01 · A quadratic
Divide x² + 5x + 6 by x + 2.
Hint
The first quotient term is x.
Worked solution
x² + 5x + 6 − x(x + 2) = 3x + 6
3x + 6 = 3(x + 2)
Quotient x + 3; remainder 0.
Understand · explore · practise
Polynomial division
Divide polynomials by linear factors using selectable worked steps. Handle missing powers, non-unit divisors and remainders, with full solutions to practice questions.
Before you startIndex laws and expanding brackets
01 / Polynomials
Polynomial powers are non-negative integers.
A polynomial in x is a finite sum of terms axⁿ, where n is a non-negative integer and each coefficient a is constant. Constants are allowed: x⁰ = 1. Coefficients can be negative, fractions or irrational numbers.
3x⁴ − √2x + 7 is a polynomial.
1/x, √x and x⁻² are not polynomials in x.
The degree of a non-zero polynomial is its highest power with a non-zero coefficient. Write the terms in descending powers. Insert zero coefficients for any missing powers when dividing.
02 / Quotient and remainder
Multiplying back must recover the original polynomial.
Dividing f(x) by a non-zero polynomial d(x) gives a quotient q(x) and remainder r(x):
f(x) = d(x)q(x) + r(x)
The remainder is zero or has degree less than the divisor. For a linear divisor, the remainder is a constant. Remainder zero means the divisor is a factor.
This polynomial identity holds for every x, including roots of d. The fraction f/d = q + r/d only makes sense where d ≠ 0.
03 / Worked steps
Divide, multiply, subtract — then repeat.
Divide the leading term of what remains by the leading term of the divisor. Add that result to the quotient. Multiply the whole divisor by the new quotient term and subtract the whole product.
For (x³ − 2x² − 5x + 6) ÷ (x − 3), the quotient terms are x², then x, then −2. Select any stage in the model. Every line remains available, so you can compare how one subtraction leads to the next.
Stop when the remaining polynomial has lower degree than the divisor. Here it is zero.
Divide by x − 3Explore at your pace
Start
x³ − 2x² − 5x + 6
Quotient so far: 0. Nothing has been subtracted.
1 · Divide x³ by x
x³ − 2x² − 5x + 6
− (x³ − 3x²)
= x² − 5x + 6
New quotient term x²; subtract x²(x − 3).
2 · Divide x² by x
x² − 5x + 6
− (x² − 3x)
= −2x + 6
New quotient term x; subtract x(x − 3).
3 · Divide −2x by x
−2x + 6 − (−2x + 6) = 0
New quotient term −2; subtract −2(x − 3).
Quotient so far: 0. Remaining: x³ − 2x² − 5x + 6. All stages stay visible.
Watch the quotient build one term at a time
Pause, replay or seek freely. The notes explain the same idea and stay in view.
04 / Missing powers
Leave a place for every power.
Write x⁴ − 81 as x⁴ + 0x³ + 0x² + 0x − 81 before dividing by x − 3. Each new product lines up with the correct power.
Divide a quarticWorked example
x⁴ − (x⁴ − 3x³)
= 3x³
The first quotient term is x³; bring down the remaining zero terms and −81.
3x³ − (3x³ − 9x²)
= 9x²
The next quotient term is 3x².
9x² − (9x² − 27x)
= 27x
The next quotient term is 9x.
27x − 81 − (27x − 81) = 0
The final term is 27.
x⁴ − 81 = (x − 3)
× (x³ + 3x² + 9x + 27)
Multiply back to check every coefficient.
05 / Other linear divisors
Divide by the actual leading coefficient.
For divisor 2x − 1, each new quotient term comes from dividing by 2x. Subtract carefully: removing a negative term adds its opposite.
A divisor beginning with 2xWorked example
4x⁴ − 9x² + 2
÷ (2x − 1)
Include 0x³ and 0x.
First term: 2x³
Remaining: 2x³ − 9x² + 2
Subtract 4x⁴ − 2x³.
Next term: x²
Remaining: −8x² + 2
Subtract 2x³ − x².
Next term: −4x
Remaining: −4x + 2
Subtract −8x² + 4x.
Final term: −2
Remaining: 0
Subtract −4x + 2. Quotient: 2x³ + x² − 4x − 2.
06 / Non-zero remainders
A constant remainder stays outside the product.
Changing the constant in the main example from 6 to 8 leaves remainder 2:
x³ − 2x² − 5x + 8
= (x − 3)(x² + x − 2) + 2
Do not put the 2 into the quotient. Dividing both sides by x − 3 gives:
(x³ − 2x² − 5x + 8)/(x − 3)
= x² + x − 2 + 2/(x − 3)
x ≠ 3
At x = 3, the polynomial identity gives f(3) = 2. This is the basis of the remainder theorem in the next lesson.
07 / Your turn
Check with divisor × quotient + remainder.
Write the full identity first. If you also give a rational form, state where its denominator is non-zero.
02 · A cubic
Divide x³ − 4x² + x + 6 by x − 2.
Hint
Begin with x², then −2x.
Worked solution
After subtracting x²(x − 2):
−2x² + x + 6
After subtracting −2x(x − 2):
−3x + 6
After subtracting −3(x − 2): 0
Quotient x² − 2x − 3.
03 · Missing powers
Divide x³ + 8 by x + 2.
Hint
Write the dividend as x³ + 0x² + 0x + 8.
Worked solution
Subtract x²(x + 2): −2x² + 8
Subtract −2x(x + 2): 4x + 8
Subtract 4(x + 2): 0
Quotient x² − 2x + 4.
04 · Remainder
Divide x³ + 2x² − x + 5 by x + 1.
Hint
The quotient terms begin x², x.
Worked solution
Remaining after x²(x + 1): x² − x + 5
After x(x + 1): −2x + 5
After −2(x + 1): 7
f(x) = (x + 1)(x² + x − 2) + 7
05 · Leading coefficient
Divide 6x³ + x² − 7x − 2 by 3x + 2.
Hint
6x³ ÷ 3x = 2x².
Worked solution
After 2x²(3x + 2): −3x² − 7x − 2
After −x(3x + 2): −5x − 2
After −5(3x + 2)/3: 4/3
Quotient 2x² − x − 5/3;
remainder 4/3.
06 · Degree five
Divide 6x⁵ + x⁴ + 7x³ − x + 2 by 3x + 2.
Hint
Include 0x². The first term is 2x⁴.
Worked solution
After 2x⁴: −3x⁴ + 7x³ − x + 2
After −x³: 9x³ − x + 2
After 3x²: −6x² − x + 2
After −2x: 3x + 2
After 1: 0
Quotient 2x⁴ − x³ + 3x² − 2x + 1.
Each “after” means subtract the stated term multiplied by the divisor.
07 · Interpret an identity
If f(x) = (2x − 3)(x² + 1) − 4, state the quotient and remainder on division by 2x − 3, and find f(3/2).
Hint
The product vanishes at x = 3/2.
Worked solution
Quotient x² + 1, remainder −4, and f(3/2) = −4. The identity holds there even though f(x)/(2x − 3) is undefined there.
08 / Recap
Each subtraction removes the current leading term.
- Arrange descending powers, inserting zeros where needed.
- Divide the leading terms to find the next quotient term.
- Multiply the entire divisor and subtract the entire product.
- Stop when the remainder has lower degree than the divisor.
- Verify f = dq + r by multiplying back.
Section 1 of 8 · Polynomials