The Timpson Lyceum · Intermediate Algebra (2023)

Lecture 13 (Intermediate Algebra) Factoring

Factoring (continued) – Lecture 13

Overview

This lecture introduces the special-product formulas that let you factor certain polynomials on sight, without hunting for "magic numbers" every time. You will learn the difference of squares, why a sum of squares is prime, perfect-square trinomials, and the sum and difference of cubes. The lecture also revisits factor-by-grouping to confirm that these special forms drop out of the general method automatically. Finally, it connects factoring to solving equations through the Zero Product Property, the key tool for solving polynomial equations. The recurring theme: always pull out the greatest common factor (GCF) first, then look for a special form.

Key Concepts

Difference of squares. A binomial of the form $a^2 - b^2$ always factors as the conjugate pair: $a^2 - b^2 = (a - b)(a + b)$ Recognize each term as a perfect square, then write "first minus second, times first plus second." Example pieces: $x^2 - 16 = (x-4)(x+4)$; $9a^2 - 25b^2 = (3a - 5b)(3a + 5b)$.

Sum of squares is prime. A sum of two squares $a^2 + b^2$ cannot be factored over the real numbers — there are no two numbers that multiply to $+1$ (the coefficient relationship) and add to $0$. So $x^2 + 36$ is already fully reduced (prime). Slogan: difference of squares we can factor; sum of squares we cannot.

Factor completely. Keep factoring until nothing factors further. For example: $a^4 - b^4 = (a^2 - b^2)(a^2 + b^2) = (a - b)(a + b)(a^2 + b^2)$ The $a^2 + b^2$ factor is a prime sum of squares and stops there.

Perfect-square trinomials. Squaring a binomial gives: $(a + b)^2 = a^2 + 2ab + b^2 \qquad (a - b)^2 = a^2 - 2ab + b^2$ You can spot one by checking that the middle term equals twice the product of the square roots of the outer terms. Either match the pattern directly or just factor by grouping — grouping always reproduces the perfect square, e.g. $4x^2 + 20xy + 25y^2 = (2x + 5y)^2$.

Sum and difference of cubes. These factor every time into a binomial times a trinomial: $a^3 + b^3 = (a + b)(a^2 - ab + b^2) \qquad a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ Memory aid: the binomial keeps the original sign; the trinomial's middle term takes the opposite sign of the binomial, and the squared terms are always positive ("Same, Opposite, Always Positive"). Recognizing perfect cubes (1, 8, 27, 64, 125, ...) is the only hard part. Examples: $n^3 - 27 = (n - 3)(n^2 + 3n + 9)$; $125p^3 + 8r^3 = (5p + 2r)(25p^2 - 10pr + 4r^2)$.

General strategy. Always factor out the GCF first, then identify a special form or factor by grouping. Watch for hidden perfect squares like $1$ ($1 = 1^2 = 1^3$), which is a perfect power of every degree.

Zero Product Property. If a product equals zero, at least one factor is zero: $ab = 0 \ \Longrightarrow\ a = 0 \ \text{ or } \ b = 0$ To solve a polynomial equation: move everything to one side so it equals $0$, factor completely, set each factor equal to zero, and solve each. A bare factor like $2x$ gives the solution $x = 0$; a constant factor like $3$ is never zero and contributes no solution.

Worked Example

Solve $4x^2 + x - 3 = 0$ by factoring.

Split the middle term using two numbers that multiply to $(4)(-3) = -12$ and add to $+1$ (namely $+4$ and $-3$), factor by grouping, then apply the Zero Product Property:

Summary

• $a^2 - b^2 = (a - b)(a + b)$; a sum of squares $a^2 + b^2$ is prime and cannot be factored.
• Factor completely — re-examine each factor, as with $a^4 - b^4 = (a-b)(a+b)(a^2+b^2)$.
• Perfect-square trinomials follow $(a \pm b)^2 = a^2 \pm 2ab + b^2$; grouping reproduces them automatically.
• Sum/difference of cubes: $a^3 \pm b^3 = (a \pm b)(a^2 \mp ab + b^2)$ — "Same, Opposite, Always Positive."
• Always pull out the GCF first, and remember $1$ is a perfect square and a perfect cube.
• The Zero Product Property turns a factored equation into separate linear equations: set each factor to $0$ and solve.