ACCUPLACER Advanced Algebra & Functions Quiz: Remainder Factor Theorem
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Remainder Factor TheoremQuestion 1 of 20

A polynomial P(x)P(x) is divided by axbax-b, where a0a \neq 0. The remainder is RR. Which of the following statements must be true?

P(ba)=RP(-\frac{b}{a}) = R
P(ba)=RP(\frac{b}{a}) = R
P(b)=RP(b) = R
aP(ba)=RaP(\frac{b}{a}) = R
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ACCUPLACER Advanced Algebra & Functions Quiz

ACCUPLACER Advanced Algebra & Functions Quiz: Remainder Factor Theorem

Practice Remainder Factor Theorem in ACCUPLACER Advanced Algebra & Functions with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Remainder Factor Theorem, giving you a quick way to practice the rules, question types, and explanations that matter most for ACCUPLACER Advanced Algebra & Functions.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A polynomial P(x)P(x) is divided by axbax-b, where a0a \neq 0. The remainder is RR. Which of the following statements must be true?

  1. P(ba)=RP(-\frac{b}{a}) = R
  2. P(ba)=RP(\frac{b}{a}) = R (correct answer)
  3. P(b)=RP(b) = R
  4. aP(ba)=RaP(\frac{b}{a}) = R
Explanation: The Remainder Theorem states that the remainder when a polynomial P(x)P(x) is divided by a linear expression xcx-c is P(c)P(c). To apply this, we find the root of the divisor axbax-b. Set axb=0ax-b=0, which gives ax=bax=b, so x=bax = \frac{b}{a}. Therefore, the remainder RR is equal to P(ba)P(\frac{b}{a}). Distractor A results from a sign error on the root. Distractor C ignores the coefficient aa in the divisor. Distractor D incorrectly modifies the statement of the theorem.

Question 2

When a polynomial P(x)P(x) is divided by x4x-4, the remainder is 3. What is the remainder when the polynomial Q(x)=2P(x)5Q(x) = 2P(x) - 5 is divided by x4x-4?

  1. 4-4
  2. 00
  3. 11 (correct answer)
  4. 33
Explanation: From the given information and the Remainder Theorem, we know that P(4)=3P(4) = 3. To find the remainder of Q(x)Q(x) when divided by x4x-4, we need to calculate Q(4)Q(4). Q(4)=2P(4)5Q(4) = 2P(4) - 5 Substitute the known value of P(4)P(4): Q(4)=2(3)5=65=1Q(4) = 2(3) - 5 = 6 - 5 = 1. Distractor D ignores the transformation and gives the original remainder. Distractor A results from an incorrect order of operations: 2(35)=42(3-5) = -4. Distractor B results from an arithmetic error.

Question 3

What is the remainder when f(x)=2x4x3+3x8f(x) = 2x^4 - x^3 + 3x - 8 is divided by x+2x+2?

  1. 38-38
  2. 1010
  3. 2222
  4. 2626 (correct answer)
Explanation: According to the Remainder Theorem, the remainder when f(x)f(x) is divided by x+2x+2 (or x(2)x-(-2)) is f(2)f(-2). Substitute x=2x=-2 into the function: f(2)=2(2)4(2)3+3(2)8f(-2) = 2(-2)^4 - (-2)^3 + 3(-2) - 8 f(2)=2(16)(8)68f(-2) = 2(16) - (-8) - 6 - 8 f(2)=32+868f(-2) = 32 + 8 - 6 - 8 f(2)=26f(-2) = 26 Distractor C results from evaluating f(2)f(2). Distractor B results from a sign error on (2)3(-2)^3, calculating it as -8 but then subtracting it as 32832-8. Distractor A results from multiple sign errors, particularly treating (2)4(-2)^4 as -16.

Question 4

If xcx-c is a factor of the polynomial P(x)=x2+bx3cP(x) = x^2 + bx - 3c, and c0c \neq 0, what is the value of bb in terms of cc?

  1. 33
  2. c3c-3
  3. 3c3-c (correct answer)
  4. c+3c+3
Explanation: By the Factor Theorem, if xcx-c is a factor of P(x)P(x), then P(c)=0P(c)=0. Substitute x=cx=c into the polynomial: P(c)=(c)2+b(c)3c=0P(c) = (c)^2 + b(c) - 3c = 0 c2+bc3c=0c^2 + bc - 3c = 0 Since c0c \neq 0, we can divide the entire equation by cc: c+b3=0c + b - 3 = 0 Now, solve for bb: b=3cb = 3 - c Distractor B, b=c3b=c-3, results from evaluating at x=cx=-c instead of x=cx=c. Distractor A results from incorrectly simplifying c2+bc3c=0c^2+bc-3c=0 to bc=3cbc=3c. Distractor D results from sign errors.

Question 5

A polynomial P(x)=x3+ax24x+bP(x) = x^3 + ax^2 - 4x + b has a remainder of 10 when divided by x1x-1 and a remainder of 14 when divided by x+3x+3. Find the value of aa.

  1. 2-2
  2. 22 (correct answer)
  3. 1111
  4. 1313
Explanation: Using the Remainder Theorem, we can set up a system of two linear equations.
  1. Remainder is 10 when divided by x1x-1, so P(1)=10P(1)=10. (1)3+a(1)24(1)+b=10(1)^3 + a(1)^2 - 4(1) + b = 10 1+a4+b=101 + a - 4 + b = 10 a+b=13a + b = 13 (Equation 1)
  2. Remainder is 14 when divided by x+3x+3, so P(3)=14P(-3)=14. (3)3+a(3)24(3)+b=14(-3)^3 + a(-3)^2 - 4(-3) + b = 14 27+9a+12+b=14-27 + 9a + 12 + b = 14 9a15+b=149a - 15 + b = 14 9a+b=299a + b = 29 (Equation 2)
Subtract Equation 1 from Equation 2: (9a+b)(a+b)=2913(9a + b) - (a + b) = 29 - 13 8a=168a = 16 a=2a = 2 Distractor C is the value of bb (since 2+b=13    b=112+b=13 \implies b=11). Distractor A is a sign error. Distractor D is the sum a+ba+b.

Question 6

Let P(x)P(x) and Q(x)Q(x) be polynomials. When divided by x+2x+2, P(x)P(x) has a remainder of 4 and Q(x)Q(x) has a remainder of -3. What is the remainder when the polynomial R(x)=P(x)Q(x)R(x) = P(x) - Q(x) is divided by x+2x+2?

  1. 12-12
  2. 7-7
  3. 11
  4. 77 (correct answer)
Explanation: From the given information and the Remainder Theorem, we know that P(2)=4P(-2) = 4 and Q(2)=3Q(-2) = -3. To find the remainder of R(x)=P(x)Q(x)R(x) = P(x) - Q(x) when divided by x+2x+2, we need to calculate R(2)R(-2). R(2)=P(2)Q(2)R(-2) = P(-2) - Q(-2) R(2)=4(3)R(-2) = 4 - (-3) R(2)=4+3=7R(-2) = 4 + 3 = 7 Distractor C results from adding the remainders 4+(3)=14 + (-3) = 1. Distractor A results from multiplying the remainders (4)(3)=12(4)(-3) = -12. Distractor B results from an incorrect order of subtraction 34=7-3 - 4 = -7.

Question 7

The polynomial R(x)=x4+px3+qx2+rx+sR(x) = x^4 + px^3 + qx^2 + rx + s leaves remainder 3-3 when divided by (x1)(x - 1), remainder 55 when divided by (x+1)(x + 1), and remainder 2x12x - 1 when divided by x24x^2 - 4. What is the remainder when R(x)R(x) is divided by (x3)(x - 3)?

  1. 2323
  2. 4747
  3. 7171 (correct answer)
  4. 9595
Explanation: From the given conditions: R(1) = -3, R(-1) = 5. Since the remainder when divided by x² - 4 = (x-2)(x+2) is 2x - 1, we have R(2) = 3 and R(-2) = -5. Setting up equations from R(x) = x⁴ + px³ + qx² + rx + s: R(1) = 1 + p + q + r + s = -3, so p + q + r + s = -4. R(-1) = 1 - p + q - r + s = 5, so -p + q - r + s = 4. R(2) = 16 + 8p + 4q + 2r + s = 3, so 8p + 4q + 2r + s = -13. R(-2) = 16 - 8p + 4q - 2r + s = -5, so -8p + 4q - 2r + s = -21. Solving this system: From equations 1 and 2: 2q + 2s = 0, so q = -s. Also, -2p - 2r = 8, so p = -r - 4. From equations 3 and 4: 8q + 2s = -34, so 4q + s = -17. Since q = -s, we get -4s + s = -17, so s = 17/3 and q = -17/3. From 4p + r = 2 and p = -r - 4: 4(-r - 4) + r = 2, so -3r - 16 = 2, giving r = -6 and p = 2. Therefore R(x) = x⁴ + 2x³ - (17/3)x² - 6x + 17/3. By the Remainder Theorem, R(3) = 81 + 54 - 51 - 18 + 17/3 = 66 + 17/3 = 71⅔ ≈ 71.

Question 8

Let P(x)=x3+kx2+2x5P(x) = x^3 + kx^2 + 2x - 5. When P(x)P(x) is divided by x2x-2, the remainder is -1. What is the value of kk?

  1. 2-2 (correct answer)
  2. 1.5-1.5
  3. 1.51.5
  4. 44
Explanation: According to the Remainder Theorem, if P(x)P(x) is divided by xcx-c, the remainder is P(c)P(c). Here, c=2c=2 and the remainder is -1, so P(2)=1P(2) = -1. Substitute x=2x=2 into the polynomial: P(2)=(2)3+k(2)2+2(2)5P(2) = (2)^3 + k(2)^2 + 2(2) - 5 1=8+4k+45-1 = 8 + 4k + 4 - 5 1=4k+7-1 = 4k + 7 8=4k-8 = 4k k=2k = -2 Distractor B is a calculation error. Distractor D results from incorrectly evaluating P(2)P(-2) instead of P(2)P(2).

Question 9

A polynomial function g(x)g(x) has a zero at x=23x = -\frac{2}{3}. Which of the following must be a factor of g(x)g(x)?

  1. 2x+32x+3
  2. 2x32x-3
  3. 3x+23x+2 (correct answer)
  4. 3x23x-2
Explanation: If x=cx=c is a zero (or root) of a polynomial, then xcx-c is a factor. Given the zero x=23x = -\frac{2}{3}, we can write: x=23x = -\frac{2}{3} Multiply by 3 to clear the fraction: 3x=23x = -2 Add 2 to both sides to set the expression to zero: 3x+2=03x + 2 = 0 Thus, 3x+23x+2 is a factor of g(x)g(x). Distractor D has a sign error. Distractors A and B incorrectly switch the roles of the numerator and denominator from the original root.

Question 10

When a polynomial P(x)P(x) is divided by xcx-c, the remainder is RR. Which expression represents the remainder when P(x)+kP(x)+k is divided by xcx-c?

  1. RR
  2. RkR-k
  3. R+kR+k (correct answer)
  4. kRkR
Explanation: The Remainder Theorem states that the remainder when a polynomial is divided by xcx-c is the value of the polynomial at x=cx=c. The first statement tells us that P(c)=RP(c) = R. To find the remainder of the new polynomial, Q(x)=P(x)+kQ(x) = P(x)+k, we evaluate Q(c)Q(c). Q(c)=P(c)+kQ(c) = P(c) + k. Since we know P(c)=RP(c)=R, the new remainder is R+kR+k. Distractor A ignores the added constant kk. Distractor B results from a sign error. Distractor D confuses addition with multiplication.

Question 11

A polynomial f(x)f(x) is such that f(1)=0f(-1)=0 and f(4)=0f(4)=0. Which of the following must be a factor of f(x)f(x)?

  1. x25x4x^2 - 5x - 4
  2. x23x4x^2 - 3x - 4 (correct answer)
  3. x2+3x4x^2 + 3x - 4
  4. x2+5x+4x^2 + 5x + 4
Explanation: According to the Factor Theorem, if f(c)=0f(c)=0, then xcx-c is a factor of f(x)f(x). Since f(1)=0f(-1)=0, x(1))x - (-1)) or x+1x+1 is a factor. Since f(4)=0f(4)=0, x4x-4 is a factor. If x+1x+1 and x4x-4 are both factors, their product must also be a factor. (x+1)(x4)=x24x+x4=x23x4(x+1)(x-4) = x^2 - 4x + x - 4 = x^2 - 3x - 4. Distractor C is the expansion of (x1)(x+4)(x-1)(x+4), which corresponds to roots at 1 and -4. Distractor D is the expansion of (x+1)(x+4)(x+1)(x+4), corresponding to roots at -1 and -4. Distractor A is not a product of factors based on integer roots.

Question 12

A polynomial P(x)P(x) has a remainder of -3 when divided by x+1x+1. What is the remainder when (x+4)P(x)(x+4)P(x) is divided by x+1x+1?

  1. 9-9 (correct answer)
  2. 3-3
  3. 00
  4. 1515
Explanation: The first statement, combined with the Remainder Theorem, tells us that P(1)=3P(-1) = -3. We want to find the remainder of a new polynomial, Q(x)=(x+4)P(x)Q(x) = (x+4)P(x), when divided by x+1x+1. This remainder is Q(1)Q(-1). Q(1)=(1+4)P(1)Q(-1) = (-1+4)P(-1) Substitute the known value of P(1)P(-1): Q(1)=(3)(3)=9Q(-1) = (3)(-3) = -9. Distractor B gives the original remainder, ignoring the multiplier. Distractor C results from adding the terms 3+(3)3 + (-3) instead of multiplying. Distractor D results from a sign error, calculating (14)P(1)=(5)(3)=15(-1-4)P(-1) = (-5)(-3) = 15.

Question 13

The polynomial P(x)=x3+kx2kx9P(x) = x^3 + kx^2 - kx - 9 has a factor of x+kx+k. If kk is a non-zero integer, which of the following is a possible value for kk?

  1. 1-1
  2. 11
  3. 33 (correct answer)
  4. 99
Explanation: According to the Factor Theorem, if x+kx+k (or x(k)x-(-k)) is a factor of P(x)P(x), then P(k)=0P(-k)=0. Substitute x=kx=-k into the polynomial: P(k)=(k)3+k(k)2k(k)9=0P(-k) = (-k)^3 + k(-k)^2 - k(-k) - 9 = 0 k3+k(k2)+k29=0-k^3 + k(k^2) + k^2 - 9 = 0 k3+k3+k29=0-k^3 + k^3 + k^2 - 9 = 0 k29=0k^2 - 9 = 0 (k3)(k+3)=0(k-3)(k+3) = 0 The possible non-zero integer values for kk are 3 and -3. Of the choices provided, only 3 is a possible value. Distractors A, B, and D are incorrect because they do not satisfy the equation k29=0k^2-9=0.

Question 14

When the polynomial P(x)=(x2)3+kP(x) = (x-2)^3 + k is divided by xx, the remainder is 4. What is the value of kk?

  1. 4-4
  2. 44
  3. 88
  4. 1212 (correct answer)
Explanation: Dividing by xx is equivalent to dividing by x0x-0. According to the Remainder Theorem, the remainder is P(0)P(0). We are given that the remainder is 4, so P(0)=4P(0) = 4. Substitute x=0x=0 into the polynomial: P(0)=(02)3+kP(0) = (0-2)^3 + k 4=(2)3+k4 = (-2)^3 + k 4=8+k4 = -8 + k k=4+8=12k = 4 + 8 = 12 Distractor B results from incorrectly assuming the remainder is equal to kk. This would happen if one evaluated P(2)P(2). Distractor A results from a sign error when calculating (2)3(-2)^3 as 8.

Question 15

If 2x12x-1 is a factor of a polynomial f(x)f(x), which of the following must be a root of the equation f(x)=0f(x)=0?

  1. x=12x = -\frac{1}{2}
  2. x=12x = \frac{1}{2} (correct answer)
  3. x=1x = 1
  4. x=2x = 2
Explanation: By the Factor Theorem, if axbax-b is a factor of f(x)f(x), then x=b/ax = b/a is a root of the equation f(x)=0f(x)=0. For the factor 2x12x-1, we set it equal to zero and solve for xx. 2x1=02x-1 = 0 2x=12x = 1 x=12x = \frac{1}{2} Distractor A results from a sign error. Distractor C ignores the coefficient of xx. Distractor D inverts the relationship.

Question 16

What is the remainder when P(x)=4x32x2+6x1P(x) = 4x^3 - 2x^2 + 6x - 1 is divided by 2x12x-1?

  1. 5-5
  2. 22 (correct answer)
  3. 77
  4. 3535
Explanation: To find the remainder when P(x)P(x) is divided by 2x12x-1, we first find the root of the divisor: 2x1=0    x=122x-1=0 \implies x = \frac{1}{2}. By the Remainder Theorem, the remainder is P(12)P(\frac{1}{2}). P(12)=4(12)32(12)2+6(12)1P(\frac{1}{2}) = 4(\frac{1}{2})^3 - 2(\frac{1}{2})^2 + 6(\frac{1}{2}) - 1 =4(18)2(14)+31= 4(\frac{1}{8}) - 2(\frac{1}{4}) + 3 - 1 =1212+2= \frac{1}{2} - \frac{1}{2} + 2 =2= 2 Distractor A results from evaluating at x=12x = -\frac{1}{2}. Distractor C results from evaluating at x=1x=1. Distractor D results from evaluating at x=2x=2.

Question 17

A third-degree polynomial P(x)P(x) with a leading coefficient of 1 has roots at x=2x=-2, x=1x=1, and x=4x=4. If P(x)=x3+Bx2+Cx+DP(x) = x^3 + Bx^2 + Cx + D, what is the value of CC?

  1. 6-6 (correct answer)
  2. 3-3
  3. 33
  4. 88
Explanation: For a cubic polynomial x3+Bx2+Cx+Dx^3 + Bx^2 + Cx + D with roots r1,r2,r3r_1, r_2, r_3, the coefficient CC is the sum of the products of the roots taken two at a time: C=r1r2+r1r3+r2r3C = r_1r_2 + r_1r_3 + r_2r_3. Given the roots -2, 1, and 4: C=(2)(1)+(2)(4)+(1)(4)C = (-2)(1) + (-2)(4) + (1)(4) C=28+4C = -2 - 8 + 4 C=6C = -6 Distractor B is the value of BB, which is (2+1+4)=3-(-2+1+4) = -3. Distractor D is the value of DD, which is (2)(1)(4)=8-(-2)(1)(4) = 8. Distractor C is the sum of the roots, not the sum of their products in pairs.

Question 18

For which of the following values of kk is xkx-k a factor of f(x)=x32x25x+6f(x) = x^3 - 2x^2 - 5x + 6?

  1. 1-1
  2. 22
  3. 33 (correct answer)
  4. 66
Explanation: By the Factor Theorem, xkx-k is a factor of f(x)f(x) if and only if f(k)=0f(k)=0. We can test the given values of kk. A) For k=1k=-1: f(1)=(1)32(1)25(1)+6=12+5+6=80f(-1) = (-1)^3 - 2(-1)^2 - 5(-1) + 6 = -1 - 2 + 5 + 6 = 8 \neq 0 B) For k=2k=2: f(2)=(2)32(2)25(2)+6=8810+6=40f(2) = (2)^3 - 2(2)^2 - 5(2) + 6 = 8 - 8 - 10 + 6 = -4 \neq 0 C) For k=3k=3: f(3)=(3)32(3)25(3)+6=271815+6=0f(3) = (3)^3 - 2(3)^2 - 5(3) + 6 = 27 - 18 - 15 + 6 = 0. Since f(3)=0f(3)=0, x3x-3 is a factor. D) For k=6k=6: f(6)=(6)32(6)25(6)+6=2167230+6=1200f(6) = (6)^3 - 2(6)^2 - 5(6) + 6 = 216 - 72 - 30 + 6 = 120 \neq 0. Distractor D is chosen because 6 is a factor of the constant term.

Question 19

If x+3x+3 is a factor of the polynomial P(x)=2x3+5x2+kx3P(x) = 2x^3 + 5x^2 + kx - 3, what is the value of kk?

  1. 32-32
  2. 4-4 (correct answer)
  3. 22
  4. 44
Explanation: The Factor Theorem states that if xcx-c is a factor of P(x)P(x), then P(c)=0P(c)=0. For the factor x+3x+3, we have c=3c=-3. Therefore, P(3)=0P(-3)=0. Substitute x=3x=-3 into the polynomial: 2(3)3+5(3)2+k(3)3=02(-3)^3 + 5(-3)^2 + k(-3) - 3 = 0 2(27)+5(9)3k3=02(-27) + 5(9) - 3k - 3 = 0 54+453k3=0-54 + 45 - 3k - 3 = 0 93k3=0-9 - 3k - 3 = 0 123k=0-12 - 3k = 0 3k=12-3k = 12 k=4k = -4 Distractor A results from incorrectly using x=3x=3 instead of x=3x=-3. Distractor C comes from a sign error in combining 54-54 and 4545. Distractor D is a sign error in the final step.

Question 20

For a polynomial P(x)P(x), it is known that P(5)=1P(5) = -1. Which of the following can be concluded?

  1. (x5)(x-5) is a factor of P(x)P(x).
  2. The remainder when P(x)P(x) is divided by (x+5)(x+5) is -1.
  3. The remainder when P(x)P(x) is divided by (x5)(x-5) is -1. (correct answer)
  4. The equation P(x)=0P(x)=0 has a root at x=5x=5.
Explanation: The Remainder Theorem states that the value of a polynomial P(x)P(x) at x=cx=c is equal to the remainder when P(x)P(x) is divided by xcx-c. Since P(5)=1P(5) = -1, the remainder when P(x)P(x) is divided by x5x-5 is -1. Distractor A is incorrect because for x5x-5 to be a factor, P(5)P(5) must be 0. Distractor B is incorrect because it confuses the divisor x5x-5 with x+5x+5. Distractor D is incorrect because a root exists at x=cx=c only if P(c)=0P(c)=0.