Applying the Remainder Theorem - Algebra 2

1. What is the remainder when dividing $p(x)=2x^4-x^2+5$ by $(x+1)$? Use the Remainder Theorem (remainder $=p(a)$ for divisor $(x-a)$).​
   Answer: The remainder is $6$. 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 + 1) = (x - (-1)), a = -1; p(-1) = 2*$(-1)^4$ - $(-1)^2$ + 5 = 2*1 - 1 + 5 = 2 - 1 + 5 = 6. Choice A correctly evaluates p(-1) as 6. Choice B says 4, which could happen if you mistakenly computed $(-1)^4$ as -1 instead of 1, leading to 2*(-1) -1 +5 = -2 -1 +5 =2, but remember even powers are positive! 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.
2. Is $(x+2)$ a factor of $p(x)=x^3+3x^2-4x-12$? Use the Remainder Theorem / Factor Theorem (i.e., $(x-a)$ is a factor iff $p(a)=0$).
   Answer: Yes, because $p(-2)=0$. 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)), a = -2; p(-2) = $(-2)^3$ + 3*$(-2)^2$ - 4*(-2) - 12 = -8 + 3*4 + 8 - 12 = -8 + 12 + 8 - 12, which is (-8 - 12) + (12 + 8) = -20 + 20 = 0. Choice A correctly determines that yes, it is a factor because p(-2) = 0. Choice B says no because p(-2) = 8, which might come from forgetting to include the -12 or mishandling signs, but actually it's 0—keep track of all terms! 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.
3. A polynomial $f(x)$ has the property that $f(-3) = 8$ and $f(1) = 0$. When $f(x)$ is divided by $x^2 + 2x - 3$, what is the remainder?
   Answer: $-2x + 2$ Since $x^2 + 2x - 3 = (x + 3)(x - 1)$, when $f(x)$ is divided by this quadratic, the remainder must be linear: $R(x) = ax + b$. By the Remainder Theorem, $R(-3) = f(-3) = 8$ and $R(1) = f(1) = 0$. So $-3a + b = 8$ and $a + b = 0$. Solving: $b = -a$ and $-3a + (-a) = 8$, so $-4a = 8$ and $a = -2,b = 2$. Therefore $R(x) = -2x + 2$. The other choices don't satisfy both conditions simultaneously.
4. Which of the following are factors of $p(x)=x^3-6x^2+11x-6$? Use the Factor Theorem (test $p(a)=0$).​
   Answer: $(x-1)$, $(x-2)$, and $(x-3)$ 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. Test each: p(1) = 1 - 6 + 11 - 6 = (1 + 11) - (6 + 6) = 12 - 12 = 0; p(2) = 8 - 24 + 22 - 6 = (8 + 22) - (24 + 6) = 30 - 30 = 0; p(3) = 27 - 54 + 33 - 6 = (27 + 33) - (54 + 6) = 60 - 60 = 0, so all are factors. Choice C correctly identifies (x-1), (x-2), and (x-3) as factors. Choice D omits (x-2), maybe from a calculation error like p(2) = 8 - 24 + 22 - 6 as 30 - 30 = 0, but wait, it is 0—perhaps misadding 8 - 24 = -16, -16 + 22 = 6, 6 - 6 = 0, still 0; double-check arithmetic! 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.
5. Use the Remainder Theorem to find the remainder when $p(x)=3x^3-x^2-7x+2$ is divided by $(x+2)$.​
   Answer: The remainder is $-12$. 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.
6. A cubic polynomial $f(x)$ has leading coefficient 1 and satisfies $f(1) = f(3) = f(5) = 0$. When $f(x)$ is divided by $(x - 7)$, the remainder is:
   Answer: 48 Since $f(x)$ is cubic with leading coefficient 1 and has zeros at $x = 1, 3, 5$, we have $f(x) = (x-1)(x-3)(x-5)$. By the Remainder Theorem, the remainder when divided by $(x-7)$ is $f(7) = (7-1)(7-3)(7-5) = 6 \cdot 4 \cdot 2 = 48$.
7. Use the Remainder Theorem to find the remainder when $p(x)=3x^3-x^2-7x+2$ is divided by $(x+2)$.
   Answer: The remainder is $-12$. 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.
8. Use the Remainder Theorem to find the remainder when $p(x)=x^3+4x^2-x-12$ is divided by $(x+3).$
   Answer: The remainder is $0$. 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 p(x) = x³ + 4x² - x - 12 divided by (x + 3) = (x - (-3)), we need p(-3): p(-3) = (-3)³ + 4(-3)² - (-3) - 12 = -27 + 4(9) + 3 - 12 = -27 + 36 + 3 - 12 = 0. Choice A correctly identifies the remainder as 0, which also means (x + 3) is a factor of p(x)! Sign safety for negatives: when testing (x + 3) = (x - (-3)), you're evaluating at a = -3. Substitute carefully using parentheses: (-3)³ = -27 and (-3)² = 9. The zero remainder reveals that this polynomial is exactly divisible by (x + 3)!
9. Consider the polynomial $R(x) = 3x^3 - 7x^2 + 4x - 12$. If $(x - k)$ is a factor of $R(x)$, which of the following could be the remainder when $R(x)$ is divided by $(x - 2k)$?
   Answer: Equal to $R(2k)$ which depends on $k$ If $(x - k)$ is a factor of $R(x)$, then $R(k) = 0$. By the Remainder Theorem, the remainder when $R(x)$ is divided by $(x - 2k)$ is $R(2k)$. Since $k$ and $2k$ are generally different values, $R(2k)$ will depend on the specific value of $k$. Choice A is wrong because there's no reason $R(2k)$ must equal zero. Choice C incorrectly assumes the remainder is always the constant term. Choice D confuses the relationship between $R(k)$ and $R(2k)$.
10. What is the remainder when dividing $p(x)=2x^4-x^2+5$ by $(x+1)$? Use the Remainder Theorem (remainder $=p(a)$ for divisor $(x-a)$).
    Answer: The remainder is $6$. 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 + 1) = (x - (-1)), a = -1; p(-1) = 2*$(-1)^4$ - $(-1)^2$ + 5 = 2*1 - 1 + 5 = 2 - 1 + 5 = 6. Choice A correctly evaluates p(-1) as 6. Choice B says 4, which could happen if you mistakenly computed $(-1)^4$ as -1 instead of 1, leading to 2*(-1) -1 +5 = -2 -1 +5 =2, but remember even powers are positive! 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.
11. Given that $W(x) = x^5 - 2x^4 + 3x^3 - x^2 + 4x - 6$ and the remainder when $W(x)$ is divided by $(x^2 - 4)$ is $ax + b$, what is the value of $a + b$?
    Answer: $\frac{W(2) - W(-2)}{4} + W(2)$ Since $x^2 - 4 = (x-2)(x+2)$, the remainder when dividing by this quadratic is linear: $R(x) = ax + b$. By the Remainder Theorem, $R(2) = W(2)$ and $R(-2) = W(-2)$. So $2a + b = W(2)$ and $-2a + b = W(-2)$. Solving: $4a = W(2) - W(-2)$, so $a = \frac{W(2) - W(-2)}{4}$. And $b = W(2) - 2a = W(2) - \frac{W(2) - W(-2)}{2}$. Therefore $a + b = \frac{W(2) - W(-2)}{4} + W(2) - \frac{W(2) - W(-2)}{2} = \frac{W(2) - W(-2)}{4} + W(2)$. The other choices give incorrect combinations of these values.
12. Use the Remainder Theorem to find the remainder when $p(x)=x^3-4x^2+5x-2$ is divided by $(x-3)$. (Recall: remainder $=p(a)$ when dividing by $(x-a)$.)
    Answer: The remainder is $4$. 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)=$3^3$ -4*$3^2$ +5*3 -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.
13. If $g(x) = x^4 - 3x^3 + 2x^2 - 5x + 1$ and the remainder when $g(x)$ is divided by $(x - c)$ is $15$, which equation must be satisfied by $c$?
    Answer: $c^4 - 3c^3 + 2c^2 - 5c + 1 = 15$ By the Remainder Theorem, when $g(x)$ is divided by $(x - c)$, the remainder equals $g(c)$. Since the remainder is $15$, we have $g(c) = 15$. Therefore $c^4 - 3c^3 + 2c^2 - 5c + 1 = 15$. Choice B incorrectly sets the expression equal to zero after subtracting 15. Choice C adds 15 instead of setting the original equal to 15. Choice D incorrectly modifies the constant term before setting equal to 15.
14. Is $(x+2)$ a factor of $p(x)=x^3+3x^2-4x-12$? Use the Remainder Theorem / Factor Theorem (i.e., $(x-a)$ is a factor iff $p(a)=0$).​
    Answer: Yes, because $p(-2)=0$. 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)), a = -2; p(-2) = $(-2)^3$ + 3*$(-2)^2$ - 4*(-2) - 12 = -8 + 3*4 + 8 - 12 = -8 + 12 + 8 - 12, which is (-8 - 12) + (12 + 8) = -20 + 20 = 0. Choice A correctly determines that yes, it is a factor because p(-2) = 0. Choice B says no because p(-2) = 8, which might come from forgetting to include the -12 or mishandling signs, but actually it's 0—keep track of all terms! 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.
15. Determine if $x=2$ is a zero of $p(x)=x^3-6x^2+11x-6$. (Use: $x=a$ is a zero iff $p(a)=0$.)
    Answer: Yes, because $p(2)=0$. 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)=$2^3$ -6*$2^2$ +11*2 -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.
16. Determine if $x=2$ is a zero of $p(x)=x^3-6x^2+11x-6$. (Hint: $x=2$ is a zero iff $p(2)=0$, which also means $(x-2)$ is a factor.)
    Answer: Yes, because $p(2)=0$. 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, -6*$2^2$ = -6*4 = -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.
17. If the remainder when $p(x)=x^3-2x^2+kx-4$ is divided by $(x-2)$ is $6$, what is the value of $k$? (Use remainder $=p(2)$.)​
    Answer: $k=5$ 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! Here, divisor (x - 2), a = 2, and remainder = 6, so p(2) = 6: $(2)^3$ - 2*$(2)^2$ + k*(2) - 4 = 8 - 2*4 + 2k - 4 = 8 - 8 + 2k - 4 = 2k - 4 = 6, so 2k = 10, k = 5. Choice D correctly finds k = 5. Choice A says k = -1, perhaps from setting 2k - 4 = 0 instead of 6 or a solving error like 2k = 2, but always solve the equation carefully! 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.
18. Verify the claim using the Remainder Theorem: When dividing $p(x)=x^4-3x^2+2x+1$ by $(x-1)$, the remainder is $1$.
    Answer: True, because $p(1)=1$. 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, -3*$1^2$ = -3, +2*1 = +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.
19. Use $p(a)$ to find the remainder for division by $(x-a).$ If $p(x)=x^4-x^2+2x-6,$ what is the remainder when dividing by $(x-2)?$
    Answer: The remainder is $10$. 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 p(x) = x⁴ - x² + 2x - 6 divided by (x - 2), we need p(2): p(2) = (2)⁴ - (2)² + 2(2) - 6 = 16 - 4 + 4 - 6 = 10. Choice D correctly identifies the remainder as 10. Using the Remainder Theorem: (1) Identify the divisor (x - 2) so a = 2, (2) Substitute 2 for every x in p(x), (3) Calculate carefully: 16 - 4 + 4 - 6 = 10, (4) That result is your remainder. This method is dramatically faster than polynomial long division, especially for higher-degree polynomials!
20. Given that $(x - 4)$ is a factor of $h(x) = x^3 - 6x^2 + kx + 8$, which statement about $h(x + 4)$ is necessarily true?
    Answer: $h(x + 4)$ has $x$ as a factor Since $(x - 4)$ is a factor of $h(x)$, we know $h(4) = 0$ by the Remainder Theorem. For the transformed function $h(x + 4)$, we evaluate at $x = 0$: $h(0 + 4) = h(4) = 0$. Therefore, by the Remainder Theorem, $x$ is a factor of $h(x + 4)$. Choice A is incorrect because the horizontal shift changes the factor. Choice C incorrectly suggests $(x + 4)$ is a factor. Choice D incorrectly identifies $(x - 8)$ as the factor.

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