🎯⭐ INTERACTIVE LESSON

Fundamental Theorem of Algebra and Factoring

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Fundamental Theorem of Algebra and Factoring - Complete Interactive Lesson

Part 1: Zeros, Factors & the Connection Between Them

🧩 The Fundamental Theorem of Algebra & Factoring

Part 1 of 7 — Zeros, Factors & the Connection Between Them

Topics in This Part

Section
What Is a Polynomial Zero?
The Factor Theorem
Reading Zeros Off a Factored Polynomial
Going Backward: Factors from Zeros

🔑 Key Concept: A number $c$ is a zero of a polynomial $p(x)$ exactly when $(x - c)$ is a factor of $p(x)$ . Factoring and finding zeros are two views of the same idea — and that idea drives this entire lesson.

What Is a Polynomial Zero?

A polynomial is a sum of terms $a_n x^n + \dots + a_1 x + a_0$ with whole-number exponents. A (or ) of is any value that makes the polynomial equal :

The Factor Theorem

This is the engine of the whole lesson:

$\textbf{Factor Theorem: } \quad (x - c) \text{ is a factor of } p(x) \iff p(c) = 0$

The double arrow means it works :

Concept Check 🎯

Factors from Zeros 🧮

Each value below is a zero of some polynomial. Write the matching factor in the form $(x \,?\, k)$ — enter just the constant with its sign that goes inside the factor after the $x$ .

Example: zero $= 6$ → factor $(x - 6)$ , so you would enter .

Is It a Zero? 🔽

For $p(x) = x^2 - 5x + 6$ , decide whether each value is a zero by evaluating $p$ at it.

Why This Matters

Everything ahead is about turning a polynomial into a product of factors, because a factored polynomial displays its zeros:

$p(x) = a(x - r_1)(x - r_2)\cdots(x - r_n) \quad\Longrightarrow\quad \text{zeros } r_1, r_2, \dots, r_n$

Part 2: Core Factoring Techniques

🧩 The Fundamental Theorem of Algebra & Factoring

Part 2 of 7 — Core Factoring Techniques

🔑 Strategy: Before any fancy theorem, always try these in order: (1) pull out a GCF, (2) recognize a special pattern, (3) factor a trinomial, (4) try grouping. This part trains all four.

Step 1 — Greatest Common Factor (GCF)

Always pull out the GCF first. It simplifies everything that follows.

$6x^3 - 9x^2 = 3x^2(2x - 3)$

Part 3: The Rational Root Theorem

🧩 The Fundamental Theorem of Algebra & Factoring

Part 3 of 7 — The Rational Root Theorem

🔑 The Problem: A cubic like $2x^3 - 3x^2 - 11x + 6$ won't factor by inspection. The hands you a short list of rational zeros to test — turning guessing into a finite search.

Part 4: Synthetic Division & Peeling Off Factors

🧩 The Fundamental Theorem of Algebra & Factoring

Part 4 of 7 — Synthetic Division & Peeling Off Factors

🔑 The Payoff: Found a zero $c$ ? Synthetic division divides $p(x)$ by $(x - c)$ in seconds, leaving a lower-degree quotient you can finish factoring. This is how cubics and quartics get fully cracked.

How Synthetic Division Works

To divide by :

Part 5: The Fundamental Theorem of Algebra

🧩 The Fundamental Theorem of Algebra & Factoring

Part 5 of 7 — The Fundamental Theorem of Algebra

🔑 The Big Theorem: Every polynomial of degree $n \ge 1$ has exactly $n$ roots — if you count complex roots and count repeated roots by their multiplicity. No degree- $n$ polynomial is ever "missing" a root.

The Fundamental Theorem of Algebra (FTA)

FTA: Every polynomial of degree $n \geq 1$ with complex coefficients has .

Part 6: Complex Conjugate Roots & Building Polynomials

🧩 The Fundamental Theorem of Algebra & Factoring

Part 6 of 7 — Complex Conjugate Roots & Building Polynomials

🔑 The Pairing Rule: For a polynomial with real coefficients, complex roots always come in conjugate pairs: if $a + bi$ is a root, so is $a - bi$ . This lets you account for non-real roots and build polynomials from a known root set.

The Complex Conjugate Root Theorem

If $p(x)$ has and (with ) is a root, then its conjugate is a root.

Part 7: Mixed Practice & Mastery Check

🧩 The Fundamental Theorem of Algebra & Factoring

Part 7 of 7 — Mixed Practice & Mastery Check

You've built the full machine: zeros ↔ factors, a factoring toolkit, the Rational Root Theorem, synthetic division, the Fundamental Theorem of Algebra, and conjugate pairs. Time to put it together.

Quick Reference

Tool	What it does
Factor Theorem	$(x - c)$ is a factor $\iff p(c) = 0$

Graphically, the real zeros are the $x$ -intercepts — where the curve crosses or touches the $x$ -axis.

For $p(x) = x^2 - 5x + 6$ , test $x = 2$ :

$p(2) = (2)^2 - 5(2) + 6 = 4 - 10 + 6 = 0 \;\checkmark$

So $2$ is a zero. Testing $x = 3$ gives $9 - 15 + 6 = 0$ , so $3$ is a zero too.

💡 The degree $n$ (highest exponent) caps how many zeros a polynomial can have. A degree- $2$ polynomial has at most $2$ zeros, degree- $3$ at most $3$ , and so on. Part 5 sharpens "at most" into an exact count.

If $(x - c)$ is a factor, then $c$ is a zero.
If $c$ is a zero, then $(x - c)$ is a factor.

We found that $x^2 - 5x + 6$ has zeros $2$ and $3$ . The Factor Theorem promises:

$x^2 - 5x + 6 = (x - 2)(x - 3)$

Check by expanding: $(x-2)(x-3) = x^2 - 3x - 2x + 6 = x^2 - 5x + 6 \;\checkmark$

⚠️ Watch the sign: a zero of $c = 5$ gives the factor $(x - 5)$ , and a zero of $c = -4$ gives the factor $(x - (-4)) = (x + 4)$ .

1) zero $= 9$ → factor $(x \,?)$ , enter the constant: $\,?$ 2) zero $= -5$ → factor $(x \,?)$ , enter the constant: $\,?$ 3) zero $= 0$ → factor $(x \,?)$ , enter the constant: $\,?$

Part 2 builds your factoring toolkit (GCF, grouping, quadratic patterns).
Parts 3–4 find zeros you can't spot by eye, using the Rational Root Theorem and division.
Parts 5–6 count every zero — real and complex — using the Fundamental Theorem of Algebra.

Let's stock the toolbox.

The GCF of $6x^3$ and $9x^2$ is $3x^2$ (largest coefficient $3$ , lowest power $x^2$ ).

Step 2 — Special Patterns

Pattern	Factored form
Difference of squares: $a^2 - b^2$	$(a - b)(a + b)$
Perfect square: $a^2 + 2ab + b^2$	$(a + b)^2$
Sum of cubes: $a^3 + b^3$	$(a + b)(a^2 - ab + b^2)$
Difference of cubes: $a^3 - b^3$	$(a - b)(a^2 + ab + b^2)$

Example (difference of squares)

$x^2 - 16 = x^2 - 4^2 = (x - 4)(x + 4)$

⚠️ A sum of squares like $x^2 + 16$ does not factor over the real numbers — but it does factor with complex numbers (Part 6).

Step 3 — Factoring Trinomials

To factor $x^2 + bx + c$ , find two numbers that multiply to $c$ and add to $b$ .

Example: $x^2 + 7x + 12$

Need a pair multiplying to $12$ , adding to $7$ → that's $3$ and $4$ :

$x^2 + 7x + 12 = (x + 3)(x + 4)$

Example with a negative: $x^2 - x - 12$

Need a pair multiplying to $-12$ , adding to $-1$ → that's $-4$ and $3$ :

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

💡 Sign logic: product negative → the two numbers have opposite signs. Product positive → same sign (both match the sign of $b$ ).

Match Each Factorization 🔽

Choose the correct factored form for each expression.

Step 4 — Factoring by Grouping

For four-term polynomials, group in pairs and pull a GCF from each:

Example: $x^3 + 3x^2 + 2x + 6$

$\underbrace{x^3 + 3x^2}_{x^2(x + 3)} + \underbrace{2x + 6}_{2(x + 3)} = x^2(x + 3) + 2(x + 3)$

Both pieces share $(x + 3)$ , so factor it out:

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

✅ Check: $(x+3)(x^2+2) = x^3 + 2x + 3x^2 + 6 = x^3 + 3x^2 + 2x + 6 \;\checkmark$

The key signal that grouping worked: both groups leave the same binomial behind.

Concept Check 🎯

Factoring Drill 🧮

Factor each. Enter the answer with no spaces, in the exact form shown.

1) $x^2 - 25$ . Enter as (x-5)(x+5): $\,?$ 2) $x^2 + 9x + 20$ . Enter as (x+a)(x+b) with the smaller number first: $\,?$ 3) Factor out the GCF of $5x^3 - 10x^2$ . Enter as 5x^2(x-2): $\,?$

Rational Root Theorem

The Rational Root Theorem

For a polynomial with integer coefficients

$p(x) = a_n x^n + \dots + a_1 x + a_0,$

every rational zero has the form

$\frac{p}{q} \quad\text{where}\quad p \mid a_0 \;(\text{the constant}) \quad\text{and}\quad q \mid a_n \;(\text{the leading coefficient}).$

In words: $p$ is a factor of the constant term, and $q$ is a factor of the leading coefficient.

⚠️ This lists every possible rational zero. It does not guarantee any of them actually work — you still have to test. And it says nothing about irrational or complex zeros.

Building the Candidate List

Example: $p(x) = 2x^3 - 3x^2 - 11x + 6$

Constant term $a_0 = 6$ → factors $p$ : $\pm 1, \pm 2, \pm 3, \pm 6$

Form every $\dfrac{p}{q}$ :

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

That's the complete candidate list. Now test one — try $x = 3$ :

$p(3) = 2(27) - 3(9) - 11(3) + 6 = 54 - 27 - 33 + 6 = 0 \;\checkmark$

So $x = 3$ is a genuine zero, and $(x - 3)$ is a factor.

💡 Test the small integers ( $\pm 1, \pm 2$ ) first — they're the easiest to evaluate and often work.

Concept Check 🎯

Identify $p$ and $q$ 🧮

For $p(x) = 4x^3 + x^2 - 7x + 5$ , the rational-zero candidates are $\dfrac{p}{q}$ .

1) The values of $p$ are the positive factors of the constant term. The constant term is: $\,?$ 2) The values of $q$ are the positive factors of the leading coefficient. The leading coefficient is: $\,?$ 3) The largest possible denominator $q$ in any candidate is:

Test the Candidates 🔽

For $p(x) = x^3 + 2x^2 - 5x - 6$ , evaluate $p$ at each candidate to decide if it's an actual zero.

Where This Leads

The Rational Root Theorem gives you a zero to test. Once you confirm a zero $c$ , the Factor Theorem says $(x - c)$ divides $p(x)$ evenly. But dividing it out by hand is slow.

In Part 4 you'll learn synthetic division — a fast way to divide out $(x - c)$ and collapse the polynomial to a smaller, easier one. Chaining these tools lets you fully factor cubics and quartics.

Write $c$ on the left and the coefficients of $p(x)$ in a row (include $0$ for any missing power).
Bring down the first coefficient.
Multiply by $c$ , add to the next coefficient, and repeat.
The last number is the remainder; the others are the quotient coefficients.

Example: divide $x^3 - 6x^2 + 11x - 6$ by $(x - 1)$ , so $c = 1$

\begin{array}{c|cccc} 1 & 1 & -6 & 11 & -6 \\ & & 1 & -5 & 6 \\ \hline & 1 & -5 & 6 & 0 \end{array}

Read the bottom row: quotient $x^2 - 5x + 6$ , remainder $0$ .

✅ Remainder $0$ confirms $(x - 1)$ is a factor (and $x = 1$ is a zero). So $x^3 - 6x^2 + 11x - 6 = (x - 1)(x^2 - 5x + 6)$ .

The Remainder Theorem

The remainder from dividing $p(x)$ by $(x - c)$ equals $p(c)$ :

$\text{remainder} = p(c)$

That's why a remainder of $0$ means $c$ is a zero — it's the Factor Theorem in action.

Finish the Factorization

From the example, the quotient was $x^2 - 5x + 6$ , which factors further:

$x^2 - 5x + 6 = (x - 2)(x - 3)$

So the complete factorization is

$x^3 - 6x^2 + 11x - 6 = (x - 1)(x - 2)(x - 3),$

with zeros $x = 1, 2, 3$ .

💡 The full pipeline: Rational Root Theorem finds a zero → synthetic division peels off its factor → factor (or repeat on) the smaller quotient.

Concept Check 🎯

Walk the Division 🔽

You're dividing $x^3 - 6x^2 + 11x - 6$ by $(x - 1)$ . The coefficients are $1,\ -6,\ 11,\ -6$ and $c = 1$ . Fill in each step.

Synthetic Division Drill 🧮

Divide $x^3 + 2x^2 - 5x - 6$ by $(x + 1)$ , so $c = -1$ . The coefficients are $1, 2, -5, -6$ .

1) Bring down the $1$ , multiply by $-1$ , add to $2$ . The result is: $\,?$ 2) Continuing, the next bottom-row entry (added to $-5$ ) is: $\,?$ The remainder is:

at least one complex root

Applying it repeatedly (peel off a root, repeat on the quotient) gives the headline consequence:

$\textbf{A degree-}n\textbf{ polynomial has exactly }n\textbf{ complex roots, counted with multiplicity.}$

Equivalently, over the complex numbers it factors completely into linear pieces:

$p(x) = a_n(x - r_1)(x - r_2)\cdots(x - r_n)$

Here "complex" includes the real numbers (a real number is a complex number with zero imaginary part), so real roots count too.

💡 Degree is a promise: a quartic ( $n = 4$ ) has exactly $4$ roots — some may be real, some complex, some repeated, but the total is always $4$ .

Multiplicity — Counting Repeats

A root's multiplicity is how many times its factor appears.

Example

$p(x) = (x - 2)^3(x + 5)$

$x = 2$ has multiplicity $3$
$x = -5$ has multiplicity $1$

Total roots: $3 + 1 = 4$ , matching the degree $4$ . ✓

Graph behavior of real roots

Multiplicity	At that $x$ -intercept the graph…
Odd (1, 3, …)	crosses the $x$ -axis
Even (2, 4, …)	touches and turns around (bounces)

💡 A double root ( $x-r)^2$ touches the axis like the vertex of a parabola; a triple root flattens as it crosses.

Concept Check 🎯

Multiplicity & Degree Drill 🧮

Consider $p(x) = (x - 3)^2(x + 1)(x - 7)^4$ .

1) The multiplicity of the root $x = 7$ is: $\,?$ 2) The degree of $p(x)$ is (add the exponents): $\,?$ 3) The number of distinct real roots is: $\,?$

Cross or Touch? 🔽

For $p(x) = (x + 2)^2(x - 5)^3$ , decide how the graph behaves at each real $x$ -intercept.

real coefficients

So non-real roots are never lonely — they arrive two at a time. A direct consequence:

$\textbf{A polynomial of odd degree with real coefficients has at least one real root.}$

(There's an odd number of roots, complex ones pair off evenly, so at least one must be real.)

Why $x^2 + 1$ has no real zeros

$x^2 + 1 = 0 \;\Rightarrow\; x^2 = -1 \;\Rightarrow\; x = \pm i$

The roots are $i$ and $-i$ — a conjugate pair. Over the complex numbers it factors as $(x - i)(x + i)$ .

💡 Recall $i = \sqrt{-1}$ , so $i^2 = -1$ .

Building a Polynomial from Its Roots

Reverse the process: turn each root $r$ into a factor $(x - r)$ and multiply.

Example: build a polynomial with roots $2$ and $-3$

$p(x) = (x - 2)(x + 3) = x^2 + x - 6$

Example: a conjugate pair — roots $3i$ and $-3i$

Multiply the conjugate factors using $(x - 3i)(x + 3i)$ , a difference of squares:

$(x - 3i)(x + 3i) = x^2 - (3i)^2 = x^2 - 9i^2 = x^2 + 9$

The imaginary parts cancel, leaving real coefficients — exactly why conjugate pairs keep a real polynomial real.

✅ Check: $x^2 + 9 = 0 \Rightarrow x^2 = -9 \Rightarrow x = \pm 3i$ . ✓

Concept Check 🎯

Build the Polynomial 🧮

Multiply out each factored form into standard form $x^2 + bx + c$ (monic, leading coefficient $1$ ).

1) Roots $5$ and $-2$ : expand $(x - 5)(x + 2)$ . Enter as x^2-3x-10: $\,?$ 2) Conjugate roots $2i$ and $-2i$ : expand $(x - 2i)(x + 2i)$ . Enter as x^2+4: $\,?$ 3) Double root $x = 7$ : expand $(x - 7)^2$ . Enter as x^2-14x+49: $\,?$

Conjugate Pairs 🔽

Each polynomial has real coefficients. Match the given non-real root to its required partner and the consequence.

Putting It All Together

You can now account for every root of a real polynomial:

Real rational roots — found via the Rational Root Theorem + synthetic division (Parts 3–4).
Real irrational roots — e.g. from the quadratic formula on a leftover quadratic.
Complex roots — always in conjugate pairs (this part).

Total count always equals the degree, exactly as the Fundamental Theorem of Algebra guarantees. Part 7 mixes everything into one final challenge.

⚠️ Strategy reminder: GCF first → special patterns / trinomials → for higher degree, find a zero (RRT), divide it out (synthetic), repeat until you reach a quadratic you can finish.

Mixed Practice 🔽

Apply the right tool to each.

Full Factorization Challenge 🧮

You're factoring $p(x) = x^3 - 4x^2 + x + 6$ completely.

1) Test $x = -1$ . Compute $p(-1) = (-1)^3 - 4(-1)^2 + (-1) + 6$ . The value is: Since that confirms is a factor, dividing gives the quotient . Factor it as : How many zeros does have in total?

Exit Quiz ✅

Answer all three to finish the lesson.

an(the leading coefficient).

Leading coefficient

a_n = 2

→ factors

q

\pm 1, \pm 2

Fundamental Theorem of Algebra and Factoring

Fundamental Theorem of Algebra and Factoring - Complete Interactive Lesson

Part 1: Zeros, Factors & the Connection Between Them

🧩 The Fundamental Theorem of Algebra & Factoring

Topics in This Part

What Is a Polynomial Zero?

The Factor Theorem

Why This Matters

Part 2: Core Factoring Techniques

🧩 The Fundamental Theorem of Algebra & Factoring

Step 1 — Greatest Common Factor (GCF)

Part 3: The Rational Root Theorem

🧩 The Fundamental Theorem of Algebra & Factoring

Part 4: Synthetic Division & Peeling Off Factors

🧩 The Fundamental Theorem of Algebra & Factoring

How Synthetic Division Works

Part 5: The Fundamental Theorem of Algebra

🧩 The Fundamental Theorem of Algebra & Factoring

The Fundamental Theorem of Algebra (FTA)

Part 6: Complex Conjugate Roots & Building Polynomials

🧩 The Fundamental Theorem of Algebra & Factoring

The Complex Conjugate Root Theorem

Part 7: Mixed Practice & Mastery Check

🧩 The Fundamental Theorem of Algebra & Factoring

Quick Reference

Example

Example

Step 2 — Special Patterns

Example (difference of squares)

Step 3 — Factoring Trinomials

Example: x2+7x+12x^2 + 7x + 12x2+7x+12

Example with a negative: x2−x−12x^2 - x - 12x2−x−12

Step 4 — Factoring by Grouping

Example: x3+3x2+2x+6x^3 + 3x^2 + 2x + 6x3+3x2+2x+6

The Rational Root Theorem

Building the Candidate List

Example: p(x)=2x3−3x2−11x+6p(x) = 2x^3 - 3x^2 - 11x + 6p(x)=2x3−3x2−11x+6

Where This Leads

Example: divide x3−6x2+11x−6x^3 - 6x^2 + 11x - 6x3−6x2+11x−6 by (x−1)(x - 1)(x−1), so c=1c = 1c=1

The Remainder Theorem

Finish the Factorization

Multiplicity — Counting Repeats

Example

Graph behavior of real roots

Why x2+1x^2 + 1x2+1 has no real zeros

Building a Polynomial from Its Roots

Example: build a polynomial with roots 222 and −3-3−3

Example: a conjugate pair — roots 3i3i3i and −3i-3i−3i

Putting It All Together

Example: $x^2 + 7x + 12$

Example with a negative: $x^2 - x - 12$

Example: $x^3 + 3x^2 + 2x + 6$

Example: $p(x) = 2x^3 - 3x^2 - 11x + 6$

Example: divide $x^3 - 6x^2 + 11x - 6$ by $(x - 1)$ , so $c = 1$

Why $x^2 + 1$ has no real zeros

Example: build a polynomial with roots $2$ and $-3$

Example: a conjugate pair — roots $3i$ and $-3i$