Polynomials & Factoring | Math Learning Guide

📋 Prerequisites

Quadratics & Parabolas

Polynomials are the next step up from quadratics — they can have any degree. In competition math, you won't often need to solve a degree-5 polynomial directly, but you will need to factor them, multiply them, and use key theorems like the Factor Theorem and Remainder Theorem.

📚 Key Concepts

Polynomial Basics

A polynomial in $x$ has the form:

\[ a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0 \]

Degree: highest power of $x$ (the $n$ in $x^n$)
Leading coefficient: $a_n$ (coefficient of highest degree term)
Constant term: $a_0$ (no $x$)
Standard form: terms written from highest degree to lowest

Special names by degree: linear (1), quadratic (2), cubic (3), quartic (4), quintic (5)

Adding & Subtracting Polynomials

Combine like terms (same power of $x$).

Example: $ (3x^2 + 2x - 1) + (x^2 - 4x + 5) = 4x^2 - 2x + 4 $

For subtraction, distribute the negative sign first: every term in the second polynomial flips sign.

Multiplying Polynomials

$Distribute each term of the first polynomial through the second (FOIL is just the 2 \\times 2 case).$

$Monomial \\times polynomial: distribute:$ $ 2x(x^2 + 3x - 1) = 2x^3 + 6x^2 - 2x $
$Binomial \\times binomial: FOIL (First, Outer, Inner, Last)$
Larger polynomials: multiply each term by each term, then combine

Key Identities (Memorize These!)

These save you enormous time. Every competition has problems that become trivial if you recognize the pattern.

$ (a + b)^2 = a^2 + 2ab + b^2 $ — perfect square trinomial
$ (a - b)^2 = a^2 - 2ab + b^2 $ — perfect square trinomial
$ (a + b)(a - b) = a^2 - b^2 $ — difference of squares (most important!)
$ (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 $
$ (a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3 $
$ a^3 + b^3 = (a + b)(a^2 - ab + b^2) $ — sum of cubes
$ a^3 - b^3 = (a - b)(a^2 + ab + b^2) $ — difference of cubes

Factoring Strategy

Always follow this order when factoring:

GCF first: Factor out the greatest common factor from all terms
Count terms:
- $2 terms \\to check for difference of squares, sum/difference of cubes$
- $3 terms \\to trinomial factoring (AC method, perfect square)$
- $4 terms \\to try grouping (factor pairs, then factor out common binomial)$
Factor completely: Check if any factor can be factored further

Remainder Theorem & Factor Theorem

Remainder Theorem: When polynomial $P(x)$ is divided by $(x - a)$, the remainder equals $P(a)$.

Factor Theorem: $(x - a)$ is a factor of $P(x)$ if and only if $P(a) = 0$.

These are huge for competition problems. To test if $(x - a)$ divides a polynomial, just plug in $x = a$ and see if you get zero.

Rational Root Theorem: Any rational root $\frac{p}{q}$ of $P(x) = 0$ has $p$ dividing the constant term and $q$ dividing the leading coefficient.

⚠️ Common Mistake: $ (a + b)^2 \neq a^2 + b^2 $. Don't forget the middle term $2ab$! This is the most common algebra error in competition math. The difference of squares $a^2 - b^2$ factors to $(a+b)(a-b)$, but $a^2 + b^2$ does NOT factor over the reals.

💡 Key Insight: Difference of squares is the single most useful factoring pattern. Always be on the lookout for it, even when it's disguised. $x^4 - 16$ is $(x^2)^2 - 4^2$, which factors as $(x^2 + 4)(x^2 - 4)$ — and then $x^2 - 4$ factors again! Factoring completely means keep going until you can't factor anymore.

✏️ Example Problems

📝 Example 1 (Difference of squares)

Factor completely: $ 16x^2 - 81 $
What is the sum of the constant terms in the two binomial factors?
(For $(ax + b)(cx + d)$, add $b$ and $d$)

Recognize: This is a difference of squares: $ a^2 - b^2 = (a+b)(a-b) $

Step 1: Identify $a$ and $b$.

$ 16x^2 = (4x)^2 $, so $ a = 4x $

$ 81 = 9^2 $, so $ b = 9 $

Step 2: Apply the formula.

\[ 16x^2 - 81 = (4x + 9)(4x - 9) \]

Step 3: The constant terms are +9 and -9.

Sum = $ 9 + (-9) = 0 $

Check (FOIL): $(4x+9)(4x-9) = 16x^2 - 36x + 36x - 81 = 16x^2 - 81$. ✓

Answer: 0

📝 Example 2 (Remainder Theorem)

When $ P(x) = x^3 - 4x^2 + 2x + 5 $ is divided by $(x - 3)$, what is the remainder?

Remainder Theorem: Remainder = $ P(3) $

Plug in $x = 3$:

\[ P(3) = 3^3 - 4(3)^2 + 2(3) + 5 \\ = 27 - 4(9) + 6 + 5 \\ = 27 - 36 + 6 + 5 \\ = -9 + 11 \\ = 2 \]

The remainder is 2.

Note: Since the remainder is not 0, $(x - 3)$ is NOT a factor of $P(x)$.

Answer: 2

📝 Example 3 (Factor by grouping)

Factor: $ x^3 + 3x^2 + 2x + 6 $
How many linear factors with integer coefficients does it have?

Strategy: Factor by grouping $(4 terms \\to group in pairs).$

Step 1: Group first two and last two terms.

\[ (x^3 + 3x^2) + (2x + 6) \]

Step 2: Factor out GCF from each group.

$ x^2(x + 3) + 2(x + 3) $

Step 3: Factor out the common binomial $(x + 3)$.

\[ (x + 3)(x^2 + 2) \]

Step 4: Check if we can factor further.

$(x + 3)$ is already linear. $x^2 + 2$ is a sum of squares — doesn't factor over integers.

So there are 2 factors with integer coefficients: one linear ($x + 3$) and one quadratic ($x^2 + 2$).

The question asks for linear factors: only $(x + 3)$ is linear. Wait — let me re-read.

"How many linear factors with integer coefficients?" $\\to Only$ $(x + 3)$ is linear. That would be 1.

Hmm, let me reconsider. The answer should be 2 total factors (one linear, one quadratic). But the question specifically says "linear factors."

Correction: Only $(x + 3)$ is linear. $x^2 + 2$ is quadratic and doesn't factor further over integers.

Number of linear factors with integer coefficients = 1

Answer: 1