Polynomial Theorems and Roots

🌳 Polynomial Theorems and Roots

Part 1 of 5 — Roots, Factors, and Zeros

Topics in This Part

🔑 Key Concept: A "root" (or zero) of a polynomial is a value of $x$ that makes it equal $0$. Every theorem in this lesson is just a faster way to find those roots. Part 1 builds the vocabulary the theorems rely on.

What Counts as a Polynomial?

A polynomial in $x$ is a sum of terms of the form $a x^n$, where every exponent $n$ is a whole number ($0, 1, 2, 3, \dots$):

$P(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0$

• The degree is the highest exponent, $n$.
• The leading coefficient is $a_n$ (the number on the highest-power term).
• The constant term is $a_0$ (the term with no $x$).

Polynomial or Not?

💡 Why degree matters: A polynomial of degree $n$ can have at most $n$ roots. That single fact controls how many solutions you go looking for.

Concept Check 🎯

The Roots–Factors–Intercepts Connection

These three ideas are the same fact wearing three hats:

$\underbrace{P(c) = 0}_{c \text{ is a root}} \;\Longleftrightarrow\; \underbrace{(x - c) \text{ is a factor}}_{\text{divides evenly}} \;\Longleftrightarrow\; \underbrace{(c, 0) \text{ is an } x\text{-intercept}}_{\text{on the graph}}$

Reading It Off a Factored Polynomial

For $P(x) = (x - 2)(x + 3)(x - 5)$:

⚠️ Watch the sign. The factor $(x + 3)$ gives the root $x = -3$, not $+3$. A factor $(x - c)$ produces the root $x = c$, so flip the sign you see inside the parentheses.

Read the Roots 🧮

Find the roots of each factored polynomial. Enter them from smallest to largest.

1) $P(x) = (x - 4)(x + 1)$ → roots $= \,?$ and $\,?$ 2) $P(x) = (x + 6)(x - 2)(x)$ → roots $= \,?$, $\,?$, and $\,?$

Counting Roots by Degree

A degree-$n$ polynomial has at most $n$ real roots — equivalently, its graph crosses or touches the $x$-axis at most $n$ times.

A degree-$3$ polynomial might have $3$, $2$, or even $1$ real root — but never $4$. This ceiling tells you when you're done hunting for roots.

🔑 Big picture: The rest of this lesson is a toolkit for actually finding those roots — the Remainder Theorem (Part 2), the Factor Theorem (Part 2), the Rational Root Theorem (Part 3), and the Fundamental Theorem of Algebra (Part 4).

Vocabulary & Degree 🔽

Pin down the language before the theorems arrive.

🌳 Polynomial Theorems and Roots

Part 2 of 5 — The Remainder & Factor Theorems

🔑 The Idea: You don't have to do long division to find the remainder when you divide by $(x - c)$. Just plug in $c$. And if that remainder is $0$, you've found a root and a factor at the same time.

The Remainder Theorem

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

So to find the remainder, you evaluate instead of dividing.

Worked Example: divide $P(x) = 2x^3 - 3x^2 + 4x - 5$ by $(x - 2)$

Substitute $x = 2$:

$P(2) = 2(2)^3 - 3(2)^2 + 4(2) - 5$ $= 2(8) - 3(4) + 8 - 5 = 16 - 12 + 8 - 5 = 7$

So the remainder is $\boxed{7}$ — no long division required.

💡 Why it works: Division writes $P(x) = (x - c)\,Q(x) + R$. Setting $x = c$ kills the $(x - c)$ part, leaving $P(c) = R$.

Concept Check 🎯

A Tidy Way to Evaluate

When the numbers get large, substitute carefully one power at a time — or use synthetic substitution (the same row of arithmetic as synthetic division). Either way, the answer you get is the remainder.

💡 Spot the shortcut: if you only need the remainder, never set up long division. One substitution does it. And if that single number turns out to be $0$, you've just discovered a factor for free.

Find the Remainder 🧮

Use the Remainder Theorem (evaluate — don't divide).

1) Remainder of $\dfrac{x^3 + 4x^2 - 7x + 2}{x - 1}$ $= \,?$ 2) Remainder of $\dfrac{2x^4 - x^3 + 3x - 1}{x + 1}$ $= \,?$

The Factor Theorem

The Factor Theorem is the special case where the remainder is zero:

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

In other words: factor $\Longleftrightarrow$ zero remainder $\Longleftrightarrow$ root.

Worked Example: Is $(x + 2)$ a factor of $P(x) = x^3 + 3x^2 - 4$?

Here $(x + 2) = (x - (-2))$, so test $c = -2$:

$P(-2) = (-2)^3 + 3(-2)^2 - 4 = -8 + 12 - 4 = 0$

Since $P(-2) = 0$, yes — $(x + 2)$ is a factor, and $x = -2$ is a root. ✓

⚠️ Don't confuse the two theorems. The Remainder Theorem gives you any remainder. The Factor Theorem only fires when that remainder is exactly $0$.

Factor or Not? 🔽

For each, decide whether the given binomial is a factor by checking the right value.

🌳 Polynomial Theorems and Roots

Part 3 of 5 — The Rational Root Theorem

🔑 Why it's powerful: The Factor Theorem tells you how to test a candidate root, but not which numbers to try. The Rational Root Theorem hands you the complete shortlist of possible rational roots — so you test a few numbers instead of infinitely many.

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.

So every possible rational root has the form:

$\frac{p}{q} = \frac{\text{factor of constant term}}{\text{factor of leading coefficient}}$

Building the Candidate List for $P(x) = 2x^3 + x^2 - 13x + 6$

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

All candidates $\dfrac{p}{q}$:

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

💡 The theorem doesn't promise a rational root exists — it only says that if one does, it's on this list. You still have to test them (use the Factor Theorem: plug in and look for $0$).

Concept Check 🎯

Finding a Root, Then Factoring

Once you find one root with the candidate list, divide it out (synthetic or long division) to shrink the polynomial, then finish factoring.

Worked Example: solve $P(x) = 2x^3 + x^2 - 13x + 6 = 0$

Step 1 — Test candidates. Try $x = 2$: $P(2) = 2(8) + 4 - 26 + 6 = 16 + 4 - 26 + 6 = 0 \;\checkmark$ So $(x - 2)$ is a factor.

Step 2 — Divide it out. Dividing $2x^3 + x^2 - 13x + 6$ by $(x - 2)$ gives the quotient $2x^2 + 5x - 3$: $P(x) = (x - 2)(2x^2 + 5x - 3)$

Step 3 — Factor the quadratic. $2x^2 + 5x - 3 = (2x - 1)(x + 3)$: $P(x) = (x - 2)(2x - 1)(x + 3)$

Step 4 — Read off the roots: $x = 2,\; x = \tfrac{1}{2},\; x = -3$. ✓

⚠️ Notice $x = \tfrac{1}{2}$ comes from the factor $(2x - 1)$: set $2x - 1 = 0 \Rightarrow x = \tfrac{1}{2}$. Non-integer roots show up exactly because the leading coefficient wasn't $1$.

Test the Candidates 🧮

Use the Rational Root Theorem + Factor Theorem to find an actual root.

1) A candidate for $P(x) = x^3 - 6x^2 + 11x - 6$ is $x = 1$. Compute $P(1) = \,?$ 2) The three roots of $P(x) = x^3 - 6x^2 + 11x - 6$ are $1$, $2$, and what third value? Root $= \,?$

Putting the Workflow Together

Every "find all the roots" problem follows the same four-beat rhythm:

⚠️ Don't skip straight to dividing. You need a confirmed root first, or there's nothing valid to divide by.

Order the Strategy 🔽

You want to find all roots of a cubic with integer coefficients. Put the steps in order.

🌳 Polynomial Theorems and Roots

Part 4 of 5 — Fundamental Theorem, Multiplicity & Complex Roots

🔑 The Guarantee: The Fundamental Theorem of Algebra promises that a degree-$n$ polynomial has exactly $n$ roots — as long as you count complex roots and repeated roots correctly. This is what tells you when your hunt is complete.

The Fundamental Theorem of Algebra

Fundamental Theorem of Algebra: A polynomial of degree $n \ge 1$ has exactly $n$ roots in the complex numbers, counted with multiplicity.

Two ideas make the count come out exactly right:

1. Multiplicity — a repeated factor counts as many times as it appears. $P(x) = (x - 1)^2(x + 2)$ Degree $3$. Roots: $x = 1$ (multiplicity $2$) and $x = -2$ (multiplicity $1$). That's $2 + 1 = 3$ roots. ✓

2. Complex roots fill the gap. If a polynomial has fewer real roots than its degree, the missing ones are complex (involve $i = \sqrt{-1}$).

💡 Multiplicity and the graph: an odd multiplicity means the graph crosses the $x$-axis at that root; an even multiplicity means it just touches and turns around (a bounce).

Concept Check 🎯

The Complex Conjugate Root Theorem

Conjugate Root Theorem: If a polynomial has real coefficients and $a + bi$ is a root, then its conjugate $a - bi$ is also a root.

Complex roots always travel in pairs. So a real-coefficient polynomial always has an even number of complex roots.

Worked Example: a cubic with root $2 + i$

If $P(x)$ has real coefficients and one root is $2 + i$, then $2 - i$ must also be a root. With one more real root, say $x = 3$, the polynomial is:

$P(x) = (x - 3)\big(x - (2 + i)\big)\big(x - (2 - i)\big)$

The conjugate pair multiplies to a real quadratic: $\big(x - (2+i)\big)\big(x - (2-i)\big) = (x - 2)^2 - i^2 = x^2 - 4x + 4 + 1 = x^2 - 4x + 5$

$P(x) = (x - 3)(x^2 - 4x + 5) = x^3 - 7x^2 + 17x - 15$

⚠️ The conjugate rule needs real coefficients. If the polynomial has complex coefficients, roots need not come in conjugate pairs.

Complex Roots Come in Pairs 🧮

1) A polynomial with real coefficients has a root $5 - 2i$. Its conjugate root has real part $\,?$ and imaginary part $\,?$ (enter the coefficient of $i$, including its sign). So the conjugate is $5 + \,?\,i$ → enter the two answers: real part, then the $i$-coefficient. 2) A degree-$4$ polynomial with real coefficients has exactly $2$ real roots. How many complex (non-real) roots must it have? $\,?$

Building a Polynomial From Its Roots

Run the Factor Theorem in reverse: each root $c$ becomes a factor $(x - c)$, then multiply out.

Worked Example: a polynomial with roots $2, -3,$ and $1$

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

Multiply the first two: $(x - 2)(x + 3) = x^2 + x - 6$. Then times $(x - 1)$:

$P(x) = (x^2 + x - 6)(x - 1) = x^3 - 7x + 6$

💡 Quick checks for $x^3 + bx^2 + cx + d$ with roots $r_1, r_2, r_3$:

• Sum of roots $= -b$. Here $2 + (-3) + 1 = 0 = -b$, so $b = 0$ ✓ (no $x^2$ term).
• Product of roots $= -d$. Here $2 \cdot (-3) \cdot 1 = -6 = -d$, so $d = 6$ ✓.

Roots ⇄ Factors 🔽

🌳 Polynomial Theorems and Roots

Part 5 of 5 — Mixed Practice & Mastery Check

You now have the full toolkit: the Remainder and Factor Theorems to test values, the Rational Root Theorem to build candidate lists, and the Fundamental Theorem of Algebra (with multiplicity and conjugate pairs) to know when you're done. Let's put it together.

Quick Reference

⚠️ Remember: match the divisor to $(x - c)$ before reading off $c$ (so $(x + 5)$ means $c = -5$); count multiplicity and complex roots when checking you've found them all.

Mixed Practice 🎯

A Fast Self-Check: Sum & Product of Roots

For a monic cubic $x^3 + bx^2 + cx + d$ with roots $r_1, r_2, r_3$:

$r_1 + r_2 + r_3 = -b \qquad\text{and}\qquad r_1 \cdot r_2 \cdot r_3 = -d$

These let you sanity-check your roots without fully multiplying out — or even back out a missing coefficient.

💡 Found roots $1, 2, 3$ for a cubic? Their sum is $6$, so you expect $b = -6$, and their product is $6$, so you expect $d = -6$. If your expanded polynomial disagrees, recheck a root.

Apply It 🧮

1) Write the cubic with roots $-1$, $2$, $4$ in expanded form $x^3 + bx^2 + cx + d$. Using "sum of roots $= -b$," find $b = \,?$ 2) For that same cubic, "product of roots $= -d$," so find $d = \,?$

Exit Quiz ✅

Answer all three to finish the lesson.