Division Of Polynomials
Master polynomial division for Queensland Year 11 Mathematical Methods (QCAA). Dividing one polynomial by another splits it into a quotient plus a remainder.
You will learn to set out long division by a linear or quadratic divisor, apply the remainder theorem to find a remainder by substitution, and use the factor theorem to spot factors — supporting skills that help you write a cubic in factorised form.
Every question with a fully worked solution.
- Division Of Polynomials - Video - Division of polynomials Watch
Theory
In Year 11 Mathematical Methods (QCAA, Unit 1), dividing one polynomial by another gives a quotient and a remainder, written \(P(x)=A(x)\,Q(x)+R\). This page covers long division by a linear or simple quadratic divisor, the remainder theorem (the remainder on division by \(x-a\) is \(P(a)\)) and the factor theorem (\(x-a\) is a factor exactly when \(P(a)=0\)).
Polynomial long division works like long division of numbers: at each step you divide the leading term of what is left by the leading term of the divisor, multiply back, and subtract. You stop when the remainder has lower degree than the divisor.
The result is the division identity \(P(x)=A(x)\,Q(x)+R\), where \(A(x)\) is the divisor, \(Q(x)\) the quotient and \(R\) the remainder. An improper fraction can then be written as \(\dfrac{P(x)}{A(x)}=Q(x)+\dfrac{R}{A(x)}\).
The remainder theorem says the remainder when \(P(x)\) is divided by \((x-a)\) is just \(P(a)\). The factor theorem is the special case \(R=0\): \((x-a)\) is a factor of \(P(x)\) if and only if \(P(a)=0\).
The division identity for a divisor \(A(x)\):
As a quotient plus a proper fraction:
Remainder theorem and factor theorem (divisor \(x-a\)):
How to divide a polynomial by \((x-a)\) or a quadratic
- Set up: write \(P(x)\) with every power present (insert \(0\) coefficients for missing powers) and the divisor to the left.
- Divide, multiply, subtract: divide the leading terms, write the term on top, multiply it back through the divisor, subtract, and bring down the next term. Repeat.
- Stop and state: stop when the remainder has lower degree than the divisor, then write \(P(x)=A(x)\,Q(x)+R\). For a linear divisor you may instead just compute \(P(a)\).
First term of the quotient — divide, multiply, subtract:
| \(2x^3\div x\) | \(=\) | \(2x^2\) |
| \(2x^2(x-2)\) | \(=\) | \(2x^3-4x^2\) |
| \((2x^3-3x^2)-(2x^3-4x^2)\) | \(=\) | \(x^2\) |
Bring down \(+4x\), giving \(x^2+4x\).
Second term \(+x\):
| \(x^2\div x\) | \(=\) | \(x\) |
| \(x(x-2)\) | \(=\) | \(x^2-2x\) |
| \((x^2+4x)-(x^2-2x)\) | \(=\) | \(6x\) |
Bring down \(-5\), giving \(6x-5\).
Third term \(+6\):
| \(6x\div x\) | \(=\) | \(6\) |
| \(6(x-2)\) | \(=\) | \(6x-12\) |
| \((6x-5)-(6x-12)\) | \(=\) | \(7\) |
The remainder \(7\) is a constant, so we stop: \(2x^3-3x^2+4x-5=(x-2)(2x^2+x+6)+7\).
Quotient \(2x^2+x+6\), remainder \(7\).
First term of the quotient — divide, multiply, subtract:
| \(x^3\div x^2\) | \(=\) | \(x\) |
| \(x(x^2-x+1)\) | \(=\) | \(x^3-x^2+x\) |
| \((x^3+2x^2-3x)-(x^3-x^2+x)\) | \(=\) | \(3x^2-4x\) |
Bring down \(+4\), giving \(3x^2-4x+4\).
Second term \(+3\):
| \(3x^2\div x^2\) | \(=\) | \(3\) |
| \(3(x^2-x+1)\) | \(=\) | \(3x^2-3x+3\) |
| \((3x^2-4x+4)-(3x^2-3x+3)\) | \(=\) | \(-x+1\) |
The remainder \(-x+1\) has degree \(1\), lower than the divisor, so we stop: \(x^3+2x^2-3x+4=(x^2-x+1)(x+3)+(-x+1)\).
Quotient \(x+3\), remainder \(-x+1\).
By the remainder theorem, the remainder equals \(P(3)\) — no long division needed.
Substitute \(x=3\):
| \(P(3)\) | \(=\) | \((3)^3-4(3)^2+2(3)+7\) |
| \(=\) | \(27-36+6+7\) | |
| \(=\) | \(4\) |
The remainder is \(4\).
Factor theorem: \((x-2)\) a factor means \(P(2)=0\):
| \(P(2)\) | \(=\) | \((2)^3+a(2)^2-2-2\) |
| \(0\) | \(=\) | \(8+4a-4\) |
| \(0\) | \(=\) | \(4a+4\) |
| \(a\) | \(=\) | \(-1\) |
So \(P(x)=x^3-x^2-x-2\). Dividing by \((x-2)\) gives the quotient \(x^2+x+1\):
Check the quadratic factor's discriminant:
| \(\Delta\) | \(=\) | \((1)^2-4(1)(1)\) |
| \(=\) | \(-3\) |
Since \(\Delta<0\), \(x^2+x+1\) cannot be factorised over the reals, so it stays as one factor.
\(a=-1\) and \(P(x)=(x-2)(x^2+x+1)\).
Common pitfalls
Frequently asked questions
What is the remainder theorem?
The remainder when \(P(x)\) is divided by \((x-a)\) equals \(P(a)\). So to find a remainder on division by a linear, just substitute.
What is the factor theorem?
\((x-a)\) is a factor of \(P(x)\) exactly when \(P(a)=0\). It is the remainder theorem in the special case where the remainder is zero.
How do you do polynomial long division?
Divide the leading term of what is left by the leading term of the divisor, multiply that back through the divisor, subtract, and bring down the next term. Repeat until the remainder has lower degree than the divisor.
What is the division identity?
It is \(P(x)=A(x)\,Q(x)+R\): the polynomial equals the divisor times the quotient plus the remainder. It lets you check a division by expanding.
Do you always need long division to find a remainder?
No. For a linear divisor \((x-a)\) the remainder theorem gives the remainder as \(P(a)\), which is much faster. Long division is needed to find the quotient or to divide by a quadratic.
How big can the remainder be?
The remainder always has lower degree than the divisor. Dividing by a linear leaves a constant; dividing by a quadratic leaves at most a linear expression.