Rational Zeros Theorem - Algebra 2

Which statement correctly applies the Rational Zeros Theorem to $P(x)=6x^3-x^2+4x-5?$
Answer: Any rational zero must be of the form $\frac{p}{q}$ where $p\mid -5$ and $q\mid 6$ . This question tests your understanding of the Rational Zeros Theorem's core statement, ensuring you know exactly what divides what—fundamental for correct application, and you're nailing the basics! The theorem specifies that for rational zero p/q, p divides the constant term (-5 here) and q divides the leading coefficient (6), so possible p: $\pm 1, \pm 5$ ; q: $\pm 1, \pm 2, \pm 3, \pm 6$ . This setup guarantees the candidate list is precise. Choice B correctly states $p \mid -5$ and $q \mid 6$ . Distractors like choice A reverse p and q, a common mix-up—remember, p is from constant (end), q from leading (front)! Always recall the theorem's exact wording to avoid swaps, and note signs don't change the factors. Superb focus—this clarity will make listing and testing seamless for you!
Consider polynomials of the form $g(x) = ax^3 + bx^2 + cx + 12$ where $a$ is a positive integer. For which value of $a$ would the Rational Zero Theorem yield the fewest possible rational zero candidates?
Answer: $a = 1$ , because this minimizes the number of factors of the leading coefficient When you encounter questions about the Rational Zero Theorem, focus on how it generates candidates: any rational zero $\frac{p}{q}$ must have $p$ dividing the constant term and $q$ dividing the leading coefficient. For $g(x) = ax^3 + bx^2 + cx + 12$ , the possible rational zeros are $\frac{p}{q}$ where $p$ divides 12 and $q$ divides $a$ . The factors of 12 are $\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12$ , giving us 12 possible values for $p$ . The number of candidates depends entirely on how many factors $a$ has, since each factor of $a$ creates new denominators to pair with each factor of 12. When $a = 1$ , the only factor is 1 itself, so all candidates have the form $\frac{p}{1}$ , giving us exactly 12 candidates. This is the minimum possible. Choice A is incorrect because $a = 12$ actually maximizes candidates rather than minimizing them. The factors of 12 are $1, 2, 3, 4, 6, 12$ , creating many more rational zero candidates than $a = 1$ . 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 $a$ , not relating $a$ 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.
Consider the polynomial $f(x) = 6x^4 - 7x^3 + 2x^2 - 8x + 12$ . If $p$ is a rational zero of $f(x)$ , which of the following statements must be true about the numerator and denominator of $p$ when written in lowest terms as $\frac{a}{b}$ ?
Answer: $a$ divides 12 and $b$ divides 6, where $\gcd(a,b) = 1$ By the Rational Zero Theorem, if $p = \frac{a}{b}$ is a rational zero in lowest terms, then $a$ must divide the constant term (12) and $b$ must divide the leading coefficient (6). The condition $\gcd(a,b) = 1$ ensures the fraction is in lowest terms. Choice B reverses the roles of numerator and denominator. Choices C and D incorrectly require both $a$ and $b$ to be positive, but rational zeros can be negative.
Apply the Rational Zeros Theorem to help factor completely: $P(x)=2x^3-3x^2-8x+12.$
Answer: $(x+2)(2x-3)(x-2)$ 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 $P(x) = 2x^3 - 3x^2 - 8x + 12$ , possible zeros are p/q where p divides 12 (factors: ±1, ±2, ±3, ±4, ±6, ±12) and q divides 2 (factors: ±1, ±2). Testing likely candidates: $P(2) = 16 - 12 - 16 + 12 = 0$ , so $(x - 2)$ is a factor. Using synthetic division: $2x^3 - 3x^2 - 8x + 12 = (x - 2)(2x^2 + x - 6)$ . Now factor $2x^2 + x - 6$ : looking for factors of $2(-6) = -12$ that add to 1, we get 4 and -3, so $2x^2 + x - 6 = 2x^2 + 4x - 3x - 6 = 2x(x + 2) - 3(x + 2) = (2x - 3)(x + 2)$ . Choice C correctly shows $P(x) = (x + 2)(2x - 3)(x - 2)$ . Choice A has wrong quadratic factor, B has duplicate $(x - 2)$ factors, and D has $(x - 3)$ instead of $(2x - 3)$ . The complete factorization reveals zeros at $x = -2$ , $x = 3/2$ , and $x = 2$ .
The polynomial $h(x) = kx^4 - 8x^3 + mx^2 + nx - 6$ has $\frac{3}{2}$ as a rational zero. If all coefficients are integers and the Rational Zero Theorem is to be applied effectively, what constraint must $k$ satisfy?
Answer: $k$ must be even, so that $\frac{3}{2}$ appears in the list of possible rational zeros 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 $\frac{p}{q}$ of a polynomial must have $p$ as a factor of the constant term and $q$ as a factor of the leading coefficient. Since $\frac{3}{2}$ is a zero of $h(x) = kx^4 - 8x^3 + mx^2 + nx - 6$ , we need 3 to be a factor of the constant term (-6) and 2 to be a factor of the leading coefficient ( $k$ ). The factors of -6 are ±1, ±2, ±3, ±6, so 3 is indeed available. For $\frac{3}{2}$ to appear in our list of possible rational zeros, 2 must be a factor of $k$ , meaning $k$ must be even. Choice A is incorrect because requiring $k$ 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 $k$ 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 $k$ because for $\frac{3}{2}$ to be a possible rational zero, 2 must divide $k$ . Choice B correctly identifies that $k$ must be even so that 2 divides $k$ , allowing $\frac{3}{2}$ to appear in the theorem's list of candidates. Study tip: Always check both parts of a rational zero $\frac{p}{q}$ —the numerator must divide the constant term, and the denominator must divide the leading coefficient.
Use the Rational Zeros Theorem to list all possible rational zeros of the polynomial $P(x)=3x^3-2x^2+x-6.$ (Recall: any rational zero has the form $p/q$ in lowest terms where $p$ divides the constant term and $q$ divides the leading coefficient.)
Answer: $\pm1,\ \pm2,\ \pm3,\ \pm6,\ \pm\frac{1}{3},\ \pm\frac{2}{3}$ 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!
Use the Rational Zeros Theorem to help factor the polynomial completely over the integers: $P(x)=x^3-7x-6.$
Answer: $(x+2)(x-3)(x+1)$ 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!
Use the Rational Zeros Theorem to list all possible rational zeros of $P(x)=3x^3-2x^2+x-6.$ (Recall: any rational zero $p/q$ in lowest terms has $p\mid$ constant term and $q\mid$ leading coefficient.)
Answer: $\pm 1,\pm 2,\pm 3,\pm 6,\pm \frac{1}{3},\pm \frac{2}{3}$ 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!
Apply the Rational Zeros Theorem to help factor completely: $P(x)=2x^3-5x^2-4x+3.$ First list possible rational zeros, then test to find the actual rational zero(s) and factor.
Answer: Actual rational zeros: $\frac{1}{2},3$ ; factorization: $(2x-1)(x-3)(x+1)$ 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 aiding in complete factorization. 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) = $2x^3$ - $5x^2$ - 4x + 3, candidates are ±1, ±3, ±1/2, ±3/2; testing reveals actual zeros 1/2, -1, 3, leading to factorization (2x - 1)(x + 1)(x - 3). Choice A correctly identifies the zeros and provides the accurate factorization that multiplies back to the original. A tempting distractor like choice B might have sign errors in testing, such as mistaking -1/2 for a zero instead of 1/2—double-check your substitutions! Apply this strategy: generate the list, test starting with fractions if needed, factor out each found zero using synthetic division, and continue. You're doing fantastic—keep verifying your work for perfect factorizations!
A student applies the Rational Zero Theorem to $f(x) = 6x^3 - 13x^2 + 6x - 1$ and creates the list: $\pm 1, \pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{1}{6}$ . After testing $x = 1$ , the student finds $f(1) = -2 \neq 0$ . What should the student conclude?
Answer: The student should continue testing the remaining candidates before drawing any conclusions The Rational Zero Theorem provides a complete list of possible rational zeros, but testing must be systematic. Finding that $f(1) \neq 0$ only eliminates $x = 1$ 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 $f(1)$ as indicating the nature of zeros.
Possible rational zeros of $P(x)=x^3-4x^2+x+6$ (from the Rational Zeros Theorem) are $\pm 1,\pm 2,\pm 3,\pm 6$ . Which of these candidates are actual zeros? (Test by substitution or synthetic division.)
Answer: $x=-1,\;x=2,\;x=3$ 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. Given candidates $\pm 1, \pm 2, \pm 3, \pm 6$ for $P(x)=x^3 -4x^2 +x +6$ , testing shows $P(-1)=0$ , $P(2)=0$ , $P(3)=0$ , but others like $P(1)=4 \neq 0$ , confirming actual zeros $-1, 2, 3$ . Choice C correctly identifies all three actual zeros from the candidates, verified by substitution or synthetic division. Choice A omits $x=3$ , perhaps from an arithmetic error like miscalculating $P(3)=27-36+3+6=0$ correctly. Rational Zeros Theorem application process: with list provided, test each systematically—calculate $P(\text{candidate})$ carefully, tracking terms to avoid errors. If zero found, factor out $(x - r)$ using synthetic division and repeat on quotient—keep going, you're building polynomial mastery!
Use the Rational Zeros Theorem to list all possible rational zeros of $P(x)=4x^3-3x^2-8x+6.$ (List each candidate in lowest terms; include both signs.)
Answer: $\pm\frac{1}{4}, \pm\frac{1}{2}, \pm\frac{3}{4}, \pm1, \pm\frac{3}{2}, \pm2, \pm3, \pm6$ 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 $P(x) = 4x^3 - 3x^2 - 8x + 6$ , the constant term is 6 (factors: $\pm1, \pm2, \pm3, \pm6$ ) and leading coefficient is 4 (factors: $\pm1, \pm2, \pm4$ ), so possible rational zeros are p/q where p divides 6 and q divides 4. When q = 1: $\pm1, \pm2, \pm3, \pm6$ ; when q = 2: $\pm\frac{1}{2}, \pm\frac{2}{2} = \pm1$ (duplicate), $\pm\frac{3}{2}, \pm\frac{6}{2} = \pm3$ (duplicate); when q = 4: $\pm\frac{1}{4}, \pm\frac{2}{4} = \pm\frac{1}{2}$ (duplicate), $\pm\frac{3}{4}, \pm\frac{6}{4} = \pm\frac{3}{2}$ (duplicate). Choice D correctly lists all unique possibilities in lowest terms: $\pm\frac{1}{4}, \pm\frac{1}{2}, \pm\frac{3}{4}, \pm1, \pm\frac{3}{2}, \pm2, \pm3, \pm6$ . Choice A misses $\pm\frac{1}{4}$ and $\pm\frac{3}{4}$ (when q = 4), B only has integer candidates, and C misses $\pm2$ . Remember to check all possible denominators from the leading coefficient's factors!
Use the Rational Zeros Theorem to list all possible rational zeros of $P(x)=5x^4+2x^3-3x^2+x-4.$
Answer: $\pm 1,\pm 2,\pm 4,\pm \frac{1}{5},\pm \frac{2}{5},\pm \frac{4}{5}$ 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 $P(x) = 5x^4 + 2x^3 - 3x^2 + x - 4$ , constant -4 (factors $\pm 1, \pm 2, \pm 4$ ) and leading 5 ( $\pm 1, \pm 5$ ) give p/q: $\pm 1, \pm 2, \pm 4, \pm \frac{1}{5}, \pm \frac{2}{5}, \pm \frac{4}{5}, \pm \frac{1}{5}$ (duplicate), etc.—unique $\pm 1, \pm 2, \pm 4, \pm \frac{1}{5}, \pm \frac{2}{5}, \pm \frac{4}{5}$ . 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!
The polynomial $h(x) = 2x^3 + bx^2 + cx - 15$ has $\frac{3}{2}$ as one of its zeros. Using the Rational Zero Theorem, which of the following could be another rational zero of $h(x)$ ?
Answer: $-\frac{5}{2}$ , since 5 divides the constant term and 2 divides the leading coefficient By the Rational Zero Theorem, possible rational zeros have the form $\pm\frac{p}{q}$ where $p$ divides the constant term (-15) and $q$ divides the leading coefficient (2). For choice D: 5 divides 15 and 2 divides 2, so $-\frac{5}{2}$ is a valid candidate. Choice A is wrong because 3 doesn't divide 2. Choice B is wrong because 4 doesn't divide 2. Choice C reverses the theorem's requirements.
Use the Rational Zeros Theorem to find all rational zeros of $P(x)=2x^3+5x^2-x-6.$ Test candidates by substitution or synthetic division, and report the rational zeros you verify.
Answer: $x=-2, x=\frac{3}{2}, x=1$ 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 for $P(x) = 2x^3 + 5x^2 - x - 6$ , possible rational zeros are $p/q$ where $p$ divides $-6$ (factors: $\pm1, \pm2, \pm3, \pm6$ ) and $q$ divides $2$ (factors: $\pm1, \pm2$ ), giving candidates: $\pm1, \pm2, \pm3, \pm6, \pm \frac{1}{2}, \pm \frac{3}{2}$ . Now we test systematically: $P(-2) = 2(-8) + 5(4) - (-2) - 6 = -16 + 20 + 2 - 6 = 0$ ✓; $P(\frac{3}{2}) = 2(\frac{27}{8}) + 5(\frac{9}{4}) - \frac{3}{2} - 6 = \frac{27}{4} + \frac{45}{4} - \frac{6}{4} - \frac{24}{4} = \frac{42}{4} = 0$ ✓; $P(1) = 2 + 5 - 1 - 6 = 0$ ✓. Choice A correctly identifies all three rational zeros: $x = -2$ , $x = \frac{3}{2}$ , and $x = 1$ . Choice B has wrong values, C has $-\frac{3}{2}$ instead of $\frac{3}{2}$ , and D misses $x = 1$ . When testing candidates, work carefully with fractions—for $x = \frac{3}{2}$ , compute each term separately to avoid arithmetic errors!
Possible rational zeros of $P(x)=x^3-4x^2+x+6$ are $\pm 1,\pm 2,\pm 3,\pm 6$ . Which of these candidates are actual zeros? (You may test by substitution.)
Answer: $x=-1,2,3$ 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 $P(x)$ 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. Given possible zeros $\pm 1, \pm 2, \pm 3, \pm 6$ for $P(x) = x^3 - 4x^2 + x + 6$ , testing shows $P(-1) = 0$ , $P(2) = 0$ , $P(3) = 0$ —these are the actual zeros. Choice A correctly identifies $x = -1, 2, 3$ as the actual zeros after accurate substitution checks. A tempting distractor like choice B might miscalculate $P(-1)$ or swap signs, such as thinking $P(-2) = 0$ 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 $P(\text{candidate}) = 0$ , it's a zero—organize your work to prevent mistakes. Great job tackling this; with practice, you'll spot the zeros quickly and confidently!
Use the Rational Zeros Theorem to help factor completely: $P(x)=2x^3-3x^2-8x+12.$ (Find a rational zero by testing candidates, divide to reduce the degree, and continue factoring.)
Answer: $(x-2)(x+2)(2x-3)$ 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 $P(x)=2x^3 - 3x^2 - 8x + 12$ , candidates include ±1, ±2, ±3, ±4, ±6, ±12, ±1/2, ±3/2; testing finds $x=2$ zero, dividing gives $(x-2)(2x^2 + x - 6)$ , which factors to $(x-2)(x+2)(2x-3)$ . Choice C correctly gives the complete factorization with all linear factors over rationals. Choice A stops without factoring the quadratic fully, missing that $2x^2 + x - 6 = (2x-3)(x+2)$ . 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!
Use the Rational Zeros Theorem to help factor completely: $P(x)=x^3-3x^2-4x+12.$ (After finding a rational zero, divide to get a quadratic factor.)
Answer: $(x-3)(x-2)(x+2)$ 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 facilitating complete factorization. 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) = $x^3$ - $3x^2$ - 4x + 12, candidates ±1, ±2, ±3, ±4, ±6, ±12; testing finds -2, 2, 3 as zeros, allowing factorization (x + 2)(x - 2)(x - 3) = ( $x^2$ - 4)(x - 3), which expands correctly. Choice A accurately provides the complete factorization after verifying the zeros. A tempting distractor like choice B might swap signs, such as using +3 instead of -2—test each candidate carefully! After finding one zero, use synthetic division to get the quadratic, then factor or apply the theorem again. Awesome effort—you're getting great at factoring polynomials!
Apply the Rational Zeros Theorem to list all possible rational zeros of $P(x)=4x^3-3x^2-10x+5.$ (Give the full candidate list in lowest terms.)
Answer: $\pm 1,\pm 5,\pm \frac{1}{2},\pm \frac{5}{2},\pm \frac{1}{4},\pm \frac{5}{4}$ 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 $P(x)=4x^3 -3x^2 -10x +5$ , constant 5 (factors: $\pm1, \pm5$ ), leading 4 (factors: $\pm1, \pm2, \pm4$ ), so possibles $\pm1, \pm5, \pm \frac{1}{2}, \pm \frac{5}{2}, \pm \frac{1}{4}, \pm \frac{5}{4}$ —all unique in lowest terms. Choice A correctly lists the full set, including all combinations without duplicates or omissions. Choice C incorrectly uses factors as if leading were 1, including extras like $\pm2, \pm4$ that aren't proper $p/q$ —remember $p$ from constant only! Rational Zeros Theorem application process: list $\pm p$ from constant, $\pm q$ from leading, form all $p/q$ , simplify reducibles like $5/5=1$ (but none here). With this list, you're set to test and factor—keep up the great effort!
Use the Rational Zeros Theorem to identify the complete list of possible rational zeros of $P(x)=6x^4+x^3-7x^2-2x+2.$
Answer: $\pm1,\ \pm2,\ \pm\frac{1}{2},\ \pm\frac{1}{3},\ \pm\frac{2}{3},\ \pm\frac{1}{6}$ 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 $P(x) = 6x^4 + x^3 - 7x^2 - 2x + 2$ , the constant term is 2 (factors: $\pm1, \pm2$ ) and leading coefficient is 6 (factors: $\pm1, \pm2, \pm3, \pm6$ ), so possible rational zeros are $p/q$ in lowest terms. When $p = \pm1$ : $\pm1, \pm\frac{1}{2}, \pm\frac{1}{3}, \pm\frac{1}{6}$ ; when $p = \pm2$ : $\pm2, \pm\frac{2}{2} = \pm1$ (duplicate), $\pm\frac{2}{3}, \pm\frac{2}{6} = \pm\frac{1}{3}$ (duplicate). Choice A correctly lists all unique possibilities in lowest terms: $\pm1, \pm2, \pm\frac{1}{2}, \pm\frac{1}{3}, \pm\frac{2}{3}, \pm\frac{1}{6}$ . Choice B incorrectly includes $\pm\frac{2}{6}$ without simplifying to $\pm\frac{1}{3}$ , C has wrong candidates like $\pm3$ and $\pm6$ (these aren't factors of the constant 2!), and D includes both simplified and unsimplified forms. Remember: always reduce fractions to lowest terms to avoid listing duplicates— $\frac{2}{6} = \frac{1}{3}$ , so only list $\pm\frac{1}{3}$ !

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