← Back to the Polynomial Division worksheet
Last updated: October 22, 2026 · Written by the Math Made Easy math teachers
Each solution shows the division steps or substitution in full. Multiply the divisor by the quotient and add the remainder to confirm the original polynomial.
Standards: A-APR.B.2, A-APR.D.6, A-APR.C.4 · Level: Grade 11 Algebra 2
Problem 1 Long division with no remainder
Divide x2 by x to get x. Multiply: x(x + 4) = x2 + 4x. Subtract: 7x + 28.
Divide 7x by x to get 7. Multiply: 7(x + 4) = 7x + 28. Subtract: 0.
Quotient x + 7, remainder 0
Problem 2 Long division with a leading coefficient
Divide 2x2 by x to get 2x. Multiply: 2x(x + 4) = 2x2 + 8x. Subtract: −3x − 12.
Divide −3x by x to get −3. Multiply: −3(x + 4) = −3x − 12. Subtract: 0.
Quotient 2x − 3, remainder 0
Problem 3 A remainder appears
Divide x2 by x to get x. Multiply: x(x − 2) = x2 − 2x. Subtract: 8x − 5.
Divide 8x by x to get 8. Multiply: 8(x − 2) = 8x − 16. Subtract: 11.
Quotient x + 8, remainder 11
So (x2 + 6x − 5) ÷ (x − 2) = x + 8 + 11x − 2.
Problem 4 Completing a synthetic division
Products: 2·2 = 4, 2·(−1) = −2, 2·(−3) = −6. Sums: −5 + 4 = −1, −1 + (−2) = −3, 9 + (−6) = 3.
Quotient 2x2 − x − 3, remainder 3
Problem 5 Synthetic division with a zero remainder
Use r = 2. Bring down 1. Then multiply by 2 and add down each column.
Bottom row: 1, 4, −5, 0.
Quotient x2 + 4x − 5, remainder 0
Problem 6 Synthetic division by x plus a number
Use r = −2. Bring down 3. Then multiply by −2 and add down each column.
Bottom row: 3, −7, 18, −43.
Quotient 3x2 − 7x + 18, remainder −43
Problem 7 A missing term
The x3 term is missing, so use the coefficients 1, 0, −5, 2, 8.
Use r = 3. Bring down 1. Then multiply by 3 and add down each column.
Bottom row: 1, 3, 4, 14, 50.
Quotient x3 + 3x2 + 4x + 14, remainder 50
Problem 8 Dividing by 2x + 1
Divide 6x3 by 2x to get 3x2. Multiply: 3x2(2x + 1) = 6x3 + 3x2. Subtract: −2x2 + 7x + 9.
Divide −2x2 by 2x to get −x. Multiply: −x(2x + 1) = −2x2 − x. Subtract: 8x + 9.
Divide 8x by 2x to get 4. Multiply: 4(2x + 1) = 8x + 4. Subtract: 5.
Quotient 3x2 − x + 4, remainder 5
Problem 9 Reading a remainder from a graph
The graph passes through (4, 10) and (1, 4). By the Remainder Theorem, f(4) = 10 and f(1) = 4.
Use r = 4. Bring down 1. Then multiply by 4 and add down each column.
Bottom row: 1, 0, 1, 10.
Quotient x2 + 1, remainder 10
Remainders: 10 for x − 4 and 4 for x − 1
Problem 10 Evaluating with the Remainder Theorem
f(3) = 2(27) − 5(9) + 3 − 8 = 54 − 45 + 3 − 8 = 4.
f(−2) = 2(−8) − 5(4) − 2 − 8 = −16 − 20 − 2 − 8 = −46.
Remainders 4 and −46
Problem 11 Using the Factor Theorem
f(−3) = −27 + 9 + 30 + 8 = 20 ≠ 0.
f(1) = 1 + 1 − 10 + 8 = 0.
f(−4) = −64 + 16 + 40 + 8 = 0.
x − 1 and x + 4 are factors; x + 3 is not
Problem 12 Finding an unknown coefficient
No remainder means the value at x = 2 is 0: 8 − 4k + 18 − 14 = 0, so 12 − 4k = 0 and k = 3.
Check: x3 − 3x2 + 9x − 14 = (x − 2)(x2 − x + 7).
k = 3
Problem 13 Factoring completely from a zero
Use r = −2. Bring down 1. Then multiply by −2 and add down each column.
Bottom row: 1, −3, −4, 0.
Quotient x2 − 3x − 4, remainder 0
The quotient x2 − 3x − 4 = (x − 4)(x + 1).
f(x) = (x + 2)(x + 1)(x − 4)
Problem 14 Dividing by a quadratic
Divide x4 by x2 to get x2. Multiply: x2(x2 + 2x − 1) = x4 + 2x3 − x2. Subtract: x3 − x2 − 3x + 1.
Divide x3 by x2 to get x. Multiply: x(x2 + 2x − 1) = x3 + 2x2 − x. Subtract: −3x2 − 2x + 1.
Divide −3x2 by x2 to get −3. Multiply: −3(x2 + 2x − 1) = −3x2 − 6x + 3. Subtract: 4x − 2.
Quotient x2 + x − 3, remainder 4x − 2
Problem 15 A divisor with leading coefficient 2
Coefficients 6, −7, 5, −7 with r = 32: bring down 6; 6·32 = 9, −7 + 9 = 2; 2·32 = 3, 5 + 3 = 8; 8·32 = 12, −7 + 12 = 5.
So f(x) = (x − 32)(6x2 + 2x + 8) + 5 = (2x − 3)(3x2 + x + 4) + 5.
Quotient 3x2 + x + 4, remainder 5
Problem 16 Dimensions of a box
Divide x3 by x to get x2. Multiply: x2(x + 3) = x3 + 3x2. Subtract: 5x2 + 19x + 12.
Divide 5x2 by x to get 5x. Multiply: 5x(x + 3) = 5x2 + 15x. Subtract: 4x + 12.
Divide 4x by x to get 4. Multiply: 4(x + 3) = 4x + 12. Subtract: 0.
Quotient x2 + 5x + 4, remainder 0
Base area = x2 + 5x + 4 = (x + 1)(x + 4).
Base area x2 + 5x + 4; length x + 4 and width x + 1
Problem 17 Break-even points
Use r = 1. Bring down 1. Then multiply by 1 and add down each column.
Bottom row: 1, −6, 8, 0.
Quotient x2 − 6x + 8, remainder 0
The quotient x2 − 6x + 8 = (x − 2)(x − 4).
Break-even at 100, 200, and 400 meals (x = 1, 2, 4)
Problem 18 Proving an identity
Left side: x4 + 2x2y2 + y4.
Right side: (x4 − 2x2y2 + y4) + 4x2y2 = x4 + 2x2y2 + y4.
The sides match, so the identity holds.
With x = 3, y = 2: x2 − y2 = 5, 2xy = 12, x2 + y2 = 13, and 52 + 122 = 169 = 132.
Triple (5, 12, 13)
Problem 19 Using identities for calculation
a) x2 − y2 = 21, 2xy = 20, x2 + y2 = 29. Check: 441 + 400 = 841 = 292. Triple (20, 21, 29).
b) (1,003 − 997)(1,003 + 997) = 6 · 2,000 = 12,000.
Problem 20 Finding two unknown coefficients
p(1) = 0: 1 + a + b + 6 = 0, so a + b = −7.
p(−2) = 0: −8 + 4a − 2b + 6 = 0, so 2a − b = 1.
Add the two equations: 3a = −6, a = −2, and then b = −5.
Then p(x) = x3 − 2x2 − 5x + 6, and dividing by (x − 1)(x + 2) = x2 + x − 2 leaves x − 3.
a = −2, b = −5, third factor x − 3