Polynomial division and the factor theorem

Division gives an identity

Dividing polynomial P(x) by x − a gives

where Q is the quotient polynomial and R is a constant remainder. The remainder is constant because its degree must be less than that of the linear divisor. This is a polynomial identity, valid at every real input; it does not contain a division by x − a.

Substituting x = a removes the product and gives P(a) = R, the remainder theorem. In particular, x − a is a factor exactly when P(a) = 0. Evaluating P(a) can test a factor or find a remainder without carrying out the division, but it does not by itself give the quotient.

Long division keeps every degree aligned

Divide S(x) = x³-5x + 6 by x − 2. Write the dividend as x³ + 0x² − 5x + 6, preserving the missing x² column. At each step, choose a quotient term to cancel the highest remaining power, multiply the whole divisor, then subtract the whole product.

The first quotient term is x², since x³/x = x². Subtract x²(x − 2):

The next term is 2x. Subtract 2x(x − 2):

The final quotient term is −1. Subtract −1(x − 2):

The constant remainder has lower degree than the divisor, so stop. Assemble the identity:

Expanding the right side restores x³ + 0x² − 5x + 6. Also S(2) = 8 − 10 + 6 = 4, independently checking the remainder. The non-zero remainder shows that x − 2 is not a factor.

If the task asks for the rational quotient, write

The polynomial identity remains valid at 2, while this rational expression is undefined there. Keep that distinction when using division before partial fractions.

For a divisor ax + b with a ≠ 0, the remainder is P(−b/a), because that input makes the divisor zero. The remainder’s degree is always less than the divisor’s degree.

For a higher-degree divisor the remainder need not be constant. For example, x⁴ + x + 1 = (x² + 1)(x²-1) + (x + 2). Its remainder x + 2 has degree one, below the divisor’s degree two. A linear-divisor remainder shortcut must not be applied as though every divisor had one real zero and a constant remainder.

A zero remainder starts a full factorisation

For P(x) = x³-2x²-5x + 6, P(1) = 1 − 2 − 5 + 6 = 0, so x − 1 is a factor. Dividing gives quotient x² − x − 6. Factor that quotient too:

The real solutions of P(x) = 0 are 1, 3 and −2. Each makes an original factor zero, and expansion verifies the complete identity. Finding one factor alone would not yet have found every solution.

Cubic identities also come from expansion

The quadratic factor in a sum of cubes has a negative middle term, whereas the one in a difference of cubes has a positive middle term. Expanding checks the cancellation; changing every sign to match the first factor does not give the identity.

For x³-8, use (x-2)(x² + 2x + 4). The quadratic has discriminant −12, so x = 2 is the only real root. Similarly, x³ + 8 = (x + 2)(x²-2x + 4) has only the real root −2. Degree three does not mean three distinct real solutions.

A factorisation has a domain context

A real-number equation may finish with a quadratic factor having negative discriminant. State that it supplies no further real roots. Do not invent real solutions merely to match the polynomial’s degree.