Fundamental Theorem of Algebra and Factoring

🧩 The Fundamental Theorem of Algebra & Factoring

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

Topics in This Part

🔑 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 zero (or root) of $p(x)$ is any value $c$ that makes the polynomial equal $0$:

$p(c) = 0$

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

Example

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.

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 $\iff$ means it works both ways:

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

Example

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 $(x - (-4)) = (x + 4)$.

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 -6.

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

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 builds your factoring toolkit (GCF and the quadratic patterns).
• Parts 3–4 find the remaining zeros once a graph, a table, or a known factor gives you one, then divide that factor out.
• Parts 5–6 count every zero — real and complex — using the Fundamental Theorem of Algebra.

Let's stock the toolbox.

🧩 The Fundamental Theorem of Algebra & Factoring

Part 2 of 7 — Core Factoring Techniques

🔑 Strategy: Always try these in order: (1) pull out a GCF, (2) recognize a special pattern, (3) factor a trinomial, (4) spot a quadratic in disguise. On the AP Precalculus exam, factoring by hand is limited to common factors and the rules for quadratics, and these four steps cover exactly that.

Step 1 — Greatest Common Factor (GCF)

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

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

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

Step 2 — Special Patterns

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 — Quadratics in Disguise

Some higher-degree polynomials are really quadratics once you pull out a GCF or rename a power. The quadratic rules then finish the job.

Example 1: GCF first, then a trinomial

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

Zeros: $x = 0,\ 4,\ -3$.

Example 2: quadratic form

In $x^4 - 13x^2 + 36$, let $u = x^2$. Then it reads $u^2 - 13u + 36 = (u - 4)(u - 9)$, so

$x^4 - 13x^2 + 36 = (x^2 - 4)(x^2 - 9) = (x - 2)(x + 2)(x - 3)(x + 3)$

✅ Check: the four zeros $\pm 2, \pm 3$ match the degree, $4$.

The signal to look for: the exponents are $2k$ and $k$ (like $x^4$ and $x^2$), or every term shares a power of $x$.

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): $\,?$

🧩 The Fundamental Theorem of Algebra & Factoring

Part 3 of 7 — Zeros from Graphs, Tables & a Known Factor

🔑 The Problem: A cubic like $2x^3 - 3x^2 - 11x + 6$ won't factor by inspection. On the AP exam you are never asked to hunt for its zeros blind. Instead, the question hands you a foothold: a graph, a table of values, or one zero or factor. Your job is to turn that foothold into the full factorization.

Special root-hunting rules such as the Rational Root Theorem and Descartes' Rule of Signs belong to Algebra 2 and are not assessed on the AP Precalculus exam.

Foothold 1 — A Graph

Every real zero $a$ of $p$ puts an $x$-intercept at $(a, 0)$. So a graph lets you read real zeros and turn them straight into factors with the Factor Theorem.

Example

The graph of a cubic $p$ with leading coefficient $1$ crosses the $x$-axis at $x = -2$, $x = 1$ and $x = 3$, and nowhere else.

• Zeros $-2, 1, 3$ give factors $(x + 2)$, $(x - 1)$, $(x - 3)$.
• A cubic has exactly $3$ zeros counting multiplicity, so these are all of them.
• The leading coefficient is $1$, so

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

⚠️ A graph alone cannot show the leading coefficient: $2(x+2)(x-1)(x-3)$ has the same intercepts. Use one more point, such as the $y$-intercept, to pin it down. Here $p(0) = (2)(-1)(-3) = 6$.

Foothold 2 — A Table or a Given Zero

A table entry with output $0$ is a zero: if a table shows $p(3) = 0$, then $(x - 3)$ is a factor. A problem may also simply state one zero. Either way, verify it by substitution before you rely on it.

Example: $p(x) = 2x^3 - 3x^2 - 11x + 6$, given that $p(3) = 0$

Check the given zero:

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

So $(x - 3)$ is a factor, and

$2x^3 - 3x^2 - 11x + 6 = (x - 3)(\text{a quadratic})$

Part 4 shows how to divide out $(x - 3)$ to find that quadratic: $2x^2 + 3x - 2 = (2x - 1)(x + 2)$. The three zeros are $3$, $\tfrac{1}{2}$ and $-2$.

💡 Matching degrees is a quick sanity check: a cubic divided by a linear factor must leave a quadratic.

Concept Check 🎯

Pin Down the Polynomial 🧮

A cubic $p$ has zeros $x = -1$, $x = 2$ and $x = 5$ (each with multiplicity $1$), so $p(x) = a(x + 1)(x - 2)(x - 5)$ for some constant $a$. Its graph passes through $(0, 20)$.

1) Compute $(0 + 1)(0 - 2)(0 - 5)$: $\,?$ 2) Use $p(0) = 20$ to find the leading coefficient $a$: $\,?$ 3) Evaluate $p(1)$ for this polynomial: $\,?$

Check the Claimed Zeros 🔽

A classmate claims three zeros for $p(x) = x^3 + 2x^2 - 5x - 6$. Evaluate $p$ at each to decide whether the claim holds.

Where This Leads

A graph, a table, or a stated zero gives you one factor $(x - c)$. The Factor Theorem guarantees that $(x - c)$ divides $p(x)$ with remainder $0$, but you still need the quotient.

In Part 4 you'll divide out $(x - c)$ to collapse the polynomial to a smaller one you can finish with the quadratic rules. When a problem gives no foothold and the zeros are not visible by inspection, that is a graphing-calculator question (Part B of the exam).

🧩 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 $p(x)$ by $(x - c)$:

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

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

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: a graph, table, or given zero supplies a factor → division peels it off → factor (or repeat on) the smaller quotient. Polynomial long division (Topic 1.11) gives the same quotient; synthetic division is just a faster layout for dividing by $(x - c)$.

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: $\,?$ 3) The remainder is: $\,?$

🧩 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 $n \ge 1$ with complex coefficients has 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

💡 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.

🧩 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 real coefficients and $a + bi$ (with $b \ne 0$) is a root, then its conjugate $a - bi$ is also a root.

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:

1. Real roots you are given or can read from a graph or table, divided out with synthetic division (Parts 3–4).
2. Real irrational roots — e.g. from the quadratic formula on a leftover quadratic.
3. 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.

🧩 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, zeros from graphs and tables, synthetic division, the Fundamental Theorem of Algebra, and conjugate pairs. Time to put it together.

Quick Reference

⚠️ Strategy reminder: GCF first → special patterns / trinomials / quadratic form → for higher degree, take the zero the graph, table, or problem gives you, divide it out, and 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: $\,?$ 2) Since that confirms $(x + 1)$ is a factor, dividing gives the quotient $x^2 - 5x + 6$. Factor it as (x-2)(x-3): $\,?$ 3) How many distinct real zeros does $p(x)$ have in total? $\,?$

Exit Quiz ✅

Answer all three to finish the lesson.