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 $$P(x)=4x^3 -11x^2 -3x +18$$, $3/4$ is a candidate since $3$ divides $18$ and $4$ divides $4$, but testing $P\left(\frac{3}{4}\right)=\frac{27}{16} - \frac{99}{16} - \frac{36}{16} + \frac{288}{16} = \frac{180}{16} \ne 0$. Choice C correctly notes it's a possible candidate but not an actual zero, confirmed by direct substitution. Choice D wrongly assumes all candidates are zeros—theorem lists possibles, but testing is essential! Rational Zeros Theorem application process: check if $p/q$ qualifies, then substitute to verify—careful fraction arithmetic avoids errors. Great job distinguishing possibility from actuality—you're sharpening your skills!

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!

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!

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!

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 $P(x)=x^3 -7x -6$, candidates $±1, ±2, ±3, ±6$; testing finds $P(-2)=0$, $P(-1)=0$, $P(3)=0$, allowing complete factorization as $(x+2)(x+1)(x-3)$. Choice D correctly provides the full linear factorization using these zeros, verified by multiplication back to original. Choice A stops at linear times quadratic, failing to factor further despite quadratic splitting into rationals—a common oversight. Rational Zeros Theorem application process: list and test candidates, factor out each zero found via synthetic division, repeat on lower-degree quotient until fully factored. This method efficiently reveals all rational factors—excellent work applying it step by step!

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 $$P(x) = 6x^4 - x^3 - 7x^2 + x + 1$$, constant 1 ($\pm 1$), leading 6 ($\pm 1, \pm 2, \pm 3, \pm 6$), so candidates $\pm 1, \pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{1}{6}$ in lowest terms. Choice A correctly lists only these, as p is limited to factors of 1. A tempting distractor like choice C includes extras like $\pm \frac{2}{3}$, but 2 isn't a factor of the constant term—stick to p from constant only! Form the list by dividing each p by each q, simplify, and include signs—higher degrees just mean more testing if needed. Keep up the excellent work; this theorem will make solving polynomials much easier for you!

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!

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!

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 $P(x)$ 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 $P(x) = 2x^3 - 9x^2 + 7x + 6$, candidates $ \pm 1, \pm 2, \pm 3, \pm 6, \pm \frac{1}{2}, \pm \frac{3}{2}$; testing confirms $-1/2, 2, 3$ as zeros, yielding factorization $(2x + 1)(x - 2)(x - 3)$, 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 $2x - 1$ 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!

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!