Algebra · Learn by Concept

Algebra Help: Rewriting Rational Expressions

Review real example questions for Rewriting Rational Expressions in Algebra.

Question 1 / 10

0 of 10 answered

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 $\deg(r)<1$ .

All questions

Question 1

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 $\deg(r)<1$ .

$x^2+x-2+\frac{12}{x+3}$ (correct answer)
$x^2+x-2+\frac{0}{x+3}$
$x^2+x+2+\frac{12}{x+3}$
$x^2+7x+6+\frac{0}{x+3}$

Explanation: 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.

Question 2

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 $\deg(r)<1$ .

$x^2-x+2+\frac{0}{x-3}$
$x^2-7x+22+\frac{72}{x-3}$
$x^2-x-2+\frac{0}{x-3}$ (correct answer)
$x^2-x-2+\frac{6}{x-3}$

Explanation: 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. 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! Before dividing, check if numerator factors: $x^3 - 4x^2 + x + 6$ factors as $(x - 3)(x^2 - x - 2)$ since it divides evenly (remainder 0), so the expression simplifies to $x^2 - x - 2$ with no remainder term! Choice A correctly shows $x^2 - x - 2 + \frac{0}{x-3}$ where the quotient is quadratic and remainder has degree less than 1 (it's 0), giving the proper rewritten form. Choice B has a remainder of 6, but since it divides evenly, remainder should be 0—always check if factoring first makes the division unnecessary by plugging in the root (x=3) to see if it's zero! 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! 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). If you don't, there's an error somewhere in your division.

Question 3

Rewrite $\frac{2x^2+7x+1}{x+3}$ in the form $q(x)+\frac{r(x)}{x+3}$ (quotient plus remainder over divisor), where $\deg(r)<1$ .

$2x+1+\frac{-2}{x+3}$ (correct answer)
$2x+7+\frac{-20}{x+3}$
$2x-1+\frac{4}{x+3}$
$2x+1+\frac{2}{x+3}$

Explanation: 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 (2x² + 7x + 1)/(x + 3) and the numerator degree is just 1 more than denominator degree, the quotient will be linear. Try q(x) = 2x + a for some a, multiply (2x + a)(x + 3), match coefficients with 2x² + 7x + 1, and solve for a and the remainder. This 'educated guess and check' is faster than formal division when it works! For (2x² + 7x + 1)/(x + 3), we suspect quotient is linear: q(x) = 2x + a for some a. Multiplying: (2x + a)(x + 3) = 2x² + 6x + ax + 3a = 2x² + (6 + a)x + 3a. Matching with numerator 2x² + 7x + 1: coefficient of x gives 6 + a = 7, so a = 1. Constant term gives 3a = 3, but we have 1, so remainder is 1 - 3 = -2. Result: (2x + 1) + (-2)/(x + 3). Inspection works when you can guess the quotient form! Choice A correctly shows 2x + 1 + (-2)/(x + 3) where the quotient is linear and remainder -2 has degree 0 (less than divisor degree 1), giving the proper rewritten form. Choice B has the quotient wrong: when dividing 2x² by x, we get 2x, not something that would lead to 2x + 7. The leading term of the quotient comes from dividing the highest-degree terms of numerator and denominator. This is the first step of polynomial division! 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: (2x + 1)(x + 3) + (-2) = 2x² + 6x + x + 3 - 2 = 2x² + 7x + 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!

Question 4

Rewrite $\frac{2x^3+x^2-4x+5}{x+1}$ in the form $q(x)+\frac{r(x)}{x+1}$ using polynomial division, with $\deg(r)<\deg(x+1)$ .

$2x^2-x-3+\frac{8}{x+1}$ (correct answer)
$2x^2-x-3+\frac{-8}{x+1}$
$2x^2+x-3+\frac{8}{x+1}$
$2x^2-x+3+\frac{8}{x+1}$

Explanation: 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.

Question 5

Use polynomial division to express $\frac{x^3-2x^2+5x+1}{x-1}$ as $q(x)+\frac{r(x)}{x-1}$ , where $\deg(r)<1$ .

$x^2-x+4+\frac{5}{x-1}$ (correct answer)
$x^2-x+4+\frac{4}{x-1}$
$x^2+x+4+\frac{5}{x-1}$
$x^2-x+3+\frac{5}{x-1}$

Explanation: 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!

Question 6

Simplify and rewrite $\frac{x^2-4}{x-2}$ in the form $q(x)+\frac{r(x)}{x-2}$ , with $\deg(r)<1$ .

$x+2+\frac{x}{x-2}$
$x+2+\frac{4}{x-2}$
$x-2+\frac{0}{x-2}$
$x+2+\frac{0}{x-2}$ (correct answer)

Explanation: 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. 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! Before dividing, check if numerator factors: $x^2 - 4 = (x - 2)(x + 2)$ , so $(x - 2)(x + 2)/(x - 2)$ simplifies to $x + 2$ with remainder 0 (for $x ≠ 2$ ). Choice A correctly shows $x + 2 + 0/(x - 2)$ where the quotient is linear and remainder has degree less than 1 (it's 0), giving the proper rewritten form. Choice C has remainder 4, but since it factors and cancels, no remainder is needed—always check if factoring first makes the division unnecessary! 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 $a(x)$ . If you don't, there's an error somewhere in your division.

Question 7

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 $\deg(r) < 1$ ?

$x + 1 + \frac{5}{x - 1}$
$x + 1 + \frac{7x}{x - 1}$
$x + 1 + \frac{7}{x - 1}$ (correct answer)
$x - 1 + \frac{7}{x - 1}$

Explanation: 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!

Question 8

Use polynomial division to rewrite $\frac{2x^3+3x^2-5x+1}{x-2}$ in the form $q(x)+\frac{r(x)}{x-2}$ , where $\deg(r)<\deg(x-2)$ .

$2x^2+7x+9+\frac{19}{x-2}$ (correct answer)
$2x^2+7x+9+\frac{17}{x-2}$
$2x^2+7x-9+\frac{19}{x-2}$
$2x^2+3x+1+\frac{0}{x-2}$

Explanation: 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 remainder: the final subtraction gives (9x + 1) - (9x - 18) = 19, not 17. This is an arithmetic error in the last step of the division process. 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 + 7x + 9)(x - 2) + 19 = 2x³ - 4x² + 7x² - 14x + 9x - 18 + 19 = 2x³ + 3x² - 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!

Question 9

Rewrite $\frac{x^3+4x^2+x-6}{x+2}$ in the form $q(x)+\frac{r(x)}{x+2}$ using polynomial division, with $\deg(r)<\deg(x+2)$ .

$x^2+2x-3+\frac{0}{x+2}$ (correct answer)
$x^2+2x+3+\frac{0}{x+2}$
$x^2+2x-3+\frac{6}{x+2}$
$x^2+2x+\frac{-3}{x+2}$

Explanation: 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: attempting to factor x³ + 4x² + x - 6. Let's try grouping or synthetic division with x = -2: (-2)³ + 4(-2)² + (-2) - 6 = -8 + 16 - 2 - 6 = 0. So (x + 2) is a factor! Using synthetic division or factoring, we find x³ + 4x² + x - 6 = (x + 2)(x² + 2x - 3), so the expression simplifies to (x² + 2x - 3) with no remainder! Choice A correctly shows x² + 2x - 3 + 0/(x + 2), which we can write simply as x² + 2x - 3 since the remainder is 0. Choice C shows a remainder of 6, but that can't be right because we've shown the numerator is exactly divisible by (x + 2)—when a polynomial divides evenly, the remainder must be 0. Verification is your friend: after dividing, multiply your quotient x² + 2x - 3 by the divisor (x + 2) and add your remainder 0. You get (x² + 2x - 3)(x + 2) + 0 = x³ + 2x² + 2x² + 4x - 3x - 6 = x³ + 4x² + x - 6, which matches the original numerator perfectly! 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! (2) Is the numerator degree less than denominator degree already? Then q(x) = 0 and r(x) = a(x). (3) If neither, proceed with division. In this case, checking for factors saved us from doing long division!

Question 10

Rewrite $\frac{x^2+6}{x-1}$ in the form $q(x)+\frac{r(x)}{x-1}$ using polynomial division, where $\deg(r)<\deg(x-1)$ .

$x+1+\frac{7}{x-1}$ (correct answer)
$x-1+\frac{7}{x-1}$
$x+1+\frac{5}{x-1}$
$x+\frac{1}{x-1}+7$

Explanation: 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).