This question tests your understanding of the Remainder Theorem—a powerful shortcut that lets you find the remainder when dividing a polynomial by (x - a) without doing any long division! A special case is the Factor Theorem: (x - a) is a factor of p(x) if and only if p(a) = 0. This means the remainder is zero, so the division is exact with no remainder. We can test potential factors by just evaluating the polynomial—if p(a) = 0, we've found a factor! This beats trial-and-error factoring when testing specific values. For (x+2) = (x - (-2)), evaluate p(-2): $(-2)^3$ = -8, $+3*(-2)^2$ = +34 = +12, -4(-2) = +8, -12; combine: -8 + 12 = 4, 4 + 8 = 12, 12 - 12 = 0. Choice A correctly determines that yes, it's a factor because p(-2)=0. Choice B says no because p(-2)=4, which might happen if you forget the -4*(-2) = +8 and stop at -8 + 12 -12 = -8. Sign safety for negatives: when testing (x + 2) = (x - (-2)), you're evaluating at a = -2. Substitute carefully: if p(x) = x³ - 3x + 5, then p(-2) = (-2)³ - 3(-2) + 5 = -8 + 6 + 5 = 3. Use parentheses around negative values to avoid sign errors! This is where most mistakes happen with the Remainder Theorem.

This question tests your understanding of the Remainder Theorem—a powerful shortcut that lets you find the remainder when dividing a polynomial by (x - a) without doing any long division! The Remainder Theorem states that when you divide a polynomial p(x) by (x - a), the remainder is simply p(a)—just substitute a into the polynomial and evaluate! This works because of how polynomial division works: p(x) = (x - a)·q(x) + r, where r is the remainder. Substituting x = a: p(a) = (a - a)·q(a) + r = 0 + r = r. So the remainder r equals p(a)—brilliant! To verify, evaluate p(1): $1^4$ = 1, $-31^2$ = -3, +21 = +2, +1; combine: 1 - 3 = -2, -2 + 2 = 0, 0 + 1 = 1. Choice A correctly verifies it's true because p(1)=1 matches the claimed remainder. Choice B incorrectly claims false with p(1)=0, perhaps from miscalculating 1 -3 +2 +1 as 0 by ignoring the last +1. Using the Remainder Theorem: (1) Identify the divisor (x - a) and extract a (remember: (x + 3) = (x - (-3)), so a = -3), (2) Substitute a for every x in p(x), (3) Calculate carefully (use parentheses for negative values!), (4) That result is your remainder. If it equals 0, (x - a) is a factor! This method is dramatically faster than polynomial long division.

This question tests your understanding of the Remainder Theorem—a powerful shortcut that lets you find the remainder when dividing a polynomial by (x - a) without doing any long division! The Remainder Theorem states that when you divide a polynomial p(x) by (x - a), the remainder is simply p(a)—just substitute a into the polynomial and evaluate! This works because of how polynomial division works: $p(x) = (x - a) \cdot q(x) + r$, where r is the remainder. Substituting x = a: $p(a) = (a - a) \cdot q(a) + r = 0 + r = r$. So the remainder r equals p(a)—brilliant! Here, p(1) = 3 and we set it equal to 3: $1^3 - 2(1)^2 + k \cdot 1 - 5 = 1 - 2 + k - 5 = -6 + k = 3$, so $k = 9$. Choice B correctly finds k=9. A distractor like k=7 in choice A might come from misadding -6 + k =3 as k=-3+6=3 or similar arithmetic error. Using the Remainder Theorem: (1) Identify the divisor (x - a) and extract a (remember: $(x + 3) = (x - (-3))$, so a = -3), (2) Substitute a for every x in p(x), (3) Calculate carefully (use parentheses for negative values!), (4) That result is your remainder. If it equals 0, (x - a) is a factor! This method is dramatically faster than polynomial long division.

This question tests your understanding of the Remainder Theorem—a powerful shortcut that lets you find the remainder when dividing a polynomial by (x - a) without doing any long division! A special case is the Factor Theorem: (x - a) is a factor of p(x) if and only if p(a) = 0. This means the remainder is zero, so the division is exact with no remainder. We can test potential factors by just evaluating the polynomial—if p(a) = 0, we've found a factor! This beats trial-and-error factoring when testing specific values. To check if x=2 is a zero, evaluate p(2): $2^3$ = 8, $-62^2$ = -64 = -24, +11*2 = +22, -6; combine: 8 - 24 = -16, -16 + 22 = 6, 6 - 6 = 0. Choice A correctly determines yes, because p(2)=0. Choice B says no because p(2)=2, which might result from miscalculating -24 +22 = -2, then 8 -2 -6 =0 but wait, perhaps adding instead of subtracting terms. Using the Remainder Theorem: (1) Identify the divisor (x - a) and extract a (remember: (x + 3) = (x - (-3)), so a = -3), (2) Substitute a for every x in p(x), (3) Calculate carefully (use parentheses for negative values!), (4) That result is your remainder. If it equals 0, (x - a) is a factor! This method is dramatically faster than polynomial long division.

This question tests your understanding of the Remainder Theorem—a powerful shortcut that lets you find the remainder when dividing a polynomial by (x - a) without doing any long division! The Remainder Theorem states that when you divide a polynomial p(x) by (x - a), the remainder is simply p(a)—just substitute a into the polynomial and evaluate! This works because of how polynomial division works: p(x) = (x - a)·q(x) + r, where r is the remainder. Substituting x = a: p(a) = (a - a)·q(a) + r = 0 + r = r. So the remainder r equals p(a)—brilliant! For this problem, a=3, so $p(3)=3^3$ $-43^2$ +53 -2=27-36+15-2; step by step, 27-36=-9, -9+15=6, 6-2=4. Choice C correctly finds the remainder as 4. A distractor like B (-2) might come from sign errors in subtraction, but careful calculation confirms 4. Using the Remainder Theorem: (1) Identify the divisor (x - a) and extract a (remember: (x + 3) = (x - (-3)), so a = -3), (2) Substitute a for every x in p(x), (3) Calculate carefully (use parentheses for negative values!), (4) That result is your remainder. If it equals 0, (x - a) is a factor! This method is dramatically faster than polynomial long division.

This question tests your understanding of the Remainder Theorem—a powerful shortcut that lets you find the remainder when dividing a polynomial by (x - a) without doing any long division! A special case is the Factor Theorem: (x - a) is a factor of p(x) if and only if p(a) = 0. This means the remainder is zero, so the division is exact with no remainder. We can test potential factors by just evaluating the polynomial—if p(a) = 0, we've found a factor! This beats trial-and-error factoring when testing specific values. To check if x=2 is a zero, evaluate $p(2)=2^3$ $-62^2$ +112 -6=8-24+22-6; step by step, 8-24=-16, -16+22=6, 6-6=0, so yes. Choice A correctly determines yes because p(2)=0. A distractor like B might come from stopping at 6 before subtracting 6. Using the Remainder Theorem: (1) Identify the divisor (x - a) and extract a (remember: (x + 3) = (x - (-3)), so a = -3), (2) Substitute a for every x in p(x), (3) Calculate carefully (use parentheses for negative values!), (4) That result is your remainder. If it equals 0, (x - a) is a factor! This method is dramatically faster than polynomial long division.

This question tests your understanding of the Remainder Theorem—a powerful shortcut that lets you find the remainder when dividing a polynomial by (x - a) without doing any long division! The Remainder Theorem states that when you divide a polynomial p(x) by (x - a), the remainder is simply p(a)—just substitute a into the polynomial and evaluate! This works because of how polynomial division works: p(x) = (x - a)·q(x) + r, where r is the remainder. Substituting x = a: p(a) = (a - a)·q(a) + r = 0 + r = r. So the remainder r equals p(a)—brilliant! For (x-2), a=2, $p(2)=2^4$ $+m2^2$ -22 +1=16+4m-4+1=13+4m; set to 9: 13+4m=9, 4m=-4, m=-1. Choice A correctly finds m=-1. A distractor like C (1) might come from setting 16+4m-4+1=9 as 13+4m=9 but solving 4m=-4 incorrectly. Using the Remainder Theorem: (1) Identify the divisor (x - a) and extract a (remember: (x + 3) = (x - (-3)), so a = -3), (2) Substitute a for every x in p(x), (3) Calculate carefully (use parentheses for negative values!), (4) That result is your remainder. If it equals 0, (x - a) is a factor! This method is dramatically faster than polynomial long division.

This question tests your understanding of the Remainder Theorem—a powerful shortcut that lets you find the remainder when dividing a polynomial by (x - a) without doing any long division! A special case is the Factor Theorem: (x - a) is a factor of p(x) if and only if p(a) = 0. This means the remainder is zero, so the division is exact with no remainder. We can test potential factors by just evaluating the polynomial—if p(a) = 0, we've found a factor! This beats trial-and-error factoring when testing specific values. For I (x-2), p(2)=16-16=0; for II (x+2), p(-2)=16-16=0; for III (x-4), p(4)=256-16=240≠0, so I and II only. Choice B correctly identifies I and II only. A distractor like C might mistakenly think (x-4) works due to confusing with $x^2$-16. Sign safety for negatives: when testing (x + 2) = (x - (-2)), you're evaluating at a = -2. Substitute carefully: if p(x) = x³ - 3x + 5, then p(-2) = (-2)³ - 3(-2) + 5 = -8 + 6 + 5 = 3. Use parentheses around negative values to avoid sign errors! This is where most mistakes happen with the Remainder Theorem.

This question tests your understanding of the Remainder Theorem—a powerful shortcut that lets you find the remainder when dividing a polynomial by $(x - a)$ without doing any long division! The Remainder Theorem states that when you divide a polynomial $p(x)$ by $(x - a)$, the remainder is simply $p(a)$—just substitute $a$ into the polynomial and evaluate! This works because of how polynomial division works: $p(x) = (x - a) \cdot q(x) + r$, where $r$ is the remainder. Substituting $x = a$: $p(a) = (a - a) \cdot q(a) + r = 0 + r = r$. So the remainder $r$ equals $p(a)$—brilliant! For $(x+2)=(x-(-2))$, $a=-2$, $p(-2)=(-2)^4 + 3 \cdot(-2)^2 - 5 = 16 + 3 \cdot 4 - 5 = 16 + 12 - 5 = 23$. Choice B correctly finds the remainder as $23$. A distractor like C $(-23)$ might result from sign errors, such as treating $+3x^2$ as negative. Sign safety for negatives: when testing $(x + 2) = (x - (-2))$, you're evaluating at $a = -2$. Substitute carefully: if $p(x) = x^3 - 3x + 5$, then $p(-2) = (-2)^3 - 3(-2) + 5 = -8 + 6 + 5 = 3$. Use parentheses around negative values to avoid sign errors! This is where most mistakes happen with the Remainder Theorem.

This question tests your understanding of the Remainder Theorem—a powerful shortcut that lets you find the remainder when dividing a polynomial by (x - a) without doing any long division! The Remainder Theorem states that when you divide a polynomial p(x) by (x - a), the remainder is simply p(a)—just substitute a into the polynomial and evaluate! This works because of how polynomial division works: p(x) = (x - a)·q(x) + r, where r is the remainder. Substituting x = a: p(a) = (a - a)·q(a) + r = 0 + r = r. So the remainder r equals p(a)—brilliant! For (x + 2) = (x - (-2)), a = -2; p(-2) = $3*(-2)^3$ - $(-2)^2$ - 7*(-2) + 2 = 3*(-8) - 4 + 14 + 2 = -24 - 4 + 14 + 2, which is (-24 - 4) + (14 + 2) = -28 + 16 = -12. Choice A correctly evaluates p(-2) as -12. Choice B says 12, possibly from changing all signs incorrectly or forgetting the negative in $(-2)^3$, but use parentheses to track signs properly! Sign safety for negatives: when testing (x + 2) = (x - (-2)), you're evaluating at a = -2. Substitute carefully: if p(x) = x³ - 3x + 5, then p(-2) = (-2)³ - 3(-2) + 5 = -8 + 6 + 5 = 3. Use parentheses around negative values to avoid sign errors! This is where most mistakes happen with the Remainder Theorem.