Algebraic form and polynomial function
A polynomial in x is a sum like p(x) = aₙxⁿ + … + a₁x + a₀. The numbers a₀, a₁, … are the coefficients. The biggest power with a non-zero coefficient is the degree. For x³ − 6x² + 11x − 6 the degree is 3 and the coefficients are 1, −6, 11, −6.
You can see it two ways. As an algebraic form it is just a written sum. As a polynomial function it takes a number and gives a number: p(2) = 8 − 24 + 22 − 6 = 0. Two polynomials are equal when all their coefficients are equal. Polynomials can be added and multiplied, and the result is again a polynomial: that is why they form a ring. Division does not always stay inside the ring, so we use division with remainder.
Division with remainder and Horner's scheme
For any polynomial p and any non-zero divisor d there are unique polynomials q (quotient) and r (remainder) with
p = d · q + r, and degree of r < degree of d.
Like long division with numbers: 17 = 5 × 3 + 2.
Horner's scheme is the fast way to divide by (x − a). Write the coefficients in a row. Bring down the first one. Then repeat: multiply the number you just wrote by a, add the next coefficient, write the result. The numbers are b₀ = a₃, b₁ = a₂ + a·b₀, b₂ = a₁ + a·b₁, and the last one is the remainder. The others are the coefficients of the quotient.
Bézout's theorem, gcd, lcm and irreducible factors
Bézout's (remainder) theorem: when p(x) is divided by (x − a), the remainder is p(a). So (x − a) divides p if and only if p(a) = 0. Then a is called a root (or zero) of p. You can see this in the 3D: the last purple bar equals p(a).
The gcd of two polynomials is the common divisor of the highest degree. Find it by repeated division with remainder (the Euclid method), exactly like gcd of numbers. The lcm is p·q ÷ gcd(p, q) (up to a constant). Example: x² − 1 = (x − 1)(x + 1) and x² − 3x + 2 = (x − 1)(x − 2), so gcd = x − 1 and lcm = (x − 1)(x + 1)(x − 2).
A polynomial is irreducible if it cannot be written as a product of two polynomials of smaller degree. Over the reals every irreducible polynomial has degree 1, or degree 2 with a negative discriminant (like x² + 1). Over the rationals, x² − 2 is irreducible but over the reals it splits as (x − √2)(x + √2).
Roots and Viète relations
If a polynomial has roots x₁, x₂, … then p(x) = aₙ(x − x₁)(x − x₂)… Multiplying out links the roots to the coefficients. These are the Viète relations.
For ax² + bx + c = 0: x₁ + x₂ = −b/a and x₁x₂ = c/a.
For ax³ + bx² + cx + d = 0: x₁ + x₂ + x₃ = −b/a, x₁x₂ + x₂x₃ + x₁x₃ = c/a, x₁x₂x₃ = −d/a.
Tip for whole-number roots: any whole-number root must divide the constant term. For x³ − 6x² + 11x − 6 try ±1, ±2, ±3, ±6. Test each with Horner. More on roots: Roots of polynomials and identities.
Special equations: binomial, reciprocal and biquadratic
Algebraic equation: p(x) = 0. Find a root by Horner, divide it out, and repeat on the smaller quotient.
Binomial equation: xⁿ = c. Example x⁴ = 16 gives x = ±2 over the reals.
Biquadratic: ax⁴ + bx² + c = 0. Put t = x², solve the quadratic in t, then x = ±√t. Example x⁴ − 5x² + 4 = 0: t = 1 or 4, so x = ±1, ±2.
Reciprocal equation: the coefficients read the same forwards and backwards, like x⁴ + 2x³ − 6x² + 2x + 1 = 0. Divide by x² and put t = x + 1/x. Then x² + 1/x² = t² − 2 and you get a quadratic in t.
Try it: find the roots yourself
In the 3D, go to the last step and slide a from 0 to 5. First guess which values will give remainder 0 (look at the constant term −6: whole-number roots must divide 6). Then check. You should find 1, 2 and 3. Add them: 6. That equals −b/a = 6, the Viète sum. Multiply them: 6. That equals −d/a = 6.
Key formulas and definitions
- p = d · q + r, degree r < degree d
- Bézout: remainder of p ÷ (x − a) is p(a)
- Horner: b₀ = a₃; bₖ = aₙ₋ₖ + a·bₖ₋₁; last b = remainder
- Viète (quadratic): x₁ + x₂ = −b/a, x₁x₂ = c/a
- Viète (cubic): Σx = −b/a, Σxx = c/a, xxx = −d/a
- lcm(p, q) = p·q / gcd(p, q)
Worked examples
1. Divide x³ − 6x² + 11x − 6 by x − 2 with Horner's scheme.
Coefficients 1, −6, 11, −6 and a = 2. Bring down 1. Next: 2×1 + (−6) = −4. Next: 2×(−4) + 11 = 3. Last: 2×3 + (−6) = 0. Quotient x² − 4x + 3, remainder 0, so (x − 2) is a factor.
2. Find the remainder when x³ + 2x² − 5x + 1 is divided by x − 2.
By Bézout the remainder is p(2) = 8 + 8 − 10 + 1 = 7.
3. For which m is x − 3 a factor of x³ − m x² + 2x + 3?
We need p(3) = 0: 27 − 9m + 6 + 3 = 36 − 9m = 0, so m = 4.
4. Solve x³ − 6x² + 11x − 6 = 0.
Try 1: Horner gives remainder 0 and quotient x² − 5x + 6. Factor: (x − 2)(x − 3). Roots are 1, 2, 3.
5. The roots of 2x² − 7x + 3 = 0 are α and β. Find α² + β².
α + β = 7/2, αβ = 3/2. α² + β² = (α + β)² − 2αβ = 49/4 − 3 = 37/4 = 9.25.
6. Solve x⁴ − 5x² + 4 = 0, and find gcd(x² − 1, x² − 3x + 2).
Put t = x²: t² − 5t + 4 = 0, so t = 1 or 4, giving x = ±1, ±2. For the gcd, x² − 1 = (x − 1)(x + 1) and x² − 3x + 2 = (x − 1)(x − 2), so the gcd is x − 1.
Common mistakes
- Forgetting zero coefficients in Horner's row. For x³ − 4x + 1 write 1, 0, −4, 1.
- Using the wrong sign: dividing by (x + 2) means a = −2, not 2.
- Thinking the remainder must have degree 0 always. It only has to be of smaller degree than the divisor.
- Mixing the signs in Viète's formulas: the sum of roots is −b/a, with a minus sign.