Algebra 2 Quiz: Rational Zeros Theorem
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Rational Zeros TheoremQuestion 1 of 20

Use the Rational Zeros Theorem to list all possible rational zeros of P(x)=3x3−2x2+x−6.P(x)=3x^3-2x^2+x-6. (Remember: any rational zero pq\frac{p}{q} in lowest terms has p∣p\mid constant term and q∣q\mid leading coefficient.)​

±1,±2,±3,±6\pm 1,\pm 2,\pm 3,\pm 6
±1,±2,±3,±6,±13,±23,±43\pm 1,\pm 2,\pm 3,\pm 6,\pm \frac{1}{3},\pm \frac{2}{3},\pm \frac{4}{3}
±1,±2,±3,±6,±13,±23,±12,±32\pm 1,\pm 2,\pm 3,\pm 6,\pm \frac{1}{3},\pm \frac{2}{3},\pm \frac{1}{2},\pm \frac{3}{2}
±1,±2,±3,±6,±13,±23\pm 1,\pm 2,\pm 3,\pm 6,\pm \frac{1}{3},\pm \frac{2}{3}
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Algebra 2 Quiz

Algebra 2 Quiz: Rational Zeros Theorem

Practice Rational Zeros Theorem in Algebra 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Rational Zeros Theorem, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Use the Rational Zeros Theorem to list all possible rational zeros of P(x)=3x3−2x2+x−6.P(x)=3x^3-2x^2+x-6. (Remember: any rational zero pq\frac{p}{q} in lowest terms has p∣p\mid constant term and q∣q\mid leading coefficient.)​

  1. ±1,±2,±3,±6\pm 1,\pm 2,\pm 3,\pm 6
  2. ±1,±2,±3,±6,±13,±23,±43\pm 1,\pm 2,\pm 3,\pm 6,\pm \frac{1}{3},\pm \frac{2}{3},\pm \frac{4}{3}
  3. ±1,±2,±3,±6,±13,±23,±12,±32\pm 1,\pm 2,\pm 3,\pm 6,\pm \frac{1}{3},\pm \frac{2}{3},\pm \frac{1}{2},\pm \frac{3}{2}
  4. ±1,±2,±3,±6,±13,±23\pm 1,\pm 2,\pm 3,\pm 6,\pm \frac{1}{3},\pm \frac{2}{3} (correct answer)
Explanation: This question tests your understanding of the Rational Zeros Theorem—a powerful tool for finding possible rational zeros of polynomials, dramatically narrowing what values to test. The Rational Zeros Theorem states that if polynomial P(x) with integer coefficients has a rational zero p/q (in lowest terms), then p must be a factor of the constant term and q must be a factor of the leading coefficient. For P(x) = 3x^3 - 2x^2 + x - 6, the constant term is -6 (factors: ±1, ±2, ±3, ±6), and the leading coefficient is 3 (factors: ±1, ±3), so possible rational zeros are ±1, ±2, ±3, ±6, ±1/3, ±2/3, ±3/3 (±1, already listed), ±6/3 (±2, already listed)—giving the unique list ±1, ±2, ±3, ±6, ±1/3, ±2/3. Choice B correctly lists all these possible rational zeros by properly identifying factors and forming fractions p/q without duplicates. A tempting distractor like choice A fails by omitting the fractional possibilities when the leading coefficient isn't 1, forgetting that q can be ±3, which introduces thirds like ±1/3 and ±2/3—always include all combinations! To apply the Rational Zeros Theorem effectively, list all ± factors of the constant for p and ± factors of the leading for q, form p/q, simplify to avoid repeats, and remember this list contains every possible rational zero—now test them to find actual ones. You're doing great; practicing this will make polynomial solving much easier and more efficient!

Question 2

Use the Rational Zeros Theorem to list all possible rational zeros of P(x)=3x3−2x2+x−6.P(x)=3x^3-2x^2+x-6. (Remember: any rational zero pq\frac{p}{q} in lowest terms has p∣p\mid constant term and q∣q\mid leading coefficient.)

  1. ±1,±2,±3,±6,±13,±23,±43\pm 1,\pm 2,\pm 3,\pm 6,\pm \frac{1}{3},\pm \frac{2}{3},\pm \frac{4}{3}
  2. ±1,±2,±3,±6,±13,±23\pm 1,\pm 2,\pm 3,\pm 6,\pm \frac{1}{3},\pm \frac{2}{3} (correct answer)
  3. ±1,±2,±3,±6\pm 1,\pm 2,\pm 3,\pm 6
  4. ±1,±2,±3,±6,±13,±23,±12,±32\pm 1,\pm 2,\pm 3,\pm 6,\pm \frac{1}{3},\pm \frac{2}{3},\pm \frac{1}{2},\pm \frac{3}{2}
Explanation: This question tests your understanding of the Rational Zeros Theorem—a powerful tool for finding possible rational zeros of polynomials, dramatically narrowing what values to test. The Rational Zeros Theorem states that if polynomial P(x) with integer coefficients has a rational zero p/q (in lowest terms), then p must be a factor of the constant term and q must be a factor of the leading coefficient. For P(x) = 3x^3 - 2x^2 + x - 6, the constant term is -6 (factors: ±1,±2,±3,±6\pm1, \pm2, \pm3, \pm6), and the leading coefficient is 3 (factors: ±1,±3\pm1, \pm3), so possible rational zeros are ±1,±2,±3,±6,±13,±23,±33\pm1, \pm2, \pm3, \pm6, \pm\frac{1}{3}, \pm\frac{2}{3}, \pm\frac{3}{3} (±1\pm1, already listed), ±63\pm\frac{6}{3} (±2\pm2, already listed)—giving the unique list ±1,±2,±3,±6,±13,±23\pm1, \pm2, \pm3, \pm6, \pm\frac{1}{3}, \pm\frac{2}{3}. Choice B correctly lists all these possible rational zeros by properly identifying factors and forming fractions p/q without duplicates. A tempting distractor like choice A fails by omitting the fractional possibilities when the leading coefficient isn't 1, forgetting that q can be ±3\pm3, which introduces thirds like ±13\pm\frac{1}{3} and ±23\pm\frac{2}{3}—always include all combinations! To apply the Rational Zeros Theorem effectively, list all ±\pm factors of the constant for p and ±\pm factors of the leading for q, form p/q, simplify to avoid repeats, and remember this list contains every possible rational zero—now test them to find actual ones. You're doing great; practicing this will make polynomial solving much easier and more efficient!

Question 3

Possible rational zeros of P(x)=x3−4x2+x+6P(x)=x^3-4x^2+x+6 (from the Rational Zeros Theorem) are ±1,±2,±3,±6\pm 1,\pm 2,\pm 3,\pm 6. Which of these candidates are actual zeros? (Test by substitution or synthetic division.)

  1. x=−1x=-1 and x=2x=2 only
  2. x=1x=1 and x=−2x=-2 only
  3. x=−1,  x=2,  x=3x=-1,\;x=2,\;x=3 (correct answer)
  4. x=−2x=-2 and x=3x=3 only
Explanation: This question tests your understanding of the Rational Zeros Theorem—a powerful tool for finding possible rational zeros of polynomials, dramatically narrowing what values to test. The Rational Zeros Theorem states that if polynomial P(x)P(x) with integer coefficients has a rational zero p/qp/q (in lowest terms), then pp must be a factor of the constant term and qq must be a factor of the leading coefficient. Given candidates ±1,±2,±3,±6\pm 1, \pm 2, \pm 3, \pm 6 for P(x)=x3−4x2+x+6P(x)=x^3 -4x^2 +x +6, testing shows P(−1)=0P(-1)=0, P(2)=0P(2)=0, P(3)=0P(3)=0, but others like P(1)=4≠0P(1)=4 \neq 0, confirming actual zeros −1,2,3-1, 2, 3. Choice C correctly identifies all three actual zeros from the candidates, verified by substitution or synthetic division. Choice A omits x=3x=3, perhaps from an arithmetic error like miscalculating P(3)=27−36+3+6=0P(3)=27-36+3+6=0 correctly. Rational Zeros Theorem application process: with list provided, test each systematically—calculate P(candidate)P(\text{candidate}) carefully, tracking terms to avoid errors. If zero found, factor out (x−r)(x - r) using synthetic division and repeat on quotient—keep going, you're building polynomial mastery!

Question 4

Consider polynomials of the form g(x)=ax3+bx2+cx+12g(x) = ax^3 + bx^2 + cx + 12 where aa is a positive integer. For which value of aa would the Rational Zero Theorem yield the fewest possible rational zero candidates?

  1. a=12a = 12, because this creates symmetry between the leading and constant terms
  2. a=1a = 1, because this minimizes the number of factors of the leading coefficient (correct answer)
  3. a=2a = 2, because this is the smallest prime factor of the constant term
  4. The number of candidates is independent of aa, since only the constant term matters
Explanation: When you encounter questions about the Rational Zero Theorem, focus on how it generates candidates: any rational zero pq\frac{p}{q} must have pp dividing the constant term and qq dividing the leading coefficient. For g(x)=ax3+bx2+cx+12g(x) = ax^3 + bx^2 + cx + 12, the possible rational zeros are pq\frac{p}{q} where pp divides 12 and qq divides aa. The factors of 12 are ±1,±2,±3,±4,±6,±12\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12, giving us 12 possible values for pp. The number of candidates depends entirely on how many factors aa has, since each factor of aa creates new denominators to pair with each factor of 12. When a=1a = 1, the only factor is 1 itself, so all candidates have the form p1\frac{p}{1}, giving us exactly 12 candidates. This is the minimum possible. Choice A is incorrect because a=12a = 12 actually maximizes candidates rather than minimizing them. The factors of 12 are 1,2,3,4,6,121, 2, 3, 4, 6, 12, creating many more rational zero candidates than a=1a = 1. Symmetry between coefficients doesn't reduce the candidate count. Choice C misses the point because while 2 is indeed the smallest prime factor of 12, what matters is minimizing the factors of aa, not relating aa to the constant term. Choice D is wrong because the leading coefficient absolutely matters—it determines the possible denominators in our rational zero candidates. Remember: to minimize Rational Zero Theorem candidates, choose the leading coefficient with the fewest factors, which is always 1 for positive integers.

Question 5

Consider the polynomial f(x)=4x4−12x3+9x2−27x+18f(x) = 4x^4 - 12x^3 + 9x^2 - 27x + 18. A student correctly lists all possible rational zeros using the Rational Zero Theorem, then discovers that exactly half of these candidates are actually zeros of f(x)f(x). How many zeros of f(x)f(x) are rational?

  1. 6 rational zeros, since there are 12 possible candidates from the theorem
  2. 8 rational zeros, since there are 16 possible candidates from the theorem
  3. 9 rational zeros, since there are 18 possible candidates from the theorem
  4. This scenario is impossible, since f(x)f(x) has degree 4 and cannot have more than 4 zeros (correct answer)
Explanation: The Rational Zero Theorem gives possible candidates: ±1,±2,±3,±6,±9,±18,±12,±32,±92,±14,±34,±94\pm 1, \pm 2, \pm 3, \pm 6, \pm 9, \pm 18, \pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{9}{2}, \pm \frac{1}{4}, \pm \frac{3}{4}, \pm \frac{9}{4} (24 candidates total). If half were zeros, that would be 12 zeros, but a degree 4 polynomial can have at most 4 zeros. The scenario described is mathematically impossible. Choices A, B, and C incorrectly accept the premise without recognizing the degree constraint.

Question 6

A student applies the Rational Zero Theorem to f(x)=6x3−13x2+6x−1f(x) = 6x^3 - 13x^2 + 6x - 1 and creates the list: ±1,±12,±13,±16\pm 1, \pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{1}{6}. After testing x=1x = 1, the student finds f(1)=−2≠0f(1) = -2 \neq 0. What should the student conclude?

  1. The polynomial has no rational zeros, since the most likely candidate failed
  2. The student made an error in applying the theorem, since the list is incomplete
  3. The student should continue testing the remaining candidates before drawing any conclusions (correct answer)
  4. The polynomial must have irrational zeros, since f(1)<0f(1) < 0 indicates sign changes
Explanation: The Rational Zero Theorem provides a complete list of possible rational zeros, but testing must be systematic. Finding that f(1)≠0f(1) \neq 0 only eliminates x=1x = 1 as a zero; the other candidates must still be tested. Choice A prematurely concludes no rational zeros exist. Choice B is incorrect since the list includes all valid candidates (factors of constant term 1 over factors of leading coefficient 6). Choice D incorrectly interprets the sign of f(1)f(1) as indicating the nature of zeros.

Question 7

A polynomial F(x)=20x5+ax4+bx3+cx2+dx+63F(x) = 20x^5 + ax^4 + bx^3 + cx^2 + dx + 63 has exactly three rational zeros. Using the Rational Zero Theorem, what is the maximum number of negative rational zeros that F(x)F(x) could have?

  1. At most 1 negative rational zero, due to degree constraints and sign patterns
  2. At most 2 negative rational zeros, since the remaining zero could be positive
  3. All 3 rational zeros could be negative, depending on the values of the coefficients (correct answer)
  4. The maximum depends on the specific values of a,b,c,da, b, c, d, which are not provided
Explanation: The Rational Zero Theorem gives possible candidates like ±1,±3,±7,±9,±21,±63,±12,±32,...\pm 1, \pm 3, \pm 7, \pm 9, \pm 21, \pm 63, \pm \frac{1}{2}, \pm \frac{3}{2}, ... etc. There's no theoretical constraint preventing all three rational zeros from being negative. For example, if the zeros were −1,−32,−74-1, -\frac{3}{2}, -\frac{7}{4}, this would be consistent with the theorem and the given information. Choices A and B incorrectly impose unnecessary constraints. Choice D suggests the answer is indeterminate, but the question asks for the theoretical maximum.

Question 8

The polynomial h(x)=kx4−8x3+mx2+nx−6h(x) = kx^4 - 8x^3 + mx^2 + nx - 6 has 32\frac{3}{2} as a rational zero. If all coefficients are integers and the Rational Zero Theorem is to be applied effectively, what constraint must kk satisfy?

  1. kk must be divisible by 3, so that 32\frac{3}{2} appears in the list of possible rational zeros
  2. kk must be even, so that 32\frac{3}{2} appears in the list of possible rational zeros (correct answer)
  3. kk must be positive and divisible by 2, ensuring the theorem generates valid candidates
  4. kk can be any nonzero integer, since 32\frac{3}{2} being a zero provides no constraint on kk
Explanation: When applying the Rational Zero Theorem, you need to understand how it generates the list of possible rational zeros. The theorem states that any rational zero pq\frac{p}{q} of a polynomial must have pp as a factor of the constant term and qq as a factor of the leading coefficient. Since 32\frac{3}{2} is a zero of h(x)=kx4−8x3+mx2+nx−6h(x) = kx^4 - 8x^3 + mx^2 + nx - 6, we need 3 to be a factor of the constant term (-6) and 2 to be a factor of the leading coefficient (kk). The factors of -6 are ±1, ±2, ±3, ±6, so 3 is indeed available. For 32\frac{3}{2} to appear in our list of possible rational zeros, 2 must be a factor of kk, meaning kk must be even. Choice A is incorrect because requiring kk to be divisible by 3 isn't necessary—we need the numerator 3 to divide the constant term (-6), which it already does. Choice C is wrong because kk doesn't need to be positive; it can be any even integer (positive or negative). Choice D misses the point entirely—the Rational Zero Theorem does constrain kk because for 32\frac{3}{2} to be a possible rational zero, 2 must divide kk. Choice B correctly identifies that kk must be even so that 2 divides kk, allowing 32\frac{3}{2} to appear in the theorem's list of candidates. Study tip: Always check both parts of a rational zero pq\frac{p}{q}—the numerator must divide the constant term, and the denominator must divide the leading coefficient.

Question 9

Consider the polynomial f(x)=6x4−7x3+2x2−8x+12f(x) = 6x^4 - 7x^3 + 2x^2 - 8x + 12. If pp is a rational zero of f(x)f(x), which of the following statements must be true about the numerator and denominator of pp when written in lowest terms as ab\frac{a}{b}?

  1. aa divides 12 and bb divides 6, where gcd⁡(a,b)=1\gcd(a,b) = 1 (correct answer)
  2. aa divides 6 and bb divides 12, where gcd⁡(a,b)=1\gcd(a,b) = 1
  3. aa divides 12 and bb divides 6, where aa and bb are both positive
  4. aa divides 6 and bb divides 12, where aa and bb are both positive
Explanation: By the Rational Zero Theorem, if p=abp = \frac{a}{b} is a rational zero in lowest terms, then aa must divide the constant term (12) and bb must divide the leading coefficient (6). The condition gcd⁡(a,b)=1\gcd(a,b) = 1 ensures the fraction is in lowest terms. Choice B reverses the roles of numerator and denominator. Choices C and D incorrectly require both aa and bb to be positive, but rational zeros can be negative.

Question 10

A polynomial g(x)g(x) has leading coefficient −3-3 and constant term 2020. After applying the Rational Zero Theorem and testing several candidates, a student determines that 43\frac{4}{3} and −51-\frac{5}{1} are zeros of g(x)g(x). What is the minimum possible degree of g(x)g(x)?

  1. The minimum degree is 2, since two distinct zeros were found (correct answer)
  2. The minimum degree is 3, since the leading coefficient creates additional constraints
  3. The minimum degree is 4, since 43\frac{4}{3} might be a repeated zero
  4. The minimum degree cannot be determined from the given information alone
Explanation: A polynomial of degree nn has at most nn zeros (counting multiplicity). Since we found 2 distinct zeros, the polynomial must be at least degree 2. A degree 2 polynomial like g(x)=−3(x−43)(x+5)=−3x2−11x+20g(x) = -3(x - \frac{4}{3})(x + 5) = -3x^2 - 11x + 20 satisfies all conditions. Choice B incorrectly suggests the leading coefficient affects minimum degree. Choice C assumes repetition without evidence. Choice D is incorrect since we have sufficient information.

Question 11

Use the Rational Zeros Theorem to help factor completely: P(x)=2x3−3x2−8x+12.P(x)=2x^3-3x^2-8x+12. (Find a rational zero by testing candidates, divide to reduce the degree, and continue factoring.)

  1. (x−2)(2x2+x−6)(x-2)(2x^2+x-6)
  2. (x+2)(2x2−7x+6)(x+2)(2x^2-7x+6)
  3. (x−2)(x+2)(2x−3)(x-2)(x+2)(2x-3) (correct answer)
  4. (2x−3)(x2−4)(2x-3)(x^2-4)
Explanation: This question tests your understanding of the Rational Zeros Theorem—a powerful tool for finding possible rational zeros of polynomials, dramatically narrowing what values to test. The Rational Zeros Theorem states that if polynomial P(x)P(x) with integer coefficients has a rational zero p/qp/q (in lowest terms), then pp must be a factor of the constant term and qq must be a factor of the leading coefficient. For P(x)=2x3−3x2−8x+12P(x)=2x^3 - 3x^2 - 8x + 12, candidates include ±1, ±2, ±3, ±4, ±6, ±12, ±1/2, ±3/2; testing finds x=2x=2 zero, dividing gives (x−2)(2x2+x−6)(x-2)(2x^2 + x - 6), which factors to (x−2)(x+2)(2x−3)(x-2)(x+2)(2x-3). Choice C correctly gives the complete factorization with all linear factors over rationals. Choice A stops without factoring the quadratic fully, missing that 2x2+x−6=(2x−3)(x+2)2x^2 + x - 6 = (2x-3)(x+2). Rational Zeros Theorem application process: test candidates to find one zero, divide to get quotient, repeat theorem on it for more. This step-by-step factoring is key—keep practicing, you're doing amazingly!

Question 12

Use the Rational Zeros Theorem to list all possible rational zeros of P(x)=4x3−3x2−8x+6.P(x)=4x^3-3x^2-8x+6. (List candidates p/qp/q in lowest terms.)

  1. ±1,±2,±3,±6,±24,±64\pm 1,\pm 2,\pm 3,\pm 6,\pm \frac{2}{4},\pm \frac{6}{4}
  2. ±1,±2,±3,±6\pm 1,\pm 2,\pm 3,\pm 6
  3. ±1,±2,±3,±6,±12,±32\pm 1,\pm 2,\pm 3,\pm 6,\pm \frac{1}{2},\pm \frac{3}{2}
  4. ±1,±2,±3,±6,±12,±32,±14,±34\pm 1,\pm 2,\pm 3,\pm 6,\pm \frac{1}{2},\pm \frac{3}{2},\pm \frac{1}{4},\pm \frac{3}{4} (correct answer)
Explanation: This question tests your understanding of the Rational Zeros Theorem—a powerful tool for finding possible rational zeros of polynomials, dramatically narrowing what values to test. The Rational Zeros Theorem states that if polynomial P(x) with integer coefficients has a rational zero p/q (in lowest terms), then p must be a factor of the constant term and q must be a factor of the leading coefficient. For P(x) = 4x^3 - 3x^2 - 8x + 6, constant 6 (factors ±1,±2,±3,±6\pm 1, \pm 2, \pm 3, \pm 6), leading 4 (±1,±2,±4\pm 1, \pm 2, \pm 4), yielding unique candidates ±1,±2,±3,±6,±12,±32,±14,±34\pm 1, \pm 2, \pm 3, \pm 6, \pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{1}{4}, \pm \frac{3}{4} after simplifying. Choice C correctly includes all possible fractions in lowest terms, covering q up to 4 without duplicates. A tempting distractor like choice B omits the quarters like ±14\pm \frac{1}{4} and ±34\pm \frac{3}{4}, forgetting that q includes all factors of 4—don't skip them! Systematically list p and q factors (with signs), form p/q, simplify, and eliminate duplicates for the complete list. You're mastering this theorem—keep going, and testing will be a breeze!

Question 13

Use the Rational Zeros Theorem to list all possible rational zeros of P(x)=3x3−2x2+x−6.P(x)=3x^3-2x^2+x-6. (Recall: any rational zero p/qp/q in lowest terms has p∣p\mid constant term and q∣q\mid leading coefficient.)

  1. ±1,±2,±3,±6\pm 1,\pm 2,\pm 3,\pm 6
  2. ±1,±2,±3,±6,±13,±23\pm 1,\pm 2,\pm 3,\pm 6,\pm \frac{1}{3},\pm \frac{2}{3} (correct answer)
  3. ±1,±2,±3,±6,±13,±23,±43,±53\pm 1,\pm 2,\pm 3,\pm 6,\pm \frac{1}{3},\pm \frac{2}{3},\pm \frac{4}{3},\pm \frac{5}{3}
  4. ±1,±2,±3,±6,±13,±23\pm 1,\pm 2,\pm 3,\pm 6,\pm \frac{1}{3},\pm \frac{2}{3} (and also ±63\pm \frac{6}{3} listed separately)
Explanation: This question tests your understanding of the Rational Zeros Theorem—a powerful tool for finding possible rational zeros of polynomials, dramatically narrowing what values to test. The Rational Zeros Theorem states that if polynomial P(x) with integer coefficients has a rational zero p/q (in lowest terms), then p must be a factor of the constant term and q must be a factor of the leading coefficient. For P(x) = 3x^3 - 2x^2 + x - 6, the constant term is -6 (factors: ±1, ±2, ±3, ±6), and the leading coefficient is 3 (factors: ±1, ±3), so possible rational zeros are ±1, ±2, ±3, ±6, ±1/3, ±2/3, ±3/3 (simplifies to ±1), ±6/3 (simplifies to ±2)—resulting in the unique list ±1, ±2, ±3, ±6, ±1/3, ±2/3 after removing duplicates. Choice B correctly lists all possible rational zeros by including all combinations of factors and their negatives, without duplicates or extras. A tempting distractor like choice A fails by omitting the fractional candidates like ±1/3 and ±2/3, which are essential when the leading coefficient isn't 1—remember, q can be greater than 1, so include those fractions! To apply the Rational Zeros Theorem effectively, always list positive and negative factors of the constant for p and leading for q, form all p/q, simplify to lowest terms, and remove duplicates for a clean list. This strategy ensures you don't miss any potential zeros, and testing them next can reveal the actual ones—keep practicing, you've got this!

Question 14

Use the Rational Zeros Theorem to list all possible rational zeros of the polynomial P(x)=3x3−2x2+x−6.P(x)=3x^3-2x^2+x-6. (Recall: any rational zero has the form p/qp/q in lowest terms where pp divides the constant term and qq divides the leading coefficient.)

  1. ±1, ±2, ±3, ±6\pm1,\ \pm2,\ \pm3,\ \pm6
  2. ±1, ±2, ±3, ±6, ±13, ±23\pm1,\ \pm2,\ \pm3,\ \pm6,\ \pm\frac{1}{3},\ \pm\frac{2}{3} (correct answer)
  3. ±1, ±2, ±3, ±6, ±13, ±23, ±43, ±53\pm1,\ \pm2,\ \pm3,\ \pm6,\ \pm\frac{1}{3},\ \pm\frac{2}{3},\ \pm\frac{4}{3},\ \pm\frac{5}{3}
  4. ±16, ±13, ±12, ±1, ±32, ±3, ±6\pm\frac{1}{6},\ \pm\frac{1}{3},\ \pm\frac{1}{2},\ \pm1,\ \pm\frac{3}{2},\ \pm3,\ \pm6
Explanation: This question tests your understanding of the Rational Zeros Theorem—a powerful tool for finding possible rational zeros of polynomials, dramatically narrowing what values to test. The Rational Zeros Theorem states that if polynomial P(x) with integer coefficients has a rational zero p/q (in lowest terms), then p must be a factor of the constant term and q must be a factor of the leading coefficient. For P(x) = 3x³ - 2x² + x - 6, the constant term is -6 (factors: ±1, ±2, ±3, ±6) and leading coefficient is 3 (factors: ±1, ±3), so possible rational zeros are all fractions p/q where p divides -6 and q divides 3. Choice B correctly lists ±1, ±2, ±3, ±6 (when q = 1) and ±1/3, ±2/3 (when q = 3), giving the complete set of candidates. Choice A misses the fractional possibilities when q = 3, while C incorrectly includes ±4/3 and ±5/3 (4 and 5 don't divide -6!), and D has wrong fractions like ±1/6 (6 isn't a factor of the leading coefficient 3). Remember: form all possible p/q where p comes from factors of the constant term and q from factors of the leading coefficient—don't mix these up!

Question 15

Use the Rational Zeros Theorem to list all possible rational zeros of P(x)=4x4−5x3−8x2+3x+6.P(x)=4x^4-5x^3-8x^2+3x+6.

  1. ±1,±2,±3,±6,±12,±32\pm 1,\pm 2,\pm 3,\pm 6,\pm \frac{1}{2},\pm \frac{3}{2}
  2. ±1,±2,±3,±6,±12,±32,±14,±34,±13,±23\pm 1,\pm 2,\pm 3,\pm 6,\pm \frac{1}{2},\pm \frac{3}{2},\pm \frac{1}{4},\pm \frac{3}{4},\pm \frac{1}{3},\pm \frac{2}{3}
  3. ±1,±2,±3,±6,±12,±32,±14,±34\pm 1,\pm 2,\pm 3,\pm 6,\pm \frac{1}{2},\pm \frac{3}{2},\pm \frac{1}{4},\pm \frac{3}{4} (correct answer)
  4. ±1,±2,±3,±6,±24,±64\pm 1,\pm 2,\pm 3,\pm 6,\pm \frac{2}{4},\pm \frac{6}{4}
Explanation: This question tests your understanding of the Rational Zeros Theorem for higher-degree polynomials, listing all possible rational zeros even for degree 4—keep going, you're handling complexity well! For P(x)=4x4−5x3−8x2+3x+6P(x) = 4x^4 - 5x^3 - 8x^2 + 3x + 6, constant 6 (factors ±1,±2,±3,±6\pm 1, \pm 2, \pm 3, \pm 6) and leading 4 (±1,±2,±4\pm 1, \pm 2, \pm 4) give p/q like ±1,±2,±3,±6,±12,±32,±14,±34,±24\pm 1, \pm 2, \pm 3, \pm 6, \pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{1}{4}, \pm \frac{3}{4}, \pm \frac{2}{4} (±12\pm \frac{1}{2}, duplicate), ±64\pm \frac{6}{4} (±32\pm \frac{3}{2}, duplicate)—unique list ±1,±2,±3,±6,±12,±32,±14,±34\pm 1, \pm 2, \pm 3, \pm 6, \pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{1}{4}, \pm \frac{3}{4}. Choice C accurately includes all without extras or misses. A common mistake in choice B is including unnecessary denominators like ±13\pm \frac{1}{3} or ±23\pm \frac{2}{3}, which don't come from factors of 4—stick to q from the leading coefficient only! Form the list by combining all p and q, simplify fractions, and eliminate duplicates for efficiency before testing. You're progressing wonderfully—mastering this for any degree will make finding zeros a breeze!

Question 16

Apply the Rational Zeros Theorem to help factor completely: P(x)=2x3−9x2+7x+6.P(x)=2x^3-9x^2+7x+6. (Find a rational zero, factor it out, then factor the remaining quadratic.)

  1. (x−2)(2x−1)(x+3)(x-2)(2x-1)(x+3)
  2. (x−2)(2x+1)(x−3)(x-2)(2x+1)(x-3) (correct answer)
  3. (x−1)(2x+3)(x−2)(x-1)(2x+3)(x-2)
  4. (x+2)(2x−1)(x−3)(x+2)(2x-1)(x-3)
Explanation: This question tests your understanding of the Rational Zeros Theorem—a powerful tool for finding possible rational zeros of polynomials, dramatically narrowing what values to test, and helping factor completely. The Rational Zeros Theorem states that if polynomial P(x)P(x) with integer coefficients has a rational zero p/qp/q (in lowest terms), then pp must be a factor of the constant term and qq must be a factor of the leading coefficient. For P(x)=2x3−9x2+7x+6P(x) = 2x^3 - 9x^2 + 7x + 6, candidates ±1,±2,±3,±6,±12,±32 \pm 1, \pm 2, \pm 3, \pm 6, \pm \frac{1}{2}, \pm \frac{3}{2}; testing confirms −1/2,2,3-1/2, 2, 3 as zeros, yielding factorization (2x+1)(x−2)(x−3)(2x + 1)(x - 2)(x - 3), which expands correctly. Choice A provides the accurate factorization matching the zeros found. A tempting distractor like choice B might flip signs in factors, such as 2x−12x - 1 instead—confirm by expanding or testing zeros! Find one zero, divide synthetically, factor the quadratic, and verify—repeat for full factorization. Terrific job; you're becoming a pro at this!

Question 17

Possible rational zeros of P(x)=x3−4x2+x+6P(x)=x^3-4x^2+x+6 are ±1,±2,±3,±6\pm 1,\pm 2,\pm 3,\pm 6. Which of these candidates are actual zeros? (You may test by substitution.)

  1. x=−2,1,3x=-2,1,3
  2. x=−1,2,3x=-1,2,3 (correct answer)
  3. x=−1,2x=-1,2 only
  4. x=−1,−2,3x=-1,-2,3
Explanation: This question tests your understanding of the Rational Zeros Theorem—a powerful tool for finding possible rational zeros of polynomials, dramatically narrowing what values to test, by identifying which candidates actually work. The Rational Zeros Theorem states that if polynomial P(x)P(x) with integer coefficients has a rational zero p/qp/q (in lowest terms), then pp must be a factor of the constant term and qq must be a factor of the leading coefficient. Given possible zeros ±1,±2,±3,±6\pm 1, \pm 2, \pm 3, \pm 6 for P(x)=x3−4x2+x+6P(x) = x^3 - 4x^2 + x + 6, testing shows P(−1)=0P(-1) = 0, P(2)=0P(2) = 0, P(3)=0P(3) = 0—these are the actual zeros. Choice A correctly identifies x=−1,2,3x = -1, 2, 3 as the actual zeros after accurate substitution checks. A tempting distractor like choice B might miscalculate P(−1)P(-1) or swap signs, such as thinking P(−2)=0P(-2) = 0 when it doesn't—always compute each term carefully to avoid errors! To master this, list candidates, test via substitution or synthetic division, and if P(candidate)=0P(\text{candidate}) = 0, it's a zero—organize your work to prevent mistakes. Great job tackling this; with practice, you'll spot the zeros quickly and confidently!

Question 18

Use the Rational Zeros Theorem to list all possible rational zeros of P(x)=5x4+2x3−3x2+x−4.P(x)=5x^4+2x^3-3x^2+x-4.

  1. ±1,±2,±4,±5,±10,±20\pm 1,\pm 2,\pm 4,\pm 5,\pm 10,\pm 20
  2. ±1,±2,±4,±12,±14,±24\pm 1,\pm 2,\pm 4,\pm \frac{1}{2},\pm \frac{1}{4},\pm \frac{2}{4}
  3. ±1,±2,±4,±15,±25,±45\pm 1,\pm 2,\pm 4,\pm \frac{1}{5},\pm \frac{2}{5},\pm \frac{4}{5} (correct answer)
  4. ±1,±2,±4,±15,±25,±45,±54\pm 1,\pm 2,\pm 4,\pm \frac{1}{5},\pm \frac{2}{5},\pm \frac{4}{5},\pm \frac{5}{4}
Explanation: This question tests your understanding of the Rational Zeros Theorem for a quartic with leading coefficient not 1, focusing on fractional candidates—wonderful, you're getting comfortable with fractions! For P(x)=5x4+2x3−3x2+x−4P(x) = 5x^4 + 2x^3 - 3x^2 + x - 4, constant -4 (factors ±1,±2,±4\pm 1, \pm 2, \pm 4) and leading 5 (±1,±5\pm 1, \pm 5) give p/q: ±1,±2,±4,±15,±25,±45,±15\pm 1, \pm 2, \pm 4, \pm \frac{1}{5}, \pm \frac{2}{5}, \pm \frac{4}{5}, \pm \frac{1}{5} (duplicate), etc.—unique ±1,±2,±4,±15,±25,±45\pm 1, \pm 2, \pm 4, \pm \frac{1}{5}, \pm \frac{2}{5}, \pm \frac{4}{5}. Choice A precisely lists them without extras. Distractors like choice B omit fractions and add integers not from factors, forgetting q includes 5 for fifths—include all combinations! List p and q fully, form simplified p/q, and skip duplicates for a clean list ready for testing. You're doing fantastically— this precision will speed up finding actual zeros in no time!

Question 19

Use the Rational Zeros Theorem to help factor the polynomial completely over the integers: P(x)=x3−7x−6.P(x)=x^3-7x-6.​

  1. (x−1)(x2+x−6)(x-1)(x^2+x-6)
  2. (x−3)(x2+3x+2)(x-3)(x^2+3x+2)
  3. (x+2)(x−3)(x+1)(x+2)(x-3)(x+1) (correct answer)
  4. (x+1)(x2−x−6)(x+1)(x^2-x-6)
Explanation: This question tests your understanding of the Rational Zeros Theorem by using it to factor a polynomial completely, showing how the theorem leads to full linear factorization over the integers—what a rewarding application! For P(x) = x^3 - 7x - 6, possible zeros ±1, ±2, ±3, ±6 yield actual zeros -2, -1, 3 via testing, allowing factorization as (x + 2)(x + 1)(x - 3). This matches the expanded form, confirming completeness. Choice C correctly provides the complete linear factorization. Distractors like choice B stop at a quadratic factor without factoring further, missing that x^2 + 3x + 2 = (x + 1)(x + 2)—always check if quotients factor more! Apply the theorem by listing candidates, testing to find zeros, dividing sequentially, and repeating on quotients until fully linear. Fantastic effort—this method will help you factor any polynomial with rational roots efficiently!

Question 20

Use the Rational Zeros Theorem to list all possible rational zeros of P(x)=4x3−3x2−8x+6.P(x)=4x^3-3x^2-8x+6. (List each candidate in lowest terms; include both signs.)

  1. ±1,±2,±3,±6\pm1, \pm2, \pm3, \pm6
  2. ±14,±12,±34,±1,±32,±3,±6\pm\frac{1}{4}, \pm\frac{1}{2}, \pm\frac{3}{4}, \pm1, \pm\frac{3}{2}, \pm3, \pm6
  3. ±1,±2,±3,±6,±12,±32\pm1, \pm2, \pm3, \pm6, \pm\frac{1}{2}, \pm\frac{3}{2}
  4. ±14,±12,±34,±1,±32,±2,±3,±6\pm\frac{1}{4}, \pm\frac{1}{2}, \pm\frac{3}{4}, \pm1, \pm\frac{3}{2}, \pm2, \pm3, \pm6 (correct answer)
Explanation: This question tests your understanding of the Rational Zeros Theorem—a powerful tool for finding possible rational zeros of polynomials, dramatically narrowing what values to test. For P(x)=4x3−3x2−8x+6P(x) = 4x^3 - 3x^2 - 8x + 6, the constant term is 6 (factors: ±1,±2,±3,±6\pm1, \pm2, \pm3, \pm6) and leading coefficient is 4 (factors: ±1,±2,±4\pm1, \pm2, \pm4), so possible rational zeros are p/q where p divides 6 and q divides 4. When q = 1: ±1,±2,±3,±6\pm1, \pm2, \pm3, \pm6; when q = 2: ±12,±22=±1\pm\frac{1}{2}, \pm\frac{2}{2} = \pm1 (duplicate), ±32,±62=±3\pm\frac{3}{2}, \pm\frac{6}{2} = \pm3 (duplicate); when q = 4: ±14,±24=±12\pm\frac{1}{4}, \pm\frac{2}{4} = \pm\frac{1}{2} (duplicate), ±34,±64=±32\pm\frac{3}{4}, \pm\frac{6}{4} = \pm\frac{3}{2} (duplicate). Choice D correctly lists all unique possibilities in lowest terms: ±14,±12,±34,±1,±32,±2,±3,±6\pm\frac{1}{4}, \pm\frac{1}{2}, \pm\frac{3}{4}, \pm1, \pm\frac{3}{2}, \pm2, \pm3, \pm6. Choice A misses ±14\pm\frac{1}{4} and ±34\pm\frac{3}{4} (when q = 4), B only has integer candidates, and C misses ±2\pm2. Remember to check all possible denominators from the leading coefficient's factors!