Division and factorisation of polynomials
Theory
Dividing a polynomial \(P(x)\) by a linear factor \((x-a)\) produces a quotient and a remainder, and the remainder theorem says that remainder equals \(P(a)\). When \(P(a)=0\) the divisor is a factor (the factor theorem), which lets you factorise cubics and quartics fully and solve polynomial equations by the null factor law.
Polynomial long division writes a dividend as \(P(x)=D(x)\,Q(x)+R(x)\), where \(D(x)\) is the divisor, \(Q(x)\) the quotient and \(R(x)\) the remainder. The process stops once the remainder has a lower degree than the divisor, so dividing by a linear \((x-a)\) always leaves a constant remainder.
The remainder theorem gives that constant directly: the remainder when \(P(x)\) is divided by \((x-a)\) is \(P(a)\). There is no need to divide — just substitute. (For a divisor \((ax-b)\) the remainder is \(P\!\left(\dfrac{b}{a}\right)\).) This turns a long division into a single evaluation.
The factor theorem is the case where that remainder is zero: \((x-a)\) is a factor of \(P(x)\) exactly when \(P(a)=0\). To factorise a cubic, test small values \(a\) (the factors of the constant term) until \(P(a)=0\), divide out \((x-a)\) to leave a quadratic, then factorise that quadratic. Setting each factor to zero solves \(P(x)=0\).
The division identity — dividend equals divisor times quotient plus remainder:
The remainder theorem and, for a general linear divisor, its version for \((ax-b)\):
The factor theorem — a zero of \(P\) is a linear factor:
Standard factorisations of a sum and difference of two cubes:
How to factorise a cubic fully and solve \(P(x)=0\)
- Find one factor. Test values \(a\) taken from the factors of the constant term (over the leading coefficient) until \(P(a)=0\); then \((x-a)\) is a factor by the factor theorem.
- Divide it out. Use long (or synthetic) division to divide \(P(x)\) by \((x-a)\), giving a quotient \(Q(x)\) one degree lower — a quadratic for a cubic.
- Factorise the quotient. Factorise the quadratic \(Q(x)\) by inspection, or use the quadratic formula; combine to write \(P(x)\) as a product of linear factors.
- Solve. Apply the null factor law: set each factor equal to zero to read off every root of \(P(x)=0\).
| \(x^{3}\div x\) | \(=\) | \(x^{2}\), and \(x^{2}(x-3)=x^{3}-3x^{2}\) |
| remainder so far | \(=\) | \(x^{2}-4x\) |
| \(x^{2}\div x\) | \(=\) | \(x\), and \(x(x-3)=x^{2}-3x\) |
| remainder so far | \(=\) | \(-x+5\) |
| \(-x\div x\) | \(=\) | \(-1\), and \(-1(x-3)=-x+3\) |
| remainder | \(=\) | \(5-3=2\) |
| \(x^{3}-2x^{2}-4x+5\) | \(=\) | \((x-3)(x^{2}+x-1)+2\) |
| \(P(2)\) | \(=\) | \(2(8)-4(4)+3(2)-1\) |
| \(=\) | \(16-16+6-1=5\) |
| \((1)^{3}+a(1)^{2}-5(1)+8\) | \(=\) | \(2\) |
| \(a+4\) | \(=\) | \(2\) |
| \(a\) | \(=\) | \(-2\) |
| \(P(1)\) | \(=\) | \(1-2-1+2=0\) |
| \(x^{3}-2x^{2}-x+2\) | \(=\) | \((x-1)(x^{2}-x-2)\) |
| \(x^{2}-x-2\) | \(=\) | \((x-2)(x+1)\) |
| \(P(x)\) | \(=\) | \((x-1)(x-2)(x+1)\) |
| \(P(3)\) | \(=\) | \(2(27)-3(9)-11(3)+6\) |
| \(=\) | \(54-27-33+6=0\) |
| \(2x^{3}-3x^{2}-11x+6\) | \(=\) | \((x-3)(2x^{2}+3x-2)\) |
| \(2x^{2}+3x-2\) | \(=\) | \((2x-1)(x+2)\) |
| \(P(x)\) | \(=\) | \((x-3)(2x-1)(x+2)\) |
| \(x-3=0\) | \(\Rightarrow\) | \(x=3\) |
| \(2x-1=0\) | \(\Rightarrow\) | \(x=\dfrac{1}{2}\) |
| \(x+2=0\) | \(\Rightarrow\) | \(x=-2\) |
Common pitfalls
Frequently asked questions
What is the remainder theorem?
When \(P(x)\) is divided by \((x-a)\), the remainder equals \(P(a)\). You can find it by substituting \(x=a\), with no division needed; for a divisor \((ax-b)\) the remainder is \(P\!\left(\dfrac{b}{a}\right)\).
What is the factor theorem?
It is the case where the remainder is zero: \((x-a)\) is a factor of \(P(x)\) if and only if \(P(a)=0\). Such an \(a\) is both a linear factor and a root of \(P(x)=0\).
How do you do polynomial long division?
Divide the leading term of the dividend by the leading term of the divisor, multiply the divisor by that, subtract, bring down the next term and repeat until the remainder has lower degree than the divisor. Use a zero placeholder for any missing power.
How do you factorise a cubic?
Use the factor theorem to find one root \(a\) (test factors of the constant term) giving a factor \((x-a)\), divide the cubic by \((x-a)\) to get a quadratic, then factorise that quadratic. The cubic is the product of the two.
How do you find the remainder without dividing?
Evaluate \(P(a)\). For example, the remainder when \(2x^{3}-4x^{2}+3x-1\) is divided by \((x-2)\) is \(P(2)=5\), found by substitution instead of long division.
How do you solve a polynomial equation by factorising?
Factorise fully, then use the null factor law: set each factor to zero. For example \((x-3)(2x-1)(x+2)=0\) gives \(x=3\), \(x=\dfrac{1}{2}\) or \(x=-2\).