like dividing numbers where $17/5 = 3 + 2/5$. The division algorithm for polynomials says that any rational expression $a(x)/b(x)$ can be rewritten as $q(x) + r(x)/b(x)$, where $q(x)$ is the quotient, and $r(x)$ is the remainder with degree less than the degree of $b(x)$—here, since $b(x) = x - 2$ is degree 1, $r(x)$ must be a constant. Dividing $(2x^3 + 3x^2 - 5x + 1)$ by $(x - 2)$ using long division: (1) Divide $2x^3/x = 2x^2$ (first quotient term). (2) Multiply: $2x^2(x - 2) = 2x^3 - 4x^2$. (3) Subtract: $(2x^3 + 3x^2) - (2x^3 - 4x^2) = 7x^2$, bring down $-5x$ to get $7x^2 - 5x$. (4) Divide $7x^2/x = 7x$. (5) Multiply: $7x(x - 2) = 7x^2 - 14x$. (6) Subtract: $(7x^2 - 5x) - (7x^2 - 14x) = 9x$, bring down $+1$ to get $9x + 1$. (7) Divide $9x/x = 9$. (8) Multiply: $9(x - 2) = 9x - 18$. (9) Subtract: $(9x + 1) - (9x - 18) = 19$. So, quotient is $2x^2 + 7x + 9$, remainder 19. Choice A correctly shows $2x^2 + 7x + 9 + 19/(x - 2)$ where the quotient is quadratic and remainder has degree 0, less than 1, giving the proper rewritten form. Choice B has a negative remainder term, but the subtraction in division leads to a positive 19, not negative—remember to double-check signs in each subtraction step! Polynomial long division steps: (1) Divide leading terms, (2) Multiply entire divisor, (3) Subtract, (4) Repeat until remainder degree $< $ divisor degree. Verification is your friend: multiply $(2x^2 + 7x + 9)(x - 2) + 19$ and you should get back $2x^3 + 3x^2 - 5x + 1$—this check catches errors every time!
2. Rewrite $\frac{2x^3+x^2-4x+5}{x+1}$ in the form $q(x)+\frac{r(x)}{x+1}$ using polynomial division, with $\deg(r)<\deg(x+1)$.
   Answer: $2x^2-x-3+\frac{8}{x+1}$ This question tests your understanding of polynomial division—rewriting a rational expression (fraction of polynomials) in the form quotient + remainder/divisor, just like how 17/5 = 3 + 2/5 in arithmetic. The division algorithm for polynomials says that any rational expression a(x)/b(x) can be rewritten as q(x) + r(x)/b(x), where q(x) is the quotient (the polynomial part), and r(x) is the remainder (what's left over in the numerator). The crucial requirement: the degree of the remainder r(x) must be less than the degree of the divisor b(x)—just like in numerical division where the remainder must be less than the divisor! Dividing (2x³ + x² - 4x + 5) by (x + 1) using long division: (1) Divide leading terms: 2x³/x = 2x² (first term of quotient). (2) Multiply: 2x²(x + 1) = 2x³ + 2x². (3) Subtract from dividend: (2x³ + x² - 4x + 5) - (2x³ + 2x²) = -x² - 4x + 5. (4) Divide again: -x²/x = -x (second term of quotient). (5) Multiply: -x(x + 1) = -x² - x. (6) Subtract: (-x² - 4x + 5) - (-x² - x) = -3x + 5. (7) Divide once more: -3x/x = -3 (third term of quotient). (8) Multiply: -3(x + 1) = -3x - 3. (9) Subtract: (-3x + 5) - (-3x - 3) = 8. Result: quotient is 2x² - x - 3, remainder is 8, giving 2x² - x - 3 + 8/(x + 1). Choice A correctly shows 2x² - x - 3 + 8/(x + 1) where the quotient is a degree-2 polynomial and remainder 8 has degree 0 (less than divisor degree 1), giving the proper rewritten form. Choice C has the wrong sign on the x term: it shows 2x² + x - 3 instead of 2x² - x - 3. When dividing -x² by x, we get -x, not +x. The signs in polynomial division require careful attention—each subtraction step can introduce sign errors! Verification is your friend: after dividing, multiply your quotient 2x² - x - 3 by the divisor (x + 1) and add your remainder 8. You get (2x² - x - 3)(x + 1) + 8 = 2x³ + 2x² - x² - x - 3x - 3 + 8 = 2x³ + x² - 4x + 5, which matches the original numerator perfectly! This check works every time and catches sign errors quickly.
3. Use polynomial division to express $\frac{2x^3+x^2-8x+7}{x+2}$ in the form $q(x)+\frac{r(x)}{x+2}$, where $\deg(r)<1$.
   Answer: $2x^2-3x-2+\frac{11}{x+2}$ This question tests your understanding of polynomial division—rewriting a rational expression like $(2x^3 + x^2 - 8x + 7)/(x + 2)$ in the form quotient + remainder/divisor, just like dividing numbers where $17/5 = 3 + 2/5$. The division algorithm for polynomials says that any rational expression $a(x)/b(x)$ can be rewritten as $q(x) + r(x)/b(x)$, where $q(x)$ is the quotient, and $r(x)$ is the remainder with degree less than the degree of $b(x)$—here, since $b(x) = x + 2$ is degree 1, $r(x)$ must be a constant. Dividing $(2x^3 + x^2 - 8x + 7)$ by $(x + 2)$: (1) $2x^3/x = 2x^2$. (2) Multiply: $2x^2(x + 2) = 2x^3 + 4x^2$. (3) Subtract: gets $-3x^2 - 8x + 7$ (after bringing down). (4) $-3x^2/x = -3x$. (5) Multiply: $-3x(x + 2) = -3x^2 - 6x$. (6) Subtract: gets $-2x + 7$. (7) $-2x/x = -2$. (8) Multiply: $-2(x + 2) = -2x - 4$. (9) Subtract: gets 11. Result: $2x^2 - 3x - 2 + 11/(x + 2)$. Choice A correctly shows $2x^2 - 3x - 2 + 11/(x + 2)$ where the quotient is quadratic and remainder has degree 0, less than 1, giving the proper rewritten form. Choice C has quotient $2x^2 - 3x + 2$, but when dividing $-2x$ by $x$, we get $-2$, not $+2$— the leading term of the quotient comes from dividing the highest-degree terms, and signs matter in each step! Before starting polynomial division, always check: (1) Can I factor the numerator and cancel with the denominator? (2) If numerator degree < denominator, it's already in form with $q(x)=0$. This pre-check saves work!
4. Rewrite $\frac{x^2+5x+7}{x+2}$ in the form $q(x)+\frac{r}{x+2}$ using polynomial division (quotient + remainder/divisor), with remainder degree less than the divisor degree.
   Answer: $x+3+\frac{1}{x+2}$ This question tests your understanding of polynomial division—rewriting a rational expression (fraction of polynomials) in the form quotient + remainder/divisor, just like how 17/5 = 3 + 2/5 in arithmetic. For simple cases, inspection can work: if you need (x² + 5x + 7)/(x + 2) and the numerator degree is just 1 more than denominator degree, the quotient will be linear. Try q(x) = x + a for some a, multiply (x + a)(x + 2), match coefficients with x² + 5x + 7, and solve for a and the remainder. For (x² + 5x + 7)/(x + 2), we suspect quotient is linear: q(x) = x + a for some a. Multiplying: (x + a)(x + 2) = x² + (2 + a)x + 2a. Matching with numerator x² + 5x + 7: coefficient of x gives 2 + a = 5, so a = 3. Constant term gives 2a = 6, but we have 7, so remainder is 7 - 6 = 1. Result: (x + 3) + 1/(x + 2). Choice B correctly shows x + 3 + 1/(x + 2) where the quotient is linear and remainder 1 has degree 0 (less than divisor degree 1), giving the proper rewritten form. Choice C has the wrong remainder: 7 instead of 1, which might come from forgetting to subtract 2a = 6 from the constant term 7—the remainder is what's left after accounting for the quotient times divisor! Polynomial long division steps: (1) Divide leading terms to get first term of quotient, (2) Multiply entire divisor by that term, (3) Subtract from dividend, (4) Repeat with what remains until remainder degree < divisor degree. Practice with (x² + 5x + 7)/(x + 2): divide x² by x → x, multiply x(x+2) = x² + 2x, subtract to get 3x + 7, divide 3x by x → 3, multiply 3(x+2) = 3x + 6, subtract to get remainder 1. Answer: x + 3 + 1/(x+2).
5. A student claims that $\frac{2x^3 + 5x^2 - 3x + 1}{x^2 + 3} = 2x + 5 + \frac{-9x - 14}{x^2 + 3}$. To verify this claim, which calculation would be most efficient?
   Answer: Multiply $(2x + 5)(x^2 + 3) + (-9x - 14)$ and compare to the original numerator To verify the polynomial division, we use the division algorithm: if $\frac{a(x)}{b(x)} = q(x) + \frac{r(x)}{b(x)}$, then $a(x) = q(x) \cdot b(x) + r(x)$. So we need $2x^3 + 5x^2 - 3x + 1 = (2x + 5)(x^2 + 3) + (-9x - 14)$. Expanding: $(2x + 5)(x^2 + 3) = 2x^3 + 6x + 5x^2 + 15 = 2x^3 + 5x^2 + 6x + 15$. Adding the remainder: $2x^3 + 5x^2 + 6x + 15 + (-9x - 14) = 2x^3 + 5x^2 - 3x + 1$, which matches the original numerator. Choice A is inefficient since we're verifying, not discovering. Choice B only checks one value, not the entire identity. Choice D only verifies a necessary but not sufficient condition.
6. When $\frac{x^3 - 2x^2 + 4x - 8}{x - 2}$ is simplified, the result can be written in the form $p(x) + \frac{k}{x - 2}$ where $p(x)$ is a polynomial and $k$ is a constant. What is the value of $k$?
   Answer: $0$ We can factor the numerator by grouping: $x^3 - 2x^2 + 4x - 8 = x^2(x - 2) + 4(x - 2) = (x - 2)(x^2 + 4)$. Therefore, $\frac{x^3 - 2x^2 + 4x - 8}{x - 2} = \frac{(x - 2)(x^2 + 4)}{x - 2} = x^2 + 4$ (for $x \neq 2$). This means $p(x) = x^2 + 4$ and $k = 0$. Alternatively, using synthetic division or polynomial long division would yield quotient $x^2 + 4$ and remainder $0$. Students might incorrectly think $k = -8$ (the constant term), $k = 4$ (coefficient of $x$), or $k = 8$ (absolute value of constant term).
7. Rewrite $\frac{2x^3-x^2+4x-3}{x+2}$ in the form $q(x)+\frac{r(x)}{x+2}$, where $\deg(r)<1$.​
   Answer: $2x^2-5x+14+\frac{-31}{x+2}$ This question tests your understanding of polynomial division—rewriting a rational expression (fraction of polynomials) in the form quotient + remainder/divisor, just like how 17/5 = 3 + 2/5 in arithmetic. Polynomial long division works exactly like the long division you learned in elementary school, just with polynomials instead of numbers: divide the leading terms, multiply back, subtract, bring down the next term, repeat. When you can't divide anymore (when what's left has smaller degree than the divisor), that leftover is your remainder, and what you've built up is your quotient. Dividing (2x³ - x² + 4x - 3) by (x + 2) using long division: (1) Divide leading terms: 2x³/x = 2x² (first term of quotient). (2) Multiply: 2x²(x + 2) = 2x³ + 4x². (3) Subtract from dividend: (2x³ - x² + 4x - 3) - (2x³ + 4x²) = -5x² + 4x - 3. (4) Divide again: -5x²/x = -5x (second term of quotient). (5) Multiply: -5x(x + 2) = -5x² - 10x. (6) Subtract: (-5x² + 4x - 3) - (-5x² - 10x) = 14x - 3. (7) Divide once more: 14x/x = 14 (third term of quotient). (8) Multiply: 14(x + 2) = 14x + 28. (9) Subtract: (14x - 3) - (14x + 28) = -31. Result: quotient is 2x² - 5x + 14, remainder is -31, giving 2x² - 5x + 14 + (-31)/(x + 2). Choice A correctly shows 2x² - 5x + 14 + (-31)/(x + 2) where the quotient is a degree 2 polynomial and remainder -31 has degree 0 (less than divisor degree 1), giving the proper rewritten form. Choice B has the wrong remainder: the final subtraction gives (14x - 3) - (14x + 28) = -31, not -15. When subtracting (14x + 28) from (14x - 3), we get 14x - 3 - 14x - 28 = -31. Polynomial division has multiple steps where arithmetic errors can creep in—subtraction of polynomials is especially tricky. Double-check each subtraction step carefully! Verification is your friend: after dividing, multiply your quotient q(x) by the divisor b(x) and add your remainder r(x). You should get back the original numerator a(x). Let's check: (2x² - 5x + 14)(x + 2) + (-31) = 2x³ + 4x² - 5x² - 10x + 14x + 28 - 31 = 2x³ - x² + 4x - 3 ✓. If you don't get the original back, there's an error somewhere in your division. This check works every time and is much faster than redoing the whole division!
8. Rewrite $\frac{x^2+6}{x-1}$ in the form $q(x)+\frac{r(x)}{x-1}$ using polynomial division, where $\deg(r)<\deg(x-1)$.
   Answer: $x+1+\frac{7}{x-1}$ This question tests your understanding of polynomial division—rewriting a rational expression (fraction of polynomials) in the form quotient + remainder/divisor, just like how 17/5 = 3 + 2/5 in arithmetic. For simple cases, inspection can work: if you need (x² + 6)/(x - 1) and the numerator degree is just 1 more than denominator degree, the quotient will be linear. Try q(x) = x + a for some a, multiply (x + a)(x - 1), match coefficients with x² + 6, and solve for a and the remainder. For (x² + 6)/(x - 1), we suspect quotient is linear: q(x) = x + a for some a. Multiplying: (x + a)(x - 1) = x² - x + ax - a = x² + (a - 1)x - a. Matching with numerator x² + 0x + 6: coefficient of x gives a - 1 = 0, so a = 1. Constant term gives -a = -1, but we need +6, so remainder is 6 - (-1) = 7. Result: (x + 1) + 7/(x - 1). Inspection works when you can guess the quotient form! Choice A correctly shows x + 1 + 7/(x - 1) where the quotient is linear and remainder 7 has degree 0 (less than divisor degree 1), giving the proper rewritten form. Choice C shows remainder 5 instead of 7, which would come from an arithmetic error: if the quotient is x + 1, then (x + 1)(x - 1) = x² - 1, and the remainder would be (x² + 6) - (x² - 1) = 7, not 5. Always double-check the arithmetic when using inspection! Polynomial long division steps: (1) Divide leading terms to get first term of quotient, (2) Multiply entire divisor by that term, (3) Subtract from dividend, (4) Repeat with what remains until remainder degree < divisor degree. For this problem: x²/x = x, multiply x(x-1) = x² - x, subtract from x² + 6 to get x + 6, then x/x = 1, multiply 1(x-1) = x - 1, subtract from x + 6 to get 7. Answer: x + 1 + 7/(x-1).
9. Use polynomial division to rewrite the rational expression $\frac{2x^3+3x^2-5x+1}{x-2}$ in the form $q(x)+\frac{r(x)}{x-2}$, where $\deg(r)<\deg(x-2)$.
   Answer: $2x^2+7x+9+\frac{19}{x-2}$ This question tests your understanding of polynomial division—rewriting a rational expression (fraction of polynomials) in the form quotient + remainder/divisor, just like how 17/5 = 3 + 2/5 in arithmetic. Polynomial long division works exactly like the long division you learned in elementary school, just with polynomials instead of numbers: divide the leading terms, multiply back, subtract, bring down the next term, repeat. When you can't divide anymore (when what's left has smaller degree than the divisor), that leftover is your remainder, and what you've built up is your quotient. Dividing (2x³ + 3x² - 5x + 1) by (x - 2) using long division: (1) Divide leading terms: 2x³/x = 2x² (first term of quotient). (2) Multiply: 2x²(x - 2) = 2x³ - 4x². (3) Subtract from dividend: (2x³ + 3x² - 5x + 1) - (2x³ - 4x²) = 7x² - 5x + 1. (4) Divide again: 7x²/x = 7x (second term of quotient). (5) Multiply: 7x(x - 2) = 7x² - 14x. (6) Subtract: (7x² - 5x + 1) - (7x² - 14x) = 9x + 1. (7) Divide once more: 9x/x = 9 (third term of quotient). (8) Multiply: 9(x - 2) = 9x - 18. (9) Subtract: (9x + 1) - (9x - 18) = 19. Result: quotient is 2x² + 7x + 9, remainder is 19, giving 2x² + 7x + 9 + 19/(x - 2). Choice A correctly shows 2x² + 7x + 9 + 19/(x - 2) where the quotient is a degree-2 polynomial and remainder 19 has degree 0 (less than divisor degree 1), giving the proper rewritten form. Choice B has the wrong sign in the quotient (shows +7x - 9 instead of +7x + 9), which would come from an arithmetic error in the division process—polynomial division has multiple steps where arithmetic errors can creep in, especially in the subtraction steps. Verification is your friend: after dividing, multiply your quotient 2x² + 7x + 9 by the divisor (x - 2) and add your remainder 19. You get (2x² + 7x + 9)(x - 2) + 19 = 2x³ - 4x² + 7x² - 14x + 9x - 18 + 19 = 2x³ + 3x² - 5x + 1, which matches the original numerator perfectly! This check works every time and is much faster than redoing the whole division.
10. Use polynomial division to rewrite $\frac{3x^2-2x+5}{x-1}$ in the form $q(x)+\frac{r}{x-1}$, with $\deg(r)<1$.
    Answer: $3x+1+\frac{6}{x-1}$ This question tests your understanding of polynomial division—rewriting a rational expression (fraction of polynomials) in the form quotient + remainder/divisor, just like how $17/5 = 3 + 2/5$ in arithmetic. Polynomial long division works exactly like the long division you learned in elementary school, just with polynomials instead of numbers: divide the leading terms, multiply back, subtract, bring down the next term, repeat. Dividing $(3x^2 - 2x + 5)$ by $(x - 1)$ using long division: (1) Divide leading terms: $3x^2 / x = 3x$ (first term of quotient). (2) Multiply: $3x(x - 1) = 3x^2 - 3x$. (3) Subtract from dividend: $(3x^2 - 2x + 5) - (3x^2 - 3x) = x + 5$. (4) Divide again: $x / x = 1$ (second term of quotient). (5) Multiply: $1(x - 1) = x - 1$. (6) Subtract: $(x + 5) - (x - 1) = 6$. Result: quotient is $3x + 1$, remainder is 6, giving $(3x + 1) + \frac{6}{x - 1}$. Choice B correctly shows $3x + 1 + \frac{6}{x - 1}$ where the quotient is linear and remainder 6 has degree 0 (less than divisor degree 1), giving the proper rewritten form. Choice C has the wrong remainder: 4 instead of 6, which suggests an arithmetic error in the final subtraction—remember that $(x + 5) - (x - 1) = x + 5 - x + 1 = 6$, not 4! The degree requirement (degree of remainder < degree of divisor) is what makes the division 'done': the remainder 6 has degree 0, which is less than the divisor's degree 1, so we're done. If your remainder still had an x-term (degree 1), you could divide further, meaning you haven't finished.
11. Simplify and rewrite $\frac{x^2-4}{x-2}$ in the form $q(x)+\frac{r(x)}{x-2}$, where $\deg(r)<\deg(x-2)$. (You may factor first.)
    Answer: $x+2+\frac{0}{x-2}$ This question tests your understanding of polynomial division—rewriting a rational expression (fraction of polynomials) in the form quotient + remainder/divisor, just like how 17/5 = 3 + 2/5 in arithmetic. Before dividing, check if numerator factors: x² - 4 = (x + 2)(x - 2) using the difference of squares pattern. Since the numerator factors as (x + 2)(x - 2) and the denominator is (x - 2), we can cancel the common factor (x - 2), leaving just (x + 2) with no remainder term needed! Choice A correctly shows x + 2 + 0/(x - 2), which simplifies to just x + 2 since the remainder is 0. Choice C incorrectly suggests a remainder of 4, but when we factor and cancel, there's no remainder at all—the division is exact! Before starting polynomial division, always check: (1) Can I factor the numerator and cancel with the denominator? If yes, simplify first—it might eliminate the division entirely! This problem demonstrates why factoring first can save you from unnecessary long division work.
12. What are the quotient $q(x)$ and remainder $r(x)$ when dividing $x^3+4x^2+x+6$ by $x+3$? Then write $\frac{x^3+4x^2+x+6}{x+3}=q(x)+\frac{r(x)}{x+3}$ with $\deg(r)<1$.​
    Answer: $x^2+x-2+\frac{12}{x+3}$ This question tests your understanding of polynomial division—rewriting a rational expression (fraction of polynomials) in the form quotient + remainder/divisor, just like how 17/5 = 3 + 2/5 in arithmetic. The division algorithm for polynomials says that any rational expression a(x)/b(x) can be rewritten as q(x) + r(x)/b(x), where q(x) is the quotient (the polynomial part), and r(x) is the remainder (what's left over in the numerator). The crucial requirement: the degree of the remainder r(x) must be less than the degree of the divisor b(x)—just like in numerical division where the remainder must be less than the divisor! Dividing (x³ + 4x² + x + 6) by (x + 3) using long division: (1) Divide leading terms: x³/x = x² (first term of quotient). (2) Multiply: x²(x + 3) = x³ + 3x². (3) Subtract from dividend: (x³ + 4x² + x + 6) - (x³ + 3x²) = x² + x + 6. (4) Divide again: x²/x = x (second term of quotient). (5) Multiply: x(x + 3) = x² + 3x. (6) Subtract: (x² + x + 6) - (x² + 3x) = -2x + 6. (7) Divide once more: -2x/x = -2 (third term of quotient). (8) Multiply: -2(x + 3) = -2x - 6. (9) Subtract: (-2x + 6) - (-2x - 6) = 12. Result: quotient is x² + x - 2, remainder is 12, giving x² + x - 2 + 12/(x + 3). Choice A correctly shows x² + x - 2 + 12/(x + 3) where the quotient is a degree 2 polynomial and remainder 12 has degree 0 (less than divisor degree 1), giving the proper rewritten form. Choice B has remainder 0, which can't be right because multiplying back doesn't give the original: (x² + x - 2)(x + 3) + 0 = x³ + 3x² + x² + 3x - 2x - 6 = x³ + 4x² + x - 6 ≠ x³ + 4x² + x + 6. The quotient and remainder must satisfy: quotient × divisor + remainder = original numerator. This verification always catches division errors! Polynomial long division steps: (1) Divide leading terms to get first term of quotient, (2) Multiply entire divisor by that term, (3) Subtract from dividend, (4) Repeat with what remains until remainder degree < divisor degree. It's exactly like numerical long division! Practice with simpler examples first, then work up to more complex ones.
13. Express $\frac{x^4-1}{x^2-1}$ in the form $q(x)+\frac{r(x)}{x^2-1}$ using polynomial division, where $\deg(r)<2$.
    Answer: $x^2+1+\frac{0}{x^2-1}$ This question tests your understanding of polynomial division—rewriting a rational expression (fraction of polynomials) in the form quotient + remainder/divisor, just like how 17/5 = 3 + 2/5 in arithmetic. Before dividing, check if numerator factors: x⁴ - 1 = (x²)² - 1² = (x² - 1)(x² + 1) using difference of squares. Since the denominator is (x² - 1), we can cancel: (x⁴ - 1)/(x² - 1) = (x² - 1)(x² + 1)/(x² - 1) = x² + 1. The expression simplifies to x² + 1 with no remainder! Choice A correctly shows x² + 1 + 0/(x² - 1), which we can write simply as x² + 1 since adding 0 doesn't change the value. Choice B would give x² - 1, but that can't be right because multiplying back doesn't give the original: (x² - 1)(x² - 1) = x⁴ - 2x² + 1 ≠ x⁴ - 1. Before starting polynomial division, always check: (1) Can I factor the numerator and cancel with the denominator? If yes, simplify first—it might eliminate the division entirely! This problem demonstrates that factoring first can turn a potentially complex division into a simple cancellation.
14. Use polynomial division to rewrite the rational expression $\frac{2x^3+3x^2-5x+1}{x-2}$ in the form $q(x)+\frac{r(x)}{x-2}$, where $\deg(r)<\deg(x-2)$.
    Answer: $2x^2+7x+9+\frac{19}{x-2}$ This question tests your understanding of polynomial division—rewriting a rational expression (fraction of polynomials) in the form quotient + remainder/divisor, just like how 17/5 = 3 + 2/5 in arithmetic. Polynomial long division works exactly like the long division you learned in elementary school, just with polynomials instead of numbers: divide the leading terms, multiply back, subtract, bring down the next term, repeat. Dividing (2x³ + 3x² - 5x + 1) by (x - 2) using long division: (1) Divide leading terms: 2x³/x = 2x² (first term of quotient). (2) Multiply: 2x²(x - 2) = 2x³ - 4x². (3) Subtract from dividend: (2x³ + 3x² - 5x + 1) - (2x³ - 4x²) = 7x² - 5x + 1. (4) Divide again: 7x²/x = 7x. (5) Multiply: 7x(x - 2) = 7x² - 14x. (6) Subtract: (7x² - 5x + 1) - (7x² - 14x) = 9x + 1. (7) Divide once more: 9x/x = 9. (8) Multiply: 9(x - 2) = 9x - 18. (9) Subtract: (9x + 1) - (9x - 18) = 19. Result: quotient is 2x² + 7x + 9, remainder is 19, giving 2x² + 7x + 9 + 19/(x - 2). Choice A correctly shows 2x² + 7x + 9 + 19/(x - 2) where the quotient is a degree-2 polynomial and remainder 19 has degree 0 (less than divisor degree 1), giving the proper rewritten form. Choice C has the wrong remainder: 1 instead of 19, which suggests an arithmetic error in the final subtraction step—polynomial division has multiple steps where arithmetic errors can creep in, especially in subtraction! Verification is your friend: after dividing, multiply your quotient 2x² + 7x + 9 by the divisor (x - 2) and add your remainder 19. You get (2x² + 7x + 9)(x - 2) + 19 = 2x³ - 4x² + 7x² - 14x + 9x - 18 + 19 = 2x³ + 3x² - 5x + 1, which matches the original numerator perfectly! This check works every time and is much faster than redoing the whole division!
15. Use polynomial division to rewrite $\frac{x^3-x^2-4x+4}{x-2}$ as $q(x)+\frac{r(x)}{x-2}$, with $\deg(r)<\deg(x-2)$.
    Answer: $x^2+x-2+\frac{0}{x-2}$ This question tests your understanding of polynomial division—rewriting a rational expression (fraction of polynomials) in the form quotient + remainder/divisor, just like how 17/5 = 3 + 2/5 in arithmetic. Before dividing, check if numerator factors: let's test if x = 2 is a root of x³ - x² - 4x + 4. Substituting: 2³ - 2² - 4(2) + 4 = 8 - 4 - 8 + 4 = 0. Yes! So (x - 2) is a factor. We can use synthetic division or factor by grouping to find the other factor. Dividing (x³ - x² - 4x + 4) by (x - 2) using synthetic division with 2: Bring down 1, multiply by 2 to get 2, add to -1 to get 1. Multiply 1 by 2 to get 2, add to -4 to get -2. Multiply -2 by 2 to get -4, add to 4 to get 0. The quotient is x² + x - 2 with remainder 0. We can verify by factoring: x³ - x² - 4x + 4 = x²(x - 1) - 4(x - 1) = (x² - 4)(x - 1) = (x - 2)(x + 2)(x - 1). Wait, that gives three linear factors, not matching our division. Let me redo the division carefully: (1) x³/x = x², multiply x²(x - 2) = x³ - 2x², subtract to get x² - 4x + 4. (2) x²/x = x, multiply x(x - 2) = x² - 2x, subtract to get -2x + 4. (3) -2x/x = -2, multiply -2(x - 2) = -2x + 4, subtract to get 0. Quotient is x² + x - 2, remainder 0. Choice A correctly shows x² + x - 2 + 0/(x - 2) where the quotient is quadratic and remainder 0 indicates exact division. Choice B has a sign error in the quotient, showing x² - x - 2 instead of x² + x - 2. When we divided x² by x in step 2, we got +x, not -x, because we were dividing the positive x² term that remained after the first subtraction. Verification is your friend: (x² + x - 2)(x - 2) = x³ - 2x² + x² - 2x - 2x + 4 = x³ - x² - 4x + 4 ✓. The degree requirement (degree of remainder < degree of divisor) is what makes the division 'done': here our remainder is 0, which has no degree (or degree -∞), certainly less than the divisor's degree 1. A zero remainder means the division is exact—the divisor is a factor of the dividend!
16. Use polynomial division to express $\frac{x^3-2x^2+5x+1}{x-1}$ as $q(x)+\frac{r(x)}{x-1}$, where $\deg(r)<1$.​
    Answer: $x^2-x+4+\frac{5}{x-1}$ This question tests your understanding of polynomial division—rewriting a rational expression (fraction of polynomials) in the form quotient + remainder/divisor, just like how 17/5 = 3 + 2/5 in arithmetic. Polynomial long division works exactly like the long division you learned in elementary school, just with polynomials instead of numbers: divide the leading terms, multiply back, subtract, bring down the next term, repeat. When you can't divide anymore (when what's left has smaller degree than the divisor), that leftover is your remainder, and what you've built up is your quotient. Dividing (x³ - 2x² + 5x + 1) by (x - 1) using long division: (1) Divide leading terms: x³/x = x² (first term of quotient). (2) Multiply: x²(x - 1) = x³ - x². (3) Subtract from dividend: (x³ - 2x² + 5x + 1) - (x³ - x²) = -x² + 5x + 1. (4) Divide again: -x²/x = -x (second term of quotient). (5) Multiply: -x(x - 1) = -x² + x. (6) Subtract: (-x² + 5x + 1) - (-x² + x) = 4x + 1. (7) Divide once more: 4x/x = 4 (third term of quotient). (8) Multiply: 4(x - 1) = 4x - 4. (9) Subtract: (4x + 1) - (4x - 4) = 5. Result: quotient is x² - x + 4, remainder is 5, giving x² - x + 4 + 5/(x - 1). Choice A correctly shows x² - x + 4 + 5/(x - 1) where the quotient is a degree 2 polynomial and remainder 5 has degree 0 (less than divisor degree 1), giving the proper rewritten form. Choice B has the wrong remainder: the final subtraction gives (4x + 1) - (4x - 4) = 5, not 4. When subtracting (4x - 4) from (4x + 1), we get 4x + 1 - 4x + 4 = 5. Polynomial division has multiple steps where arithmetic errors can creep in—subtraction of polynomials is especially tricky. Double-check each subtraction step carefully! Verification is your friend: after dividing, multiply your quotient q(x) by the divisor b(x) and add your remainder r(x). You should get back the original numerator a(x). Let's check: (x² - x + 4)(x - 1) + 5 = x³ - x² - x² + x + 4x - 4 + 5 = x³ - 2x² + 5x + 1 ✓. If you don't get the original back, there's an error somewhere in your division. This check works every time and is much faster than redoing the whole division!
17. Rewrite the rational expression $\frac{x^2+5x+7}{x+2}$ in the form $q(x)+\frac{r(x)}{x+2}$ (quotient + remainder/divisor), with $\deg(r)<\deg(x+2)$.
    Answer: $x+3+\frac{1}{x+2}$ This question tests your understanding of polynomial division—rewriting a rational expression (fraction of polynomials) in the form quotient + remainder/divisor, just like how $17/5 = 3 + 2/5$ in arithmetic. Polynomial long division works exactly like the long division you learned in elementary school, just with polynomials instead of numbers: divide the leading terms, multiply back, subtract, bring down the next term, repeat. When you can't divide anymore (when what's left has smaller degree than the divisor), that leftover is your remainder, and what you've built up is your quotient. For $(x^2 + 5x + 7)/(x + 2)$, dividing: (1) $x^2/x = x$. (2) Multiply: $x(x + 2) = x^2 + 2x$. (3) Subtract: $(x^2 + 5x + 7) - (x^2 + 2x) = 3x + 7$. (4) $3x/x = 3$. (5) Multiply: $3(x + 2) = 3x + 6$. (6) Subtract: $(3x + 7) - (3x + 6) = 1$. Result: $x + 3 + 1/(x + 2)$. Choice A correctly shows $x + 3 + 1/(x + 2)$ where the quotient is linear and remainder has degree 0, less than divisor degree 1, giving the proper rewritten form. Choice B has a negative remainder, but the subtraction gives positive 1—arithmetic errors in subtraction are common, so verify by multiplying back! Practice with $(x^2 + 5x + 7)/(x + 2)$: divide $x^2$ by $x$ → $x$, multiply $x(x+2) = x^2 + 2x$, subtract to get $3x + 7$, divide $3x$ by $x$ → $3$, multiply $3(x+2) = 3x + 6$, subtract to get remainder $1$. The degree requirement (degree of remainder < degree of divisor) is what makes the division 'done': here, remainder $1$ has degree 0 < 1, so we're good—if it had an x-term, you'd divide further!
18. Use polynomial division to rewrite $\frac{x^4+2x^3-3x^2+4x-5}{x^2+1}$ in the form $q(x)+\frac{r(x)}{x^2+1}$, where $\deg(r)<2$.
    Answer: $x^2+2x-4+\frac{2x-1}{x^2+1}$ This question tests your understanding of polynomial division—rewriting a rational expression in the form quotient + remainder/divisor, just like how 17/5 = 3 + 2/5 in arithmetic. Polynomial long division works exactly like the long division you learned in elementary school, just with polynomials instead of numbers: divide the leading terms, multiply back, subtract, bring down the next term, repeat. When you can't divide anymore (when what's left has smaller degree than the divisor), that leftover is your remainder, and what you've built up is your quotient. For $\frac{x^4 + 2x^3 - 3x^2 + 4x - 5}{x^2 + 1}$, dividing: (1) $x^4 / x^2 = x^2$. (2) Multiply: $x^2(x^2 + 1) = x^4 + x^2$. (3) Subtract: gets $2x^3 - 4x^2 + 4x - 5$. (4) $2x^3 / x^2 = 2x$. (5) Multiply: $2x(x^2 + 1) = 2x^3 + 2x$. (6) Subtract: gets $-4x^2 + 2x - 5$. (7) $-4x^2 / x^2 = -4$. (8) Multiply: $-4(x^2 + 1) = -4x^2 - 4$. (9) Subtract: gets $2x - 1$. Result: $x^2 + 2x - 4 + \frac{2x - 1}{x^2 + 1}$. Choice A correctly shows $x^2 + 2x - 4 + \frac{2x - 1}{x^2 + 1}$ where the quotient is quadratic and remainder has degree 1, less than divisor degree 2, giving the proper rewritten form. Choice D's remainder has degree 2 which is NOT less than the divisor's degree 2—if the remainder has degree ≥ divisor degree, we can divide further, so keep dividing until the remainder's degree is strictly less! The degree requirement (degree of remainder < degree of divisor) is what makes the division 'done': here, remainder $2x - 1$ has degree 1 < 2, so we're good. For Algebra 1, stick to divisors of degree 1 or simple degree 2 like $x^2 + 1$, and know that technology can handle messier cases.
19. Verify by multiplying back that the following rewriting is correct. Which expression equals $\frac{x^2 + 6}{x - 1}$ in the form $q(x) + \frac{r(x)}{x - 1}$ with $\deg(r) < 1$?
    Answer: $x + 1 + \frac{7}{x - 1}$ This question tests your understanding of polynomial division—rewriting a rational expression in the form quotient + remainder/divisor, just like how $\frac{17}{5} = 3 + \frac{2}{5}$ in arithmetic. Rewriting reveals structure: the expression $\frac{x^2 + 6}{x - 1}$ looks like one complicated fraction, but rewritten as $x + 1 + \frac{7}{x - 1}$, we see it's a linear function plus a small fraction term. For large $|x|$, the fraction part $\to 0$, so the expression behaves like $x + 1$. This separation into polynomial + fractional part is very useful for understanding the expression! Dividing $(x^2 + 0x + 6)$ by $(x - 1)$: (1) $x^2 / x = x$. (2) Multiply: $x(x - 1) = x^2 - x$. (3) Subtract: gets $x + 6$. (4) $x / x = 1$. (5) Multiply: $1(x - 1) = x - 1$. (6) Subtract: $(x + 6) - (x - 1) = 7$. Result: $x + 1 + \frac{7}{x - 1}$. Choice A correctly shows $x + 1 + \frac{7}{x - 1}$ where the quotient is linear and remainder has degree 0, less than 1, giving the proper rewritten form. Choice B has remainder 5, but multiplying back $(x + 1)(x - 1) + 5 = x^2 - 1 + 5 = x^2 + 4 \neq x^2 + 6$—the quotient and remainder must satisfy quotient $\times$ divisor + remainder = original numerator, so this verification catches errors! Polynomial long division steps: (1) Divide leading terms, (2) Multiply entire divisor, (3) Subtract, (4) Repeat until remainder degree < divisor degree. It's exactly like numerical long division!
20. Use polynomial division to rewrite $\frac{x^3+2x^2+3x+4}{x^2+1}$ in the form $q(x)+\frac{r(x)}{x^2+1}$, where $\deg(r)<2$.
    Answer: $x+2+\frac{2x+2}{x^2+1}$ This question tests your understanding of polynomial division—rewriting a rational expression in the form quotient + remainder/divisor, just like how $17/5 = 3 + 2/5$ in arithmetic. Polynomial long division works exactly like the long division you learned in elementary school, just with polynomials instead of numbers: divide the leading terms, multiply back, subtract, bring down the next term, repeat. When you can't divide anymore (when what's left has smaller degree than the divisor), that leftover is your remainder, and what you've built up is your quotient. For $(x^3 + 2x^2 + 3x + 4)/(x^2 + 1)$, dividing: (1) $x^3/x^2 = x$. (2) Multiply: $x(x^2 + 1) = x^3 + x$. (3) Subtract: gets $2x^2 + 2x + 4$. (4) $2x^2/x^2 = 2$. (5) Multiply: $2(x^2 + 1) = 2x^2 + 2$. (6) Subtract: gets $2x + 2$. Result: $x + 2 + (2x + 2)/(x^2 + 1)$. Choice A correctly shows $x + 2 + (2x + 2)/(x^2 + 1)$ where the quotient is linear and remainder has degree 1, less than divisor degree 2, giving the proper rewritten form. Choice D's remainder has degree 2 which is NOT less than the divisor's degree 2—if the remainder has degree ≥ divisor degree, we can divide further, so keep dividing until the remainder's degree is strictly less! The degree requirement (degree of remainder < degree of divisor) is what makes the division 'done': here, remainder $2x + 2$ has degree 1 < 2, so we're good. For Algebra 1, stick to divisors of degree 1 or simple degree 2 like $x^2 + 1$, and know that technology can handle messier cases.

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