Polynomial Operations and Theorems

📐 Polynomial Operations and Theorems

Part 1 of 5 — Vocabulary & Arithmetic of Polynomials

Topics in This Part

🔑 Key Concept: A polynomial is a sum of terms of the form $a x^n$, where each exponent $n$ is a whole number ($0, 1, 2, \dots$) and each coefficient $a$ is a real number. Mastering how to add, subtract, and multiply them is the foundation for everything else in this lesson.

The Language of Polynomials

A polynomial in standard form lists its terms from the highest degree to the lowest:

$P(x) = 3x^4 - 5x^2 + 7x - 2$

Why "whole-number exponents" matters

These are not polynomials:

$\frac{5}{x} = 5x^{-1} \quad (\text{negative exponent}), \qquad \sqrt{x} = x^{1/2} \quad (\text{fractional exponent})$

💡 Quick test: If you see a variable in a denominator, under a radical, or as an exponent, it is not a polynomial.

Classify the Polynomial 🔽

Look at $\;P(x) = 6x^3 - 4x^5 + 9 - x\;$ and answer each part.

Adding & Subtracting: Combine Like Terms

Like terms have the same variable raised to the same power. You add or subtract only their coefficients.

Worked Example — Addition

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

Line up like terms:

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

Worked Example — Subtraction

⚠️ The #1 mistake: when subtracting, you must distribute the minus sign to every term in the second polynomial.

$(5x^2 - 2x + 7) - (3x^2 + x - 4)$

Distribute the $-$: $\;5x^2 - 2x + 7 - 3x^2 - x + 4$

Combine: $\;(5x^2 - 3x^2) + (-2x - x) + (7 + 4) = 2x^2 - 3x + 11$

Add & Subtract 🧮

Simplify each expression. For each, enter the coefficient of $x$ (the linear term) in the result. Include the sign.

1) $(4x^2 + 6x - 3) + (2x^2 - 5x + 8)$ → coefficient of $x = \,?$ 2) $(7x^2 - x + 2) - (3x^2 - 4x + 9)$ → coefficient of $x = \,?$

Multiplying Polynomials: Distribute Everything

Multiply each term of the first polynomial by each term of the second, then combine like terms.

Worked Example — Binomial × Trinomial

$(x + 3)(x^2 - 2x + 4)$

Distribute $x$, then $3$:

$x\cdot x^2 + x\cdot(-2x) + x\cdot 4 + 3\cdot x^2 + 3\cdot(-2x) + 3\cdot 4$

$= x^3 - 2x^2 + 4x + 3x^2 - 6x + 12$

Combine like terms:

$= x^3 + x^2 - 2x + 12$

🔑 Degree shortcut: When you multiply, the degrees add. A degree-$1$ times a degree-$2$ always gives a degree-$3$ product. Use this to check your answer.

Concept Check 🎯

📐 Polynomial Operations and Theorems

Part 2 of 5 — Special Products & Polynomial Division

🔑 Why this part matters: Special-product patterns let you multiply instantly, and polynomial division is the tool that unlocks the Remainder and Factor Theorems in Part 3.

Special Products Worth Memorizing

Worked Examples

$(x + 5)^2 = x^2 + 2(5)x + 25 = x^2 + 10x + 25$

$(3x - 2)^2 = 9x^2 - 2(3x)(2) + 4 = 9x^2 - 12x + 4$

$(x + 7)(x - 7) = x^2 - 49$

⚠️ Do not write $(a+b)^2 = a^2 + b^2$. That "freshman's dream" forgets the middle term $2ab$. Always include it.

Special Products 🧮

Expand each using a pattern. Enter the requested coefficient (with sign).

1) $(x + 6)^2$ → coefficient of $x = \,?$ 2) $(2x - 3)^2$ → coefficient of $x = \,?$ 3) $(5x + 4)(5x - 4)$ → constant term $= \,?$

Polynomial Long Division

Long division works just like with numbers: divide, multiply, subtract, bring down, repeat.

Worked Example: $(2x^2 + 5x - 3) \div (x + 3)$

Step by step:

1. $2x^2 \div x = 2x$. Multiply: $2x(x+3) = 2x^2 + 6x$. Subtract: $(2x^2 + 5x) - (2x^2 + 6x) = -x$.
2. Bring down $-3$: now $-x - 3$. Divide: $-x \div x = -1$. Multiply: $-1(x+3) = -x - 3$. Subtract: $0$.

$\frac{2x^2 + 5x - 3}{x + 3} = 2x - 1 \quad(\text{remainder } 0)$

🔑 Always insert placeholders. If a power is missing (say no $x$ term), write $0x$ so columns line up. Skipping it is the most common long-division error.

Concept Check 🎯

Synthetic Division (the fast shortcut)

When the divisor is linear of the form $(x - c)$, synthetic division is faster. Use the root $c$ (the value that makes the divisor zero).

Worked Example: $(x^3 - 4x^2 + 5x - 2) \div (x - 1)$

The divisor $(x-1)$ gives $c = 1$. Bring down, multiply by $1$, add — repeat:

Read the bottom row: quotient $x^2 - 3x + 2$, remainder $\mathbf{0}$.

$\frac{x^3 - 4x^2 + 5x - 2}{x - 1} = x^2 - 3x + 2$

💡 Sign rule: For divisor $(x - c)$ you use $+c$; for $(x + c)$ you use $-c$. Example: dividing by $(x + 2)$ uses $c = -2$.

Synthetic Division 🧮

Divide $\;x^3 - 4x^2 + 5x - 2\;$ by $\;(x - 2)\;$ using synthetic division ($c = 2$).

The quotient comes out as $\;x^2 + bx + 1\;$ with some remainder.

1) Enter $b$, the coefficient of $x$ in the quotient (include the sign). 2) The remainder $= \,?$

📐 Polynomial Operations and Theorems

Part 3 of 5 — The Remainder & Factor Theorems

🔑 The big idea: You can learn things about a polynomial without graphing it by plugging numbers in and by checking which divisions come out even. Two theorems make this precise.

The Remainder Theorem

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

In other words, evaluating $P$ at $c$ gives the same number as the remainder of the division. So you can skip the division entirely and just plug in.

Worked Example

Find the remainder when $P(x) = x^3 - 2x^2 + 4x - 5$ is divided by $(x - 2)$.

Instead of dividing, evaluate $P(2)$:

$P(2) = (2)^3 - 2(2)^2 + 4(2) - 5 = 8 - 8 + 8 - 5 = 3$

So the remainder is $\mathbf{3}$ — no long division required.

💡 For a divisor like $(x + 3)$, rewrite it as $(x - (-3))$ and evaluate $P(-3)$.

Use the Remainder Theorem 🧮

Let $\;P(x) = 2x^3 + x^2 - 7x + 4$.

1) Remainder when divided by $(x - 1)$: compute $P(1) = \,?$ 2) Remainder when divided by $(x + 2)$: compute $P(-2) = \,?$

The Factor Theorem

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

This is the Remainder Theorem taken to its punchline: a remainder of $0$ means the divisor divides evenly, which means it's a factor. And $P(c) = 0$ also means $c$ is a root (a zero) of the polynomial.

Three ideas that are all the same thing

Worked Example

Is $(x - 3)$ a factor of $P(x) = x^3 - 7x + 6$?

Check $P(3) = 27 - 21 + 6 = 12 \ne 0$. Since the remainder is $12$, $(x-3)$ is NOT a factor.

Now check $(x - 2)$: $P(2) = 8 - 14 + 6 = 0$. ✓ So $(x - 2)$ IS a factor, and $x = 2$ is a root.

Concept Check 🎯

Theorem Logic 🔽

Let $\;P(x) = x^3 - 4x^2 + x + 6$. Use the theorems (no graphing).

📐 Polynomial Operations and Theorems

Part 4 of 5 — Finding All the Roots

🔑 Big payoff: Combine the Rational Root Theorem (which roots to try), synthetic division (to test and reduce), and factoring (to finish) into one repeatable strategy for fully factoring a polynomial.

The Rational Root Theorem

🔑 Rational Root Theorem: If a polynomial with integer coefficients has a rational root $\dfrac{p}{q}$ (in lowest terms), then $p$ divides the constant term and $q$ divides the leading coefficient.

It hands you a finite list of candidates to test — you no longer have to guess blindly.

Worked Example

List the possible rational roots of $P(x) = 2x^3 - 3x^2 - 8x + 12$.

• Constant term $= 12$ → factors $p$: $\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12$
• Leading coefficient $= 2$ → factors $q$: $\pm 1, \pm 2$

Possible roots $\dfrac{p}{q}$:

$\pm 1,\; \pm 2,\; \pm 3,\; \pm 4,\; \pm 6,\; \pm 12,\; \pm \tfrac{1}{2},\; \pm \tfrac{3}{2}$

💡 The theorem doesn't say a root exists — only that if a rational root exists, it's on this list. Test candidates with the Remainder Theorem (plug in) or synthetic division.

Concept Check 🎯

Strategy: Fully Factor a Cubic

Goal: factor $P(x) = x^3 - 4x^2 + x + 6$ completely.

Step 1 — Candidates (RRT). Constant $6$, leading coefficient $1$: try $\pm 1, \pm 2, \pm 3, \pm 6$.

Step 2 — Find one root. Test $x = -1$: $P(-1) = -1 - 4 - 1 + 6 = 0$ ✓. So $(x + 1)$ is a factor.

Step 3 — Divide it out (synthetic, $c = -1$).

Quotient: $x^2 - 5x + 6$.

Step 4 — Factor the quadratic. $x^2 - 5x + 6 = (x - 2)(x - 3)$.

Result:

$P(x) = (x + 1)(x - 2)(x - 3)$

The roots are $x = -1,\; 2,\; 3$.

✅ Check: the product of the roots and the structure match — a degree-$3$ polynomial has at most $3$ real roots, and here we found exactly $3$.

Finish the Factoring 🧮

You are factoring $\;P(x) = x^3 - 2x^2 - 5x + 6$. You already verified $P(1) = 0$, so $(x - 1)$ is a factor. Synthetic division by $c = 1$ gives the quotient $x^2 - x - 6$.

1) Factor $x^2 - x - 6 = (x - 3)(x + a)$. What is $a$? 2) The three roots of $P(x)$ are $1$, $3$, and what third value?

Order the Strategy 🔽

Put the "fully factor a polynomial" workflow in order.

📐 Polynomial Operations and Theorems

Part 5 of 5 — Roots, Multiplicity & Mastery Check

You can now operate on polynomials, divide them, and use the theorems to find roots. This final part ties in the Fundamental Theorem of Algebra and multiplicity, then a full mixed-mastery check.

How Many Roots? The Fundamental Theorem of Algebra

🔑 Fundamental Theorem of Algebra: A polynomial of degree $n$ (with $n \ge 1$) has exactly $n$ roots, when you count complex roots and multiplicity.

• Multiplicity = how many times a factor repeats. In $P(x) = (x - 2)^3(x + 1)$, the root $x = 2$ has multiplicity $3$ and $x = -1$ has multiplicity $1$ — that's $3 + 1 = 4$ roots for a degree-$4$ polynomial. ✓
• Complex roots (with imaginary parts) come in conjugate pairs when coefficients are real: if $2 + 3i$ is a root, so is $2 - 3i$.

Multiplicity & the graph

💡 A degree-$5$ polynomial with real coefficients must have at least one real root, because complex roots pair up and $5$ is odd — they can't all be complex.

Concept Check 🎯

Quick Reference

⚠️ Top traps: dropping the minus sign in subtraction; forgetting the middle term in $(a+b)^2$; omitting placeholder $0$ terms in division; using the wrong sign of $c$ for divisor $(x \pm c)$.

Mixed Mastery Drill 🧮

Bring the whole toolkit together.

1) $(3x^2 - x + 2) - (x^2 + 4x - 5)$ → coefficient of $x^2$ in the result $= \,?$ 2) Remainder when $P(x) = x^3 + 2x^2 - 5$ is divided by $(x - 2)$: compute $P(2) = \,?$ 3) $P(x) = (x - 1)^2(x + 6)(x - 4)$. Counting multiplicity, how many roots? $= \,?$

Exit Quiz ✅

Answer all three to finish the lesson.