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!

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 $P(x) = 4x^4 - 5x^3 - 8x^2 + 3x + 6$, constant 6 (factors $\pm 1, \pm 2, \pm 3, \pm 6$) and leading 4 ($\pm 1, \pm 2, \pm 4$) give p/q like $\pm 1, \pm 2, \pm 3, \pm 6, \pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{1}{4}, \pm \frac{3}{4}, \pm \frac{2}{4}$ ($\pm \frac{1}{2}$, duplicate), $\pm \frac{6}{4}$ ($\pm \frac{3}{2}$, duplicate)—unique list $\pm 1, \pm 2, \pm 3, \pm 6, \pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{1}{4}, \pm \frac{3}{4}$. Choice C accurately includes all without extras or misses. A common mistake in choice B is including unnecessary denominators like $\pm \frac{1}{3}$ or $\pm \frac{2}{3}$, 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!

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!

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!

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!

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: $\pm1, \pm2, \pm3, \pm6$), and the leading coefficient is 3 (factors: $\pm1, \pm3$), so possible rational zeros are $\pm1, \pm2, \pm3, \pm6, \pm\frac{1}{3}, \pm\frac{2}{3}, \pm\frac{3}{3}$ ($\pm1$, already listed), $\pm\frac{6}{3}$ ($\pm2$, already listed)—giving the unique list $\pm1, \pm2, \pm3, \pm6, \pm\frac{1}{3}, \pm\frac{2}{3}$. 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 $\pm3$, which introduces thirds like $\pm\frac{1}{3}$ and $\pm\frac{2}{3}$—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!

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!

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!