The Timpson Lyceum · College Algebra

Lecture 8 (College Algebra)

Overview

This lecture builds the core algebraic machinery for working with polynomials and rational expressions. It opens by extending square roots to monomials, then develops polynomial multiplication and long division by analogy with ordinary arithmetic. From there it catalogs the common products every algebra student should recognize on sight — difference of squares, perfect-square trinomials, and the cubic patterns — and shows how reading them backward gives factoring. The lecture also covers prime numbers, prime factorization, prime (irreducible) polynomials, Pascal's triangle for binomial expansion, and the full toolkit for simplifying, adding, and multiplying rational expressions. It closes with a first look at the imaginary unit and complex numbers.

Key Concepts

Square root of a monomial. For a monomial that is a perfect square, take the square root of the numerical coefficient and divide each variable's exponent by 2.

Example: $\sqrt{16x^2y^{10}} = 4xy^5$ , since $\sqrt{16}=4$ and the exponents $2,10$ halve to $1,5$ .
Recall the underlying identity: $\sqrt{x^2} = |x|$ . So $\sqrt{(-5)^2} = |-5| = 5$ .

Multiplying polynomials. Works exactly like long multiplication of integers: multiply every term of one polynomial by every term of the other, line up like terms, and add. Multiplying terms means multiplying coefficients and adding exponents.

Dividing polynomials (long division). Mirrors numerical long division. At each step, divide the leading term of the current remainder by the leading term of the divisor, multiply the divisor by that quotient term, subtract, and bring down the next term. A nonzero leftover is the remainder, written over the divisor.

Common products (memorize forward and backward):

Distributive: $a(x+y) = ax + ay$
Difference of squares: $(x+y)(x-y) = x^2 - y^2$
Perfect-square trinomials: $(a+b)^2 = a^2 + 2ab + b^2$ and $(a-b)^2 = a^2 - 2ab + b^2$
Product of linear binomials: $(x+a)(x+b) = x^2 + (a+b)x + ab$ and $(ax+b)(cx+d) = acx^2 + (ad+bc)x + bd$

Factoring is reading a common product right-to-left — rewriting a sum as a product. A difference of squares always factors: $x^2 - 81 = (x-9)(x+9)$ .

Primes and prime factorization.

An integer is prime if it is greater than 1 and its only positive integer factors are 1 and itself. So 1 is not prime; 2, 3, 5, 11, 29, 31 are prime.
Factors come in pairs (e.g. for 36: $1\cdot36, 2\cdot18, 3\cdot12, 4\cdot9, 6\cdot6$ ).
By the Fundamental Theorem of Arithmetic, every integer greater than 1 has a unique prime factorization, e.g. $12 = 2^2\cdot 3$ and $36 = 2^2\cdot 3^2$ . This is the cleanest way to reduce fractions: cancel shared prime powers.

Prime (irreducible) polynomials. A polynomial is prime if it cannot be factored further. A sum of squares $x^2 + y^2$ is prime; a difference of squares is not. A single term (monomial) is left whole, never "split." Expressions with non-integer exponents (e.g. $\sqrt{x}$ ) are not polynomials, so $x - y$ is prime even though $(\sqrt{x}+\sqrt{y})(\sqrt{x}-\sqrt{y})$ multiplies back to it.

Pascal's triangle and binomial expansion. The coefficients of $(x+y)^n$ are the entries of row $n$ :

\begin{aligned} &1 \\ &1\quad 1 \\ &1\quad 2\quad 1 \\ &1\quad 3\quad 3\quad 1 \\ &1\quad 4\quad 6\quad 4\quad 1 \\ &1\quad 5\quad 10\quad 10\quad 5\quad 1 \end{aligned}

As you read each term, the exponent on $x$ decreases from $n$ to $0$ while the exponent on $y$ increases from $0$ to $n$ . For $(x-y)^n$ the same coefficients apply, with signs alternating $+,-,+,-,\dots$

Sum and difference of cubes.

$a^3 + b^3 = (a+b)(a^2 - ab + b^2)$
$a^3 - b^3 = (a-b)(a^2 + ab + b^2)$

Higher even powers can be attacked as differences of squares first; e.g. $x^6 - y^6 = (x^3)^2 - (y^3)^2 = (x^3+y^3)(x^3-y^3)$ , then factor each cube.

Rational expressions.

Lowest terms: factor numerator and denominator and cancel common factors only — you may never cancel a single term out of a sum.
Adding: put all fractions over a common denominator (the least common multiple of the denominators), built by taking each distinct factor to the highest power in which it appears. Sign-flipped factors like $3-x$ and $x-3$ are the same factor up to $-1$ .
Dividing: multiply by the reciprocal of the divisor, then factor everything and cancel before expanding.

Imaginary unit and complex numbers. No real number squares to a negative, so define $i = \sqrt{-1}$ , giving $i^2 = -1$ . Powers cycle with period 4:

$i^1 = i,\quad i^2 = -1,\quad i^3 = -i,\quad i^4 = 1$ , then repeat.

A complex number has the form $a + bi$ where $a,b$ are real. Since any real number $a = a + 0i$ , the reals are a subset of the complex numbers $\mathbb{C}$ .

Worked Example

Divide $4x^3 - 8x^2 - 9x + 7$ by $2x - 3$ using polynomial long division.

At each step, divide the leading term of the current remainder by $2x$ , multiply back, and subtract.

\begin{aligned} \frac{4x^3 - 8x^2 - 9x + 7}{2x - 3} &= q(x) + \frac{r}{2x-3} \\ 4x^3 \div 2x &= 2x^2 \;\Rightarrow\; 2x^2(2x-3) = 4x^3 - 6x^2 \\ (4x^3 - 8x^2) - (4x^3 - 6x^2) &= -2x^2,\quad \text{bring down } -9x \\ -2x^2 \div 2x &= -x \;\Rightarrow\; -x(2x-3) = -2x^2 + 3x \\ (-2x^2 - 9x) - (-2x^2 + 3x) &= -12x,\quad \text{bring down } +7 \\ -12x \div 2x &= -6 \;\Rightarrow\; -6(2x-3) = -12x + 18 \\ (-12x + 7) - (-12x + 18) &= -11 \quad (\text{remainder}) \\ \frac{4x^3 - 8x^2 - 9x + 7}{2x - 3} &= 2x^2 - x - 6 - \frac{11}{2x-3} \end{aligned}

Summary

To square-root a perfect-square monomial, square-root the coefficient and halve each exponent; remember $\sqrt{x^2}=|x|$ .
Multiply polynomials term-by-term (add exponents); divide them by long division, mirroring integer arithmetic, with any leftover written as a remainder over the divisor.
Memorize the common products — difference of squares, perfect-square trinomials, and the cubic patterns — and recognize factoring as reading them backward.
Every integer above 1 has a unique prime factorization (Fundamental Theorem of Arithmetic); a sum of squares is prime, a difference of squares always factors.
Use Pascal's triangle for binomial expansions; for $(x-y)^n$ , alternate the signs.
Simplify rational expressions by factoring and canceling common factors only; add over the LCM denominator and divide by multiplying by the reciprocal.
Define $i=\sqrt{-1}$ so $i^2=-1$ ; powers of $i$ cycle with period 4, and every complex number $a+bi$ generalizes the reals.