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Learn · Mathematics · Grade 10 · Algebra

Polynomials: operations, division and factoring

Degree and coefficients, adding and multiplying, long and synthetic division, the remainder and factor theorems, and the factoring identities.

High school Lesson 17 of 31 Interactive figure Practice

First, the rules

Sums of powers of x

A polynomial is a sum of terms a·xⁿ with whole-number powers. Two numbers describe it at a glance: the degree (highest power) and the leading coefficient.

Vocabulary
Example 1
P(x) = 2x³ − 5x + 7: degree 3, leading coefficient 2, constant term 7, coefficients 2, 0, −5, 7.
Degree of a product and a sum
Example 1
deg(P · Q) = deg P + deg Q, deg(P + Q) ≤ max(deg P, deg Q)
★ Values from the polynomial
Example 1
P(1) = sum of coefficients, P(0) = constant term
P(−1) gives the alternating sum.
Division
Example 2
P(x) = (x − a) · Q(x) + R — quotient Q, remainder R (a number when dividing by a linear).
★ Remainder theorem
Example 2
remainder of P(x) ÷ (x − a) = P(a)
★ Factor theorem
Example 3
(x − a) is a factor ⇔ P(a) = 0
★ Identities for factoring
Example 3
a² − b² = (a − b)(a + b)
a² ± 2ab + b² = (a ± b)²
a³ ± b³ = (a ± b)(a² ∓ ab + b²)

See it

Bend the curve

Move a, b, c and d. The x-intercepts are the roots, the dots on the curve are turning points, and the arms show the end behaviour of the leading term.

Step by step

Worked examples

1

Reading a polynomial

degree · coefficients

P(x) = (x² − 3)(2x + 1) − x³. Find the degree, the leading coefficient and the sum of the coefficients.

Ask yourself: You do not need to expand everything: the highest power comes from x² · 2x, and the sum of coefficients is P(1).

Show the solution
  1. Expand the highest terms: x² · 2x = 2x³, minus x³ gives x³: degree 3, leading coefficient 1.
  2. Sum of coefficients: P(1) = (1 − 3)(2 + 1) − 1 = −6 − 1 = −7.
  3. Full expansion if you want it: 2x³ + x² − 6x − 3 − x³ = x³ + x² − 6x − 3. Check: 1 + 1 − 6 − 3 = −7. ✓

Answer degree 3, leading coefficient 1, coefficient sum −7

2

Remainders without dividing

remainder theorem

Find the remainder when P(x) = x³ − 4x² + x + 6 is divided by (a) x − 2, (b) x + 1. Then find the value of m for which x³ + mx − 2 leaves remainder 4 on division by x − 1.

Ask yourself: Division by x − a leaves P(a). For x + 1, a = −1.

Show the solution
  1. (a) P(2) = 8 − 16 + 2 + 6 = 0 — remainder 0, so x − 2 is a factor.
  2. (b) P(−1) = −1 − 4 − 1 + 6 = 0 — also a factor. (So P(x) = (x − 2)(x + 1)(x − 3).)
  3. (c) 1 + m − 2 = 4 ⇒ m = 5.

Answer 0, 0, m = 5

3

Factoring with the identities

factor theorem · identities

Factor completely: (a) x⁴ − 16 (b) x³ − 27 (c) 2x³ − 2x² − 12x.

Ask yourself: Look for a common factor first, then a difference of squares or cubes, then a quadratic you can factor.

Show the solution
  1. (a) (x² − 4)(x² + 4) = (x − 2)(x + 2)(x² + 4) — x² + 4 does not factor over the reals.
  2. (b) Difference of cubes with b = 3: (x − 3)(x² + 3x + 9).
  3. (c) Common factor 2x: 2x(x² − x − 6) = 2x(x − 3)(x + 2).

Answer (x−2)(x+2)(x²+4) · (x−3)(x²+3x+9) · 2x(x−3)(x+2)

Practice: polynomials

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Degree, values, remainders, coefficient sums and the factor theorem. Your best score is kept in this browser only.

About this lesson

Polynomials: degree, leading coefficient and constant term, adding, subtracting and multiplying, division with the remainder, the remainder theorem P(a) and the factor theorem, the sum of coefficients P(1), and the factoring identities (difference of squares, perfect squares, sum and difference of cubes). Worked examples and practice.

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