Understand · explore · practise
Learn algebraic division with linear and quadratic divisors, missing powers and remainders. Step through an original worked example and practise with full solutions.
Before you startExpanding polynomials, collecting like terms and index laws
01 / Quotient and remainder
Dividing a polynomial N(x) by a nonzero polynomial D(x) produces a quotient Q(x) and a remainder R(x). The central relationship is a polynomial identity:
N(x) ≡ D(x)Q(x) + R(x)
Either R is zero, or its degree is strictly smaller than the degree of D.
Degree means the highest power with a nonzero coefficient. When the divisor is linear, the remainder is a constant. For a quadratic divisor, it can be linear or constant. If the numerator already has smaller degree than the divisor, the quotient is zero and the numerator is the remainder.
N(x)/D(x) = Q(x) + R(x)/D(x)
This fraction form requires D(x) ≠ 0. The polynomial identity itself holds for every real x.
02 / The routine
The leading term cancels at every subtraction, so the degree falls. If your degree stays the same, check the leading-term division or the subtraction signs.
(2x³ + 3x² − 5x + 7) ÷ (x + 2)
Start with the original polynomial. The quotient is 0 so far.
Quotient so far: 0
Left to divide:
2x³ + 3x² − 5x + 7
Move forward or back yourself. At every stage, original polynomial = (x + 2) × quotient so far + polynomial left. The complete ledger is in the notes.
Pause, replay or seek freely. The notes explain the same idea and stay in view.
03 / Linear divisor
Divide 2x³ + 3x² − 5x + 7 by x + 2. Keep every lower-power term when carrying the current polynomial into the next row.
2x³ ÷ x = 2x²
First quotient term. Multiply: (x + 2)2x² = 2x³ + 4x².
(2x³ + 3x² − 5x + 7)
− (2x³ + 4x²)
= −x² − 5x + 7
Subtract the whole product.
−x² ÷ x = −x
Next quotient term. Multiply: (x + 2)(−x) = −x² − 2x.
(−x² − 5x + 7)
− (−x² − 2x)
= −3x + 7
The subtracted −2x becomes +2x.
−3x ÷ x = −3
Multiply: (x + 2)(−3) = −3x − 6.
(−3x + 7) − (−3x − 6) = 13
Stop: a constant has smaller degree than x + 2.
Q = 2x² − x − 3, R = 13
Check: (x + 2)(2x² − x − 3) + 13 = 2x³ + 3x² − 5x + 7.
04 / Quadratic divisor
Divide x⁴ − x² + 2x + 3 by x² + 1. Write the numerator as x⁴ + 0x³ − x² + 2x + 3. The missing x³ term is a placeholder, not permission to shift later terms into its column.
x⁴ ÷ x² = x²
Multiply the divisor by x²: x⁴ + x².
(x⁴ + 0x³ − x² + 2x + 3)
− (x⁴ + x²)
= −2x² + 2x + 3
The zero x³ coefficient stays zero.
−2x² ÷ x² = −2
Multiply the divisor by −2: −2x² − 2.
(−2x² + 2x + 3)
− (−2x² − 2) = 2x + 5
Degree 1 is smaller than degree 2. Stop.
Q = x² − 2, R = 2x + 5
Check: (x² + 1)(x² − 2) + 2x + 5 = x⁴ − x² + 2x + 3.
05 / Non-monic divisor
For divisor 2x − 1, the leading term is 2x. Dividing 6x³ by it gives 3x², not 6x². Fractions in a quotient are allowed; they are not a sign that the method has failed.
First quotient term: 3x²
Subtract (2x − 1)3x² = 6x³ − 3x². Left: 4x² − 8x + 5.
Next quotient term: 2x
Subtract (2x − 1)2x = 4x² − 2x. Left: −6x + 5.
Last quotient term: −3
Subtract (2x − 1)(−3) = −6x + 3. Left: 2.
Q = 3x² + 2x − 3, R = 2
Fraction form: 3x² + 2x − 3 + 2/(2x − 1), with x ≠ 1/2.
06 / Independent checks
Multiply D by your quotient and add your remainder. Collect every coefficient and compare with N. A single numerical substitution is useful for finding mistakes, but normally cannot verify an entire polynomial identity.
If the divisor is x − a, substitute x = a into N ≡ (x − a)Q + R. The product vanishes, so R = N(a). A zero remainder means x − a is a factor.
Division by x − a: R = N(a)
You could write N = D(Q − 1) + (R + D). That equality is true, but R + D has degree at least that of D and still needs division. The degree condition singles out the completed quotient and remainder.
For the linear example, N(−2) = −16 + 12 + 10 + 7 = 13. This confirms the remainder independently; multiply out as well to check the quotient.
07 / Your turn
Find the quotient and remainder unless the question asks otherwise. Show the subtraction stages before checking the answer.
Divide x³ + 4x² − 5 by x + 1.
Write x³ + 4x² + 0x − 5.
Subtract x²(x + 1) to leave 3x² − 5. Subtract 3x(x + 1) to leave −3x − 5. Subtract −3(x + 1) to leave −2. Q = x² + 3x − 3, R = −2. Multiplication gives (x + 1)(x² + 3x − 3) − 2 = x³ + 4x² − 5.
Divide 2x³ − 3x² − 8x + 12 by x − 2.
A zero remainder is possible.
Subtract 2x²(x − 2), leaving x² − 8x + 12. Subtract x(x − 2), leaving −6x + 12. Subtract −6(x − 2), leaving zero. Q = 2x² + x − 6, R = 0. Thus x − 2 is a factor.
Divide x⁴ + x³ + 2x² + 3 by x² + 2.
Include 0x in the numerator.
Subtract x²(x² + 2), leaving x³ + 3. Subtract x(x² + 2), leaving −2x + 3. Stop: Q = x² + x and R = −2x + 3. Expanding (x² + 2)(x² + x) − 2x + 3 recovers the numerator.
Divide x² + x + 1 by 2x + 1.
The first quotient term is x/2.
Subtract (x/2)(2x + 1), leaving x/2 + 1. Subtract (1/4)(2x + 1), leaving 3/4. Q = x/2 + 1/4, R = 3/4. The remainder fraction is (3/4)/(2x + 1), not simply 3/4.
Divide 3x + 2 by x² + 4. Explain your stopping decision.
Compare the degrees before doing any subtraction.
Q = 0 and R = 3x + 2. The numerator degree is already smaller than the divisor degree. The identity is 3x + 2 = (x² + 4)0 + (3x + 2).
x³ + kx + 4 leaves remainder 10 when divided by x − 2. Find k.
Evaluate the numerator at 2.
N(2) = 8 + 2k + 4 = 10 gives k = −1. With that value, division gives Q = x² + 2x + 3 and R = 10.
A student writes x³ = (x² + 1)x − x, so Q = x and R = −x. Another writes x³ = (x² + 1)(x − 1) + x² − x + 1. Which is the completed division?
Both equalities can be true; inspect the remainder degrees.
The first has remainder degree 1, smaller than divisor degree 2, so it is complete. The second has remainder degree 2 and still needs one more subtraction. Dividing x² − x + 1 by x² + 1 adds 1 to the quotient and leaves −x, recovering the first answer.
Find the error: (−x² − 5x + 7) − (−x² − 2x) = −7x + 7.
Remove the second bracket by changing both signs.
The left side is −x² − 5x + 7 + x² + 2x = −3x + 7. The error was treating the subtracted −2x as another −2x.
08 / Recap
Section 1 of 8 · Quotient and remainder