MGU Mathematics Series — Algebra II
Part II: Quadratics, Complex Numbers, and Polynomial Functions
Factoring · Rational Root Theorem · Synthetic Division · Fundamental Theorem
Chapter 6 of 14
MGU Mathematics Series  ·  Algebra II  ·  Library Textbook Edition

Chapter 6:
Polynomial Equations
and Theorems

Factoring strategies, rational roots, synthetic division, complex conjugate pairs, the Fundamental Theorem of Algebra, and the complete structure of polynomial solutions

■ Math Level: Algebra II ■

■ 14 Sections  •  Part II of 5 ■

Follows Chapter 5: Polynomial Functions  ·  Prepares students for Chapter 7: Rational Expressions and Functions

Chapter 6 — Front Matter

Purpose and Overview

In Chapter 5, you studied polynomial functions: their degree, leading coefficient, end behavior, zeros, multiplicity, turning points, and graphs. You learned to divide polynomials and to identify connections between factors and zeros. Chapter 6 takes that structural knowledge and turns it toward solving. Given a polynomial equation, how do you find all of its solutions? How many solutions should you expect? Can those solutions be non-real? When factoring is not enough, what strategies give you a path forward?

This chapter answers those questions systematically. You will review and extend factoring strategies, learn to use the Rational Root Theorem to generate candidates for rational solutions, use synthetic division to test and reduce polynomials, discover how complex roots appear in conjugate pairs, and apply the Fundamental Theorem of Algebra to count roots with precision. You will also interpret graphs as tools for approximating solutions and learn to work with technology when exact methods become impractical.

The goal is not to memorize a list of procedures. The goal is to develop a flexible, theorem-supported approach to polynomial equations so that you understand why solutions exist, where to look for them, and how to interpret them in context.

Opening Question

A structural engineer is analyzing the load-bearing behavior of a curved support beam. The deflection of the beam at any point along its length is modeled by the polynomial \(d(x) = x^4 - 6x^3 + 11x^2 - 6x\), where \(x\) is measured in meters from one end of the beam. The engineer needs to find every point where the deflection equals zero — every point where the beam returns to its original horizontal position. How many such points are there? Are all of them real? How would you find them systematically? By the end of this chapter, you will have every tool you need to answer this question precisely.

14Sections
13Objectives
14Key Terms
15+Worked Examples

Learning Objectives

  1. 1Identify polynomial equations and state the number of roots expected from the degree.
  2. 2Apply factoring strategies including GCF, grouping, trinomial factoring, difference of squares, and sum and difference of cubes.
  3. 3Solve polynomial equations by factoring and by setting factors equal to zero.
  4. 4State and apply the Rational Root Theorem to generate a list of candidate rational roots.
  5. 5Test rational root candidates using synthetic division or direct substitution.
  6. 6Use depressed polynomials to reduce the degree after finding a root.
  7. 7Explain why complex roots appear in conjugate pairs when polynomial coefficients are real.
  8. 8State the Fundamental Theorem of Algebra and apply it to count the total number of roots.
  9. 9Count roots with multiplicity and connect multiplicity to graph behavior.
  10. 10Use graphs and graphing technology to approximate real solutions.
  11. 11Apply polynomial equations in real-world modeling contexts.
  12. 12Recognize and avoid common root-finding mistakes.
Chapter 6 — Contents

Sections at a Glance

Sections 6.1 – 6.3
Polynomial Equations, Factoring, and Solving
Standard Form · Roots and Zeros · Factoring Strategies · Zero Product Property
Section 6.1

Polynomial Equations

A polynomial equation is a statement that a polynomial expression equals a value. The most common form sets the polynomial equal to zero:

\[ a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 = 0 \]

where \(a_n \neq 0\) and the exponents are non-negative integers. You have already worked with polynomial equations in simpler forms. A linear equation such as \(3x - 5 = 0\) is a degree-1 polynomial equation. A quadratic equation such as \(x^2 - 5x + 6 = 0\) is a degree-2 polynomial equation. Chapter 6 extends this reasoning to degree 3, degree 4, and higher.

Setting a Polynomial Equal to Zero

Before solving any polynomial equation, it must be written in standard form with zero on one side. The most powerful solving strategies — factoring, the Rational Root Theorem, and synthetic division — all depend on the equation being set equal to zero. If \(p(x) = q(x)\), rewrite it as \(p(x) - q(x) = 0\) before proceeding.

Roots, Zeros, and Solutions

The words root, zero, and solution all refer to the same idea from different perspectives. A root of an equation is a value of \(x\) that satisfies the equation. A zero of a function \(f(x)\) is a value where \(f(x) = 0\). A solution to an equation \(p(x) = 0\) is a value of \(x\) that makes the equation true. When the zeros are real numbers, they are simultaneously the x-intercepts of the graph.

Guiding Principle — Degree and Root Count

A polynomial equation of degree \(n\) has exactly \(n\) roots when counted with multiplicity over the complex numbers. A degree-3 equation has exactly 3 roots. A degree-4 equation has exactly 4 roots. Some roots may be real, some complex, and some repeated — but the count always equals \(n\).

Example 6.1.1
Identifying a polynomial equation and predicting its root count
  • ProblemClassify each as a polynomial equation, state its degree, and predict how many roots it has. (a) \(2x^3 - x + 7 = 0\)   (b) \(x^4 - 3x^2 + x - 9 = 14\)   (c) \(\sqrt{x} + 3 = 0\)
  • ThinkCheck whether each equation involves only non-negative integer exponents. Then rewrite in standard form and identify the degree.
  • Solve (a) \(2x^3 - x + 7 = 0\) is already in standard form. Highest-degree term is \(2x^3\). Degree 3; exactly 3 roots.

    (b) Subtract 14: \(x^4 - 3x^2 + x - 23 = 0\). Degree 4; exactly 4 roots.

    (c) \(\sqrt{x} = x^{1/2}\) is a fractional exponent. Not a polynomial equation.
  • CheckIn (a) and (b), all exponents are non-negative integers. In (c), the fractional exponent disqualifies it.
Answer: (a) degree 3, 3 roots; (b) degree 4, 4 roots; (c) not a polynomial equation.
Student Note

Always rewrite a polynomial equation with zero on one side before counting degree or solving. A constant on the right side — such as the 14 in equation (b) — changes the constant term but not the degree once moved.

Practice 6.1Concept Check
  1. Explain the difference between a root, a zero, and a solution. Are they the same thing?
  2. A student says that \(x^3 - 5x = 0\) has only two solutions because they only see two x-intercepts on the graph. What might they be missing?
  3. Is \(x^{-2} + 3x = 7\) a polynomial equation? Explain.
  4. A degree-5 polynomial is set equal to zero. How many total roots should you expect to find?
  5. Rewrite each equation in standard form: (a) \(3x^2 + 1 = 2x^3 - x\)   (b) \(x^4 = 4x^2 - 3\)
Section 6.2

Reviewing Factoring Strategies

Before solving polynomial equations by factoring, it is important to review the full toolkit of factoring strategies. Factoring is the most direct path to solutions when it applies, but it requires pattern recognition and strategic thinking.

Strategy 1 — Greatest Common Factor

Always look for a GCF first. Factoring it out reduces the degree of the remaining polynomial and simplifies every step that follows.

\[ 6x^3 - 12x^2 + 18x = 6x(x^2 - 2x + 3) \]

Strategy 2 — Difference of Squares

\[ a^2 - b^2 = (a+b)(a-b) \]
\[ x^4 - 16 = (x^2 + 4)(x^2 - 4) = (x^2 + 4)(x+2)(x-2) \]
Tip

\(x^4 - 16\) is a difference of squares in \(x^2\). After factoring \((x^2 + 4)(x^2 - 4)\), you must factor \(x^2 - 4\) further. Note that \(x^2 + 4\) does not factor over the real numbers.

Strategy 3 — Sum and Difference of Cubes

\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \] \[ a^3 - b^3 = (a - b)(a^2 + ab + b^2) \]
\[ x^3 - 8 = (x - 2)(x^2 + 2x + 4) \qquad x^3 + 27 = (x + 3)(x^2 - 3x + 9) \]
Student Warning

The second factor in each cube formula is a trinomial that does not factor further over the real numbers. Do not confuse \(a^2 - ab + b^2\) with a perfect square trinomial — it is not. Do not try to factor it again.

Strategy 4 — Trinomial Factoring

For \(x^2 + bx + c\), find two numbers that multiply to \(c\) and add to \(b\). For \(ax^2 + bx + c\) with \(a \neq 1\), use factoring by grouping or trial and error.

Strategy 5 — Factoring by Grouping

Used for four-term polynomials. Group pairs of terms, factor each group, then factor the common binomial.

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

Choosing a Strategy

StepAction
1Factor out any GCF first.
2Count terms. Two terms → difference of squares or cubes. Three terms → trinomial or perfect square. Four terms → grouping.
3Check if each resulting factor can be factored further.
4Verify by multiplying back.
Example 6.2.1
Selecting and applying factoring strategies
  • ProblemFactor completely: (a) \(x^3 + 5x^2 - 4x - 20\)   (b) \(3x^4 - 48\)   (c) \(x^3 - 125\)
  • Think(a) Four terms → grouping. (b) GCF first, then difference of squares. (c) Difference of cubes.
  • Solve (a)
    Group: \((x^3 + 5x^2) + (-4x - 20)\)
    Factor each group: \(x^2(x+5) - 4(x+5)\)
    Factor common binomial: \((x^2 - 4)(x+5)\)
    Factor difference of squares: \((x+2)(x-2)(x+5)\)

    (b)
    GCF: \(3(x^4 - 16)\)
    Difference of squares: \(3(x^2+4)(x^2-4)\)
    Factor again: \(3(x^2+4)(x+2)(x-2)\)   [Note: \(x^2+4\) is irreducible over ℝ]

    (c)
    \(x^3 - 125 = x^3 - 5^3\)
    Difference of cubes: \((x-5)(x^2+5x+25)\)
  • CheckFor (a): \((x+2)(x-2) = x^2-4\), then \((x^2-4)(x+5) = x^3+5x^2-4x-20\). ✓
Answer: (a) \((x+2)(x-2)(x+5)\)   (b) \(3(x^2+4)(x+2)(x-2)\)   (c) \((x-5)(x^2+5x+25)\)
Practice 6.2Factoring Strategies
  1. \(x^3 - 8x^2 + 15x\)
  2. \(x^3 + 27\)
  3. \(x^4 - 81\)
  4. \(2x^3 - 2x^2 - 12x\)
  5. \(x^3 + x^2 - x - 1\)
  6. \(4x^3 - 32\)
  7. \(x^4 - 5x^2 + 4\)
  8. \(x^3 - 3x^2 - 4x + 12\)
Section 6.3

Solving by Factoring

Once a polynomial is completely factored, solving the equation is straightforward. You use the Zero Product Property: if \(ab = 0\), then \(a = 0\) or \(b = 0\) (or both). This extends to any number of factors — if the product of several expressions equals zero, at least one of those expressions must equal zero.

Procedure — Solving by Factoring

1. Write the equation with zero on one side.

2. Factor the polynomial completely.

3. Set each factor equal to zero.

4. Solve each resulting equation.

5. Check each solution in the original equation.

Example 6.3.1
Solving a cubic equation by factoring
  • ProblemSolve \(x^3 - 7x^2 + 12x = 0\).
  • ThinkThe equation is already set equal to zero. Factor out the GCF first, then factor the remaining quadratic.
  • Solve x(x² − 7x + 12) = 0 x(x − 3)(x − 4) = 0 x = 0 or x − 3 = 0 or x − 4 = 0 x = 0, x = 3, x = 4
  • Check\(x=0\): \(0-0+0=0\) ✓   \(x=3\): \(27-63+36=0\) ✓   \(x=4\): \(64-112+48=0\) ✓
Answer: \(x = 0, \quad x = 3, \quad x = 4\)
Example 6.3.2
Solving a fourth-degree equation — quadratic form substitution
  • ProblemSolve \(x^4 - 5x^2 + 4 = 0\).
  • ThinkThis looks like a trinomial in \(x^2\). Let \(u = x^2\) to clarify the structure, then factor.
  • Solve Let \(u = x^2\): \(u^2 - 5u + 4 = 0\)
    Factor: \((u-1)(u-4) = 0\)
    So \(u = 1\) or \(u = 4\).
    Back-substitute: \(x^2 = 1 \Rightarrow x = \pm 1\); \(x^2 = 4 \Rightarrow x = \pm 2\)
  • Check \(x = \pm 1\): \(1 - 5 + 4 = 0\) ✓   \(x = \pm 2\): \(16 - 20 + 4 = 0\) ✓
Answer: \(x = -2, \quad x = -1, \quad x = 1, \quad x = 2\)
Student Note — Quadratic Form

Treating a polynomial as a quadratic in disguise by substituting \(u = x^2\) (or another power) is called quadratic form substitution. It is powerful whenever you see a pattern like \(ax^4 + bx^2 + c\).

Student Tip

Not every cubic or higher-degree polynomial can be factored by inspection. When grouping and trinomial factoring fail, the Rational Root Theorem (Section 6.4) gives a systematic strategy for finding rational roots and reducing the polynomial.

Practice 6.3Solving by Factoring
  1. \(x^3 - 9x = 0\)
  2. \(x^3 + 2x^2 - x - 2 = 0\)
  3. \(x^4 - 10x^2 + 9 = 0\)
  4. \(x^3 + 8 = 0\)
  5. \(2x^3 - 8x = 0\)
  6. \(x^4 - 1 = 0\)
Sections 6.4 – 6.6
Rational Root Theorem and Synthetic Division
Candidate Lists · Testing Roots · Depressed Polynomials · Successive Division
Section 6.4

Rational Root Theorem

When a polynomial cannot be factored easily by inspection, the Rational Root Theorem provides a structured way to generate a list of candidate rational roots. It does not guarantee that any rational roots exist, but it narrows the search dramatically.

Rational Root Theorem

If the polynomial \(p(x) = a_n x^n + a_{n-1}x^{n-1} + \cdots + a_0\) has integer coefficients and a rational root \(\dfrac{p}{q}\) in lowest terms, then:

\(p\) is a factor of the constant term \(a_0\), and \(q\) is a factor of the leading coefficient \(a_n\).

The list of candidate rational roots is:   \(\pm\dfrac{\text{factors of }a_0}{\text{factors of }a_n}\)

Example 6.4.1
Listing candidate rational roots
  • ProblemList all possible rational roots of \(p(x) = 2x^3 - 5x^2 - 4x + 3\).
  • ThinkConstant term is 3; leading coefficient is 2. List the factors of each.
  • Solve Factors of 3: \(\pm 1, \pm 3\)
    Factors of 2: \(\pm 1, \pm 2\)
    Possible rational roots: \(\pm 1, \pm \dfrac{1}{2}, \pm 3, \pm \dfrac{3}{2}\)
Answer: \(\pm 1,\ \pm \dfrac{1}{2},\ \pm 3,\ \pm \dfrac{3}{2}\) — eight candidates in total.
Example 6.4.2
Polynomial with leading coefficient 1
  • ProblemList all possible rational roots of \(q(x) = x^4 - 3x^3 - 2x^2 + 8x - 4\).
  • ThinkWhen the leading coefficient is 1, all denominators are \(\pm 1\), so every candidate is an integer.
  • Solve Factors of \(-4\): \(\pm 1, \pm 2, \pm 4\)
    Factors of \(1\): \(\pm 1\)
    Possible rational roots: \(\pm 1, \pm 2, \pm 4\)
Answer: \(\pm 1,\ \pm 2,\ \pm 4\)
Common Mistake

Students sometimes list only the positive candidates. The theorem requires both positive and negative versions of each fraction, because \(p\) can be any factor of \(a_0\), including negative factors.

Section 6.5

Testing Possible Rational Roots

Listing candidates is only half the process. You must then test each candidate to see whether it is actually a root. The most efficient testing method is synthetic division, which simultaneously tests a candidate and produces a depressed polynomial. You can also test by direct substitution: evaluate \(p(c)\) for each candidate \(c\). If \(p(c) = 0\), then \(c\) is a root.

Two values are especially quick to compute by substitution: \(p(1)\) is the sum of all coefficients; \(p(-1)\) is the alternating sum of coefficients. Always try these first when they are candidates.

Example 6.5.1
Testing rational root candidates by substitution
  • ProblemFind a rational root of \(p(x) = 2x^3 - 5x^2 - 4x + 3\) from the candidates in Example 6.4.1.
  • ThinkStart with the simplest candidates: \(x = 1\) and \(x = -1\).
  • Solve \(p(1) = 2 - 5 - 4 + 3 = -4 \neq 0\). Not a root.

    \(p(-1) = -2 - 5 + 4 + 3 = 0\). ✓ This equals zero, so \(x = -1\) is a root.
  • Check\(p(-1) = 0\) confirmed by substitution.
Answer: \(x = -1\) is a rational root of \(p(x)\).
Example 6.5.2
Testing candidates with synthetic division
  • ProblemSolve \(x^3 + 3x^2 - 10x - 24 = 0\).
  • ThinkLeading coefficient is 1; all candidates are integer factors of \(-24\): \(\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 24\). Test small values first.
  • Solve \(p(2) = 8 + 12 - 20 - 24 = -24 \neq 0\).
    \(p(-2) = -8 + 12 + 20 - 24 = 0\). ✓   So \(x = -2\) is a root.

    Synthetic division with \(c = -2\):
\(-2\) 13−10−24
−2−224
11−120
  • Solve (cont.) Depressed polynomial: \(x^2 + x - 12\)
    Factor: \((x+4)(x-3)\)
    Fully factored: \((x+2)(x+4)(x-3) = 0\)
    Roots: \(x = -2,\ x = -4,\ x = 3\)
  • Check\(p(-4) = -64+48+40-24=0\) ✓   \(p(3) = 27+27-30-24=0\) ✓
Answer: \(x = -4, \quad x = -2, \quad x = 3\)
Section 6.6

Depressed Polynomials

When you find a root \(c\) and divide the polynomial by \((x - c)\), the resulting quotient is called the depressed polynomial. Its degree is exactly one less than the original. The key strategy for solving higher-degree polynomial equations is to find one root, divide, then repeat until the polynomial is degree 2 or less — and finish with factoring or the quadratic formula. This process is called successive division or polynomial deflation.

Successive Division Strategy

If \(p(x)\) has degree \(n\) and \(c\) is a root, then \(p(x) = (x - c)\cdot q(x)\) where \(q(x)\) is the depressed polynomial of degree \(n-1\). The roots of \(p(x)\) are: \(c\) and all roots of \(q(x)\).

Repeat: find a root of \(q(x)\), divide again, continue until you reach degree 2.

Example 6.6.1
Using a depressed polynomial twice
  • ProblemSolve \(p(x) = x^4 - 2x^3 - 7x^2 + 8x + 12 = 0\).
  • ThinkDegree 4, so 4 roots. Candidates: \(\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12\). Test \(x = 2\).
  • Solve \(p(2) = 16-16-28+16+12=0\) ✓   Divide by \((x-2)\):
\(2\) 1−2−7812
20−14−12
10−7−60
  • Solve (cont.) Depressed polynomial: \(x^3 - 7x - 6\). Test \(x = -1\):
    \((-1)^3 - 7(-1) - 6 = -1+7-6 = 0\) ✓   Divide by \((x+1)\):
\(-1\) 10−7−6
−116
1−1−60
  • Solve (cont.) Second depressed polynomial: \(x^2 - x - 6 = (x-3)(x+2)\)
    Roots: \(x = 3\) and \(x = -2\)
  • CheckTotal roots: \(x = 2, -1, 3, -2\). Four roots for a degree-4 polynomial. ✓
Answer: \(x = -2, \quad x = -1, \quad x = 2, \quad x = 3\)
Student Tip

Once you reduce to a quadratic, you can always finish with factoring or the quadratic formula. The goal of repeated synthetic division is to get to degree 2 as efficiently as possible.

Practice 6.4–6.6Rational Root and Synthetic Division
  1. List all candidate rational roots of \(2x^3 + x^2 - 5x + 2\). Do not solve.
  2. List all candidate rational roots of \(3x^4 - x^3 + 2x - 6\). Do not solve.
  3. Use synthetic division to test whether \(x = 3\) is a root of \(p(x) = x^3 - 2x^2 - 5x + 6\). If so, state the depressed polynomial.
  4. Use synthetic division to test whether \(x = -2\) is a root of \(p(x) = x^3 + x^2 - 8x - 12\). If so, solve the equation completely.
  5. Solve \(2x^3 - 3x^2 - 11x + 6 = 0\) using the Rational Root Theorem and synthetic division.
  6. Solve \(x^4 - 3x^3 + x^2 + 3x - 2 = 0\) using successive division.
Sections 6.7 – 6.11
Complex Roots, Fundamental Theorem, and Approximation
Conjugate Pairs · Fundamental Theorem of Algebra · Multiplicity · IVT · Technology
Section 6.7

Complex Roots and Conjugate Pairs

In Chapter 4, you learned that some quadratic equations have no real solutions but do have complex solutions of the form \(a + bi\). The same is true of polynomial equations of any degree: some roots are complex. The critical fact is when and how complex roots appear.

Complex Conjugate Root Theorem

If \(p(x)\) is a polynomial with real coefficients and \(a + bi\) is a root (with \(b \neq 0\)), then \(a - bi\) is also a root. Complex roots of polynomials with real coefficients always come in conjugate pairs.

When you multiply a conjugate pair of factors together, you recover a quadratic with real coefficients and no real roots:

\[ \bigl(x-(a+bi)\bigr)\bigl(x-(a-bi)\bigr) = (x-a)^2 + b^2 \]

This is called an irreducible quadratic factor — a quadratic that cannot be factored further using real numbers. Every polynomial with real coefficients can be completely factored into linear factors and irreducible quadratic factors over the real numbers.

Example 6.7.1
Finding complex roots by factoring
  • ProblemSolve \(x^3 - 2x^2 + 4x - 8 = 0\).
  • ThinkTry factoring by grouping.
  • Solve \(x^2(x-2) + 4(x-2) = 0\)
    \((x^2+4)(x-2) = 0\)
    From \(x - 2 = 0\): \(x = 2\)
    From \(x^2 + 4 = 0\): \(x^2 = -4 \Rightarrow x = \pm 2i\)
  • CheckDegree 3 → expect 3 roots: one real (\(x=2\)) and two complex conjugates (\(\pm 2i\)). ✓
Answer: \(x = 2, \quad x = 2i, \quad x = -2i\)
Example 6.7.2
Finding a polynomial from roots including complex roots
  • ProblemA degree-4 polynomial with real coefficients has roots \(x = 1\), \(x = -3\), and \(x = 2 + i\). Find all roots and write the polynomial in factored form.
  • ThinkSince the polynomial has real coefficients, \(2+i\) must be paired with its conjugate \(2-i\).
  • Solve All four roots: \(1,\ -3,\ 2+i,\ 2-i\)
    Conjugate pair: \(\bigl(x-(2+i)\bigr)\bigl(x-(2-i)\bigr) = (x-2)^2 + 1 = x^2-4x+5\)
    Factored form: \((x-1)(x+3)(x^2-4x+5)\)
Answer: Roots are \(1,\ -3,\ 2+i,\ 2-i\). Polynomial: \((x-1)(x+3)(x^2-4x+5)\).
Student Warning

The Complex Conjugate Root Theorem requires that the polynomial have real coefficients. If the polynomial is allowed to have complex coefficients, complex roots do not have to come in conjugate pairs.

Section 6.8

Fundamental Theorem of Algebra

One of the most profound facts in all of mathematics is the Fundamental Theorem of Algebra. It guarantees that polynomial equations always have solutions — as long as you allow complex numbers as possible answers — and it tells you exactly how many.

Fundamental Theorem of Algebra

Every polynomial \(p(x)\) of degree \(n \geq 1\) with complex coefficients has at least one complex root.

Consequently, a polynomial of degree \(n\) can be written as a product of exactly \(n\) linear factors over the complex numbers:

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

It has exactly \(n\) roots when counted with multiplicity. Not \(n-1\), not \(n+1\), but exactly \(n\).

This theorem, first rigorously proved by Carl Friedrich Gauss in 1799, ensures that no polynomial equation is unsolvable in the complex number system. For polynomials with real coefficients, combining the Fundamental Theorem with the Complex Conjugate Root Theorem gives a precise structure:

Degree typeReal rootsComplex (non-real) roots
Odd degree, real coefficientsAt least 1 (must exist)Even number (come in pairs)
Even degree, real coefficients0 or more (may be zero)Even number (come in pairs)
Example 6.8.1
Applying the Fundamental Theorem
  • ProblemHow many roots does each polynomial have? Describe the nature of those roots. (a) \(p(x) = x^5 - 3x + 2\)   (b) \(q(x) = x^6 + x^4 + 1\)
  • Solve (a) Degree 5: exactly 5 roots. Odd degree with real coefficients → at least 1 real root. Non-real roots come in pairs, so there could be 1, 3, or 5 real roots.

    (b) Degree 6: exactly 6 roots. Since all terms are positive for all real \(x\), \(q(x) > 0\) for all real \(x\) → no real roots. All 6 roots are complex, forming 3 conjugate pairs.
Answer: (a) 5 roots, at least 1 real; (b) 6 roots, all complex.
MGU Library Connection
  • The Fundamental Theorem of Algebra connects to Complex Analysis and Abstract Algebra (upper-division MGU courses)
  • Gauss's original proof used geometric arguments studied in Topology
  • For Algebra II, the theorem provides the guarantee that polynomial equations always have solutions — the course does not require proving it, only applying it
  • See: MGU Dictionary entries for "Fundamental Theorem of Algebra," "complex number," and "irreducible polynomial"
Section 6.9

Counting Roots with Multiplicity

The phrase counted with multiplicity is important and requires explanation. A root's multiplicity is the number of times its corresponding factor appears in the complete factored form of the polynomial. If \((x - r)^k\) appears but \((x - r)^{k+1}\) does not, then \(r\) has multiplicity \(k\).

MultiplicityGraph behavior at root
1 (odd)Crosses the x-axis
2 (even)Touches the x-axis and turns back
3 (odd)Crosses with an S-shaped bend (inflection)
Even > 2Flatter touch and turn back
Example 6.9.1
Counting roots with multiplicity from factored form
  • ProblemFor \(p(x) = (x-3)^2(x+1)^3(x^2+4)\), state the degree, list all roots with multiplicity, and describe graph behavior at each real root.
  • Solve Degree: \(2+3+2 = 7\)

    Roots:
    \(x = 3\), multiplicity 2 → graph touches and turns back
    \(x = -1\), multiplicity 3 → graph crosses with S-curve
    \(x = \pm 2i\), multiplicity 1 each → complex, not on the real graph

    Total count: \(2+3+1+1 = 7\) ✓
Answer: Degree 7. Roots: \(x=3\) (mult. 2, touches); \(x=-1\) (mult. 3, S-crossing); \(x = \pm 2i\) (mult. 1, complex).
Example 6.9.2
Writing a polynomial from given roots and multiplicities
  • ProblemWrite a polynomial of lowest degree with real coefficients that has \(x = 4\) with multiplicity 2 and \(x = -1\) with multiplicity 1.
  • Solve Factors: \((x-4)^2(x+1)\)
    \((x-4)^2 = x^2 - 8x + 16\)
    \((x^2 - 8x + 16)(x+1) = x^3 - 7x^2 + 8x + 16\)
Answer: \(p(x) = (x-4)^2(x+1) = x^3 - 7x^2 + 8x + 16\)
Section 6.10

Approximate Solutions from Graphs

Not every polynomial equation has rational or even expressible exact solutions. For many equations, especially those that arise in applications, the best approach is to use the graph of the polynomial function to identify the approximate location of real roots. The theoretical foundation is the Intermediate Value Theorem.

Intermediate Value Theorem (for polynomials)

If \(p(x)\) is a polynomial and \(p(a)\) and \(p(b)\) have opposite signs, then there is at least one real root between \(a\) and \(b\). When the polynomial value goes from positive to negative (or vice versa), the graph must cross the x-axis somewhere in between.

Example 6.10.1
Using a sign change to locate a root
  • ProblemShow that \(p(x) = x^3 - 3x + 1\) has a root between \(x = 0\) and \(x = 1\).
  • ThinkEvaluate \(p(0)\) and \(p(1)\). If they have opposite signs, the IVT guarantees a root in between.
  • Solve \(p(0) = 0 - 0 + 1 = 1 > 0\)
    \(p(1) = 1 - 3 + 1 = -1 < 0\)
    Sign changes from positive to negative → root exists between 0 and 1.
Answer: Sign change confirmed. Graphing reveals the root is approximately \(x \approx 0.35\).
Example 6.10.2
Narrowing a root by bisection
  • ProblemNarrow the root of \(p(x) = x^3 - 3x + 1\) between 0 and 1 to a smaller interval.
  • Solve Test midpoint \(x = 0.5\):
    \(p(0.5) = 0.125 - 1.5 + 1 = -0.375 < 0\)
    Root is between \(x = 0\) (positive) and \(x = 0.5\) (negative).

    Test \(x = 0.35\):
    \(p(0.35) \approx 0.043 - 1.05 + 1 \approx -0.007 \approx 0\)
    Root is very close to \(x \approx 0.35\).
Answer: Root is approximately \(x \approx 0.35\). Repeatedly bisecting the interval is the bisection method — a foundational numerical technique in computing.
Section 6.11

Technology for Polynomial Equations

For Algebra II, technology is not a replacement for algebraic understanding — but it is a legitimate tool for checking work, graphing, and finding approximate roots that cannot be determined by hand. Most graphing calculators and software tools (Desmos, GeoGebra, TI-84) can graph polynomial functions, identify x-intercepts numerically, evaluate polynomials at specific values, and find approximate roots using built-in solvers.

Limitation of Technology

Technology shows real roots but not complex roots. If a degree-4 polynomial has only 2 real x-intercepts on its graph, it still has exactly 4 roots — the other 2 are complex. Technology confirms the visual, but algebra must confirm the complete picture. Always ask: does the degree confirm the right number of total roots? Are there non-real roots the graph cannot show?

Student Tip

Technology is most valuable when you already understand what the answer should look like. Use algebra to build the framework, and technology to support and verify it.

Practice 6.7–6.11Theorems and Approximation
  1. Solve \(x^3 - 2x^2 + 4x - 8 = 0\). Find all real and complex roots.
  2. A polynomial with real coefficients has degree 5 and exactly 3 real roots. How many complex (non-real) roots does it have?
  3. Write a degree-4 polynomial with real coefficients that has roots \(x = 2\) (multiplicity 2) and \(x = 1 - i\).
  4. Use the Intermediate Value Theorem to show that \(x^3 - x - 1\) has a root between \(x = 1\) and \(x = 2\).
  5. For \(p(x) = (x+2)^2(x-1)^3(x^2+9)\), state the degree, list all roots with multiplicity, and describe graph behavior at each real root.
  6. A degree-6 polynomial graph shows exactly 4 real x-intercepts. How many roots are non-real? Must those non-real roots come in conjugate pairs? Why?
Sections 6.12 – 6.14
Applications, Common Mistakes, and the Road Ahead
Polynomial Models · Seven Recurring Errors · Bridge to Rational Functions
Section 6.12

Polynomial Equations in Applications

Polynomial equations arise naturally in geometry, physics, engineering, finance, and data analysis. In applied settings, the key is to build the polynomial equation from the problem's structure, then solve it and interpret only the solutions that make physical sense. Use the Read → Represent → Solve → Check → Explain method throughout.

Example 6.12.1
An open-top box with a given volume
  • ProblemA company makes open-top boxes by cutting square corners from a 12-by-16-inch piece of cardboard. If the side length of each cut-out square is \(x\) inches, find the value(s) of \(x\) that produce a volume of 105 cubic inches.
  • Represent After cutting and folding:
    Length: \(16-2x\)   Width: \(12-2x\)   Height: \(x\)
    \(V(x) = x(16-2x)(12-2x) = 105\)
  • Solve Expand: \(x(4x^2 - 56x + 192) = 105\)
    \(4x^3 - 56x^2 + 192x - 105 = 0\)
    Context requires \(0 < x < 6\). Using technology, the real roots are approximately \(x \approx 0.62\) and \(x \approx 5.15\). (The third root, \(x \approx 8.23\), lies outside the valid range.)
  • CheckBoth \(x \approx 0.62\) and \(x \approx 5.15\) produce boxes with positive dimensions and volumes near 105 in³. ✓
Answer: Two contextually valid cut sizes — approximately \(x \approx 0.62\) inch and \(x \approx 5.15\) inches.
Example 6.12.2
Returning to the opening question — beam deflection
  • ProblemSolve \(d(x) = x^4 - 6x^3 + 11x^2 - 6x = 0\), where \(x\) is position in meters along a beam.
  • Solve Factor out \(x\): \(x(x^3 - 6x^2 + 11x - 6) = 0\)
    So \(x = 0\) is one root. Solve \(x^3 - 6x^2 + 11x - 6 = 0\).

    Test \(x = 1\): \(1-6+11-6=0\) ✓   Synthetic division gives depressed polynomial \(x^2-5x+6 = (x-2)(x-3)\).

    All roots: \(x = 0,\ 1,\ 2,\ 3\)
  • ExplainThe beam returns to its original horizontal position at four locations: 0, 1, 2, and 3 meters from one end. All roots are real, positive, and physically meaningful.
Answer: The beam deflection equals zero at \(x = 0,\ 1,\ 2,\ 3\) meters.
Practice 6.12Applications
  1. A cube has side length \(s\). Its volume equals three times its surface area. Set up and solve: \(s^3 = 3(6s^2)\). Find all positive solutions.
  2. A company's monthly profit (in thousands) is modeled by \(P(t) = -t^3 + 6t^2 - 9t + 4\) where \(t\) is years since founding. Find all values of \(t \geq 0\) where profit equals zero and interpret each.
  3. A ball's height in feet is \(h(t) = -16t^3 + 48t^2 + 64t\). Find all times \(t \geq 0\) when the ball is at ground level and interpret each.
  4. A garden is surrounded by a path of uniform width \(x\). The garden is 10 ft by 12 ft and the total area (garden + path) is 168 sq ft. Set up and solve a polynomial equation to find \(x\).
Section 6.13

Common Root-Finding Mistakes

Mistake 1 — Forgetting to set the equation equal to zero
Factoring x(x² − 1) = 6 and concluding the roots are 6 or factors of 6

Correct: Rewrite as \(x^3 - x - 6 = 0\) first. The Zero Product Property only works when the product equals zero. A product equal to 6 does not split into factors equal to 6.

Mistake 2 — Listing rational root candidates with wrong denominator
For 3x³ − 5x + 2 = 0, listing only ±1, ±2 as candidates (forgetting the leading coefficient)

Correct: Candidates are \(\pm\dfrac{\text{factors of 2}}{\text{factors of 3}} = \pm 1, \pm 2, \pm \dfrac{1}{3}, \pm \dfrac{2}{3}\). The denominator comes from the leading coefficient, not 1.

Mistake 3 — Stopping after finding one root
Finding x = 2 as a root of a degree-4 polynomial and declaring the answer complete

Correct: After finding \(x = 2\), divide to get the depressed degree-3 polynomial, then continue solving. A degree-\(n\) polynomial has exactly \(n\) roots — one is a start, not a finish.

Mistake 4 — Ignoring complex roots
Saying "there are no more roots" when the depressed quadratic has negative discriminant

Correct: Use the quadratic formula to find the two complex roots and express them as \(a \pm bi\). Polynomial equations never run out of solutions — when real roots are exhausted, complex roots complete the count.

Mistake 5 — Confusing multiplicity with the number of distinct roots
Saying (x − 3)⁴ has four different roots: 3, 3, 3, 3

Correct: There is one distinct root, \(x = 3\), with multiplicity 4. "Four roots counted with multiplicity" means the factor \((x-3)\) accounts for 4 of the polynomial's degree-worth of roots — not four different values.

Mistake 6 — Using the wrong sign in synthetic division
To test x = −3, using c = 3 in synthetic division

Correct: Synthetic division by \((x - c)\) uses \(+c\) in the divisor position. To test the root \(r = -3\), divide by \((x - (-3)) = (x + 3)\), using \(c = -3\) in the synthetic division setup.

Mistake 7 — Accepting all solutions without checking context
Reporting all algebraic roots as valid answers to a physical problem

Correct: After solving algebraically, always check each solution for contextual validity. A root may be mathematically correct but physically impossible — for example, a negative length or a time before the event began.

Section 6.14

Preparing for Rational Functions

Chapter 6 has developed a complete toolkit for polynomial equations. All of these ideas carry forward directly into Chapter 7, which studies rational functions — functions formed by dividing one polynomial by another. In rational functions, the zeros of the numerator become the zeros of the function, and the zeros of the denominator create restrictions, asymptotes, and holes in the graph. Understanding polynomial roots is therefore essential for understanding rational function behavior.

Looking Ahead
Chapter 7: Rational Expressions and Functions

Every factoring strategy, every root-finding method, and every theorem in Chapter 6 will be applied again in Chapter 7 — now in service of a broader and richer class of functions.

  • Factoring the numerator identifies the zeros of the rational function
  • Factoring the denominator identifies where the function is undefined — the excluded values
  • Polynomial roots with multiplicity determine whether a point produces a hole or a vertical asymptote
  • Irreducible quadratic factors in the denominator contribute to the function's structure without producing real asymptotes

Coming up in Chapter 7: simplifying rational expressions, operating with rational expressions, solving rational equations, identifying and interpreting asymptotes and holes, and applying rational models to real-world situations such as average cost, mixture problems, and inverse variation.

Chapter 6 — Back Matter

Chapter Summary

Key Terms Review

polynomial equation
An equation in which a polynomial is set equal to a value, typically zero, with only non-negative integer exponents.
root
A value of \(x\) that satisfies a polynomial equation; also called a solution or, in the context of a function, a zero.
zero
A value where a polynomial function equals zero; corresponds to an x-intercept when the zero is a real number.
multiplicity
The number of times a root \(r\) appears as a factor \((x-r)^k\) in the complete factored form. Determines graph behavior at that root.
Rational Root Theorem
States that every rational root \(p/q\) of an integer-coefficient polynomial satisfies: \(p\) divides the constant term, \(q\) divides the leading coefficient.
candidate rational roots
The complete list of possible rational roots generated by the Rational Root Theorem; each must be tested to confirm.
synthetic division
An efficient algorithm for dividing a polynomial by \((x-c)\) and simultaneously testing whether \(c\) is a root.
depressed polynomial
The quotient after dividing out a known factor; has degree one less than the original polynomial.
Complex Conjugate Root Theorem
If a real-coefficient polynomial has \(a+bi\) as a root, then \(a-bi\) is also a root. Non-real roots come in conjugate pairs.
Fundamental Theorem of Algebra
Every polynomial of degree \(n \geq 1\) has at least one complex root; it has exactly \(n\) roots counted with multiplicity.
irreducible quadratic factor
A quadratic with real coefficients that cannot be factored further over the real numbers; its roots are complex conjugates.
Intermediate Value Theorem
For polynomials: if \(p(a)\) and \(p(b)\) have opposite signs, there is at least one root between \(a\) and \(b\).
bisection method
A numerical technique for approximating roots by repeatedly halving the interval where a sign change occurs.
quadratic form substitution
Replacing \(x^2\) (or another expression) with \(u\) to reveal a quadratic structure in a higher-degree polynomial.

Concept Review Questions

Write complete sentences in your answers.

  1. What does it mean for a polynomial equation to have a "root"? How is a root connected to a zero and to an x-intercept?
  2. Why must you set a polynomial equation equal to zero before applying the Zero Product Property?
  3. A polynomial of degree 5 has exactly 5 roots counted with multiplicity. Explain what this means and give an example where not all 5 roots are distinct real numbers.
  4. What information does the Rational Root Theorem give you? What does it not tell you?
  5. Explain the role of the depressed polynomial in root-finding. Why is it useful?
  6. A polynomial with real coefficients has a root at \(x = 3 + 2i\). What other root must it have, and why?
  7. A polynomial of degree 6 has only 2 real x-intercepts visible on its graph. How many roots does it have in total? Where are the remaining roots?
  8. How does multiplicity affect the shape of a polynomial's graph at a root?
  9. What is the Intermediate Value Theorem, and how does it help locate approximate roots?
  10. Why is technology useful in polynomial root-finding, and what important information can technology not directly provide?

Skill Practice

Part AFactoring Review

Factor each polynomial completely.

  1. \(x^3 - 8x^2 + 15x\)
  2. \(x^3 + 27\)
  3. \(x^4 - 81\)
  4. \(2x^3 - 2x^2 - 12x\)
  5. \(x^3 + x^2 - x - 1\)
  6. \(x^4 - x^3 - 6x^2\)
  7. \(4x^3 - 32\)
  8. \(x^4 - 5x^2 + 4\)
  9. \(x^3 - 3x^2 - 4x + 12\)
  10. \(6x^3 + 7x^2 - 3x\)
Part BRational Root Candidates

List all candidate rational roots. Do not solve.

  1. \(x^3 - 4x + 1\)
  2. \(2x^3 + x^2 - 5x + 2\)
  3. \(3x^4 - x^3 + 2x - 6\)
  4. \(5x^3 - 3x + 10\)
  5. \(x^4 - 7x^2 + 12\)
Part CSynthetic Division

Use synthetic division to test the given value. If it is a root, state the depressed polynomial.

  1. \(p(x) = x^3 - 2x^2 - 5x + 6;\ c = 3\)
  2. \(p(x) = x^3 + x^2 - 8x - 12;\ c = -2\)
  3. \(p(x) = 2x^3 - 3x^2 - 11x + 6;\ c = 2\)
  4. \(p(x) = x^4 - 3x^3 + x^2 + 3x - 2;\ c = 1\)
  5. \(p(x) = x^4 + 2x^3 - 7x^2 - 8x + 12;\ c = -3\)
Part DSolving Polynomial Equations

Find all roots, real and complex.

  1. \(x^3 - 6x^2 + 11x - 6 = 0\)
  2. \(x^3 + 2x^2 - x - 2 = 0\)
  3. \(x^4 - x^3 - 7x^2 + x + 6 = 0\)
  4. \(x^3 - 4x^2 + x + 6 = 0\)
  5. \(2x^3 - x^2 - 7x + 6 = 0\)
  6. \(x^4 - 10x^2 + 9 = 0\)
  7. \(x^3 - 2x^2 + 4x - 8 = 0\)
  8. \(x^4 + 3x^3 + x^2 - 3x - 2 = 0\)
  9. \(x^3 + 8 = 0\)
  10. \(x^4 - 1 = 0\)

Word Problems

Define your variables, write a polynomial equation, solve, and interpret only the contextually valid answers.

  1. A rectangular garden has a length that is 5 feet more than its width. A path of uniform width \(x\) surrounds the garden. If the garden's width is 10 feet and the total area of garden plus path is 600 square feet, find the path width \(x\).
  2. A cube has side length \(s\). Its volume is numerically equal to three times its surface area. Set up and solve the resulting equation. (Volume \(= s^3\), Surface area \(= 6s^2\).)
  3. A company's profit in thousands of dollars is \(P(t) = -t^3 + 6t^2 - 9t + 4\), where \(t\) is years since founding. Find all values of \(t \geq 0\) where profit equals zero and interpret each.
  4. A ball is thrown upward with height \(h(t) = -16t^3 + 48t^2 + 64t\) feet. Find all times \(t \geq 0\) when \(h(t) = 0\). Interpret each solution in context.
  5. A polynomial equation models the displacement of a mechanical vibration: \(d(x) = x^4 - 5x^3 + 5x^2 + 5x - 6 = 0\). Find all roots and identify which represent physically meaningful positions (take \(x \geq 0\)).

Error Analysis

Each problem shows a student's work. Identify the error and give the correct solution.

Problem 1
Solve x²(x − 4) = 0: Student says x² = 0 gives x = 0, and x − 4 = 0 gives x = 4. "Two roots: 0 and 4."

What is wrong? What should the student have said about the root \(x = 0\)?

Problem 2
For p(x) = x³ − 5x² + 6x − 2, the student lists candidates: ±1, ±2, ±5, ±6.

What is wrong with this list? Write the correct candidate list.

Problem 3
Testing x = −3 in p(x) = x³ − x² + 2, the student uses c = 3 in synthetic division and gets remainder 29, concluding x = −3 is not a root.

Identify the error. Re-test \(x = -3\) correctly and state the actual conclusion.

Problem 4
A degree-4 polynomial yields roots x = 1 and x = −2 from synthetic division. The student declares: "The solutions are x = 1 and x = −2."

What likely error has the student made? What must they still do?

Problem 5
The student says: "3 − i is a root, so 3 + i is also a root, but only if the polynomial has even degree."

Correct the student's reasoning. What is the actual condition required?

Chapter 6 Checkpoint

Chapter 6 Checkpoint
Complete without assistance — assesses all key skills and concepts from Chapter 6
Part A — Factoring and Equation Setup
  1. Factor \(x^3 + 2x^2 - 9x - 18\) completely.
  2. Solve \(x^3 - 9x = 0\).
  3. Rewrite \(x^3 + 5x^2 = 2x + 10\) in standard form and factor completely.
Part B — Rational Root Theorem
  1. List all candidate rational roots of \(3x^4 - 2x^3 + x - 6 = 0\).
  2. Determine by substitution whether \(x = 2\) is a root of \(x^3 - 2x^2 - x + 2\).
Part C — Synthetic Division and Depressed Polynomials
  1. Use synthetic division to divide \(x^4 - 3x^3 - 6x^2 + 8x + 8\) by \((x-4)\). State the quotient and remainder.
  2. Use the result from Problem 6 to continue solving \(x^4 - 3x^3 - 6x^2 + 8x + 8 = 0\) completely.
Part D — Complex Roots and Theorems
  1. A polynomial with real coefficients has degree 5 and exactly 3 real roots. How many complex (non-real) roots does it have?
  2. Write a degree-4 polynomial with real coefficients that has roots \(x = 2\) (multiplicity 2) and \(x = 1-i\). Write it in factored form, then expand.
Part E — Applications and Interpretation
  1. Find all roots of \(f(x) = x^3 - 4x^2 + x + 6\) using the Rational Root Theorem and synthetic division. Write \(f(x)\) in completely factored form.
  2. Explain in complete sentences: why does the Fundamental Theorem of Algebra guarantee that a cubic polynomial with real coefficients has at least one real root?
  3. A degree-6 polynomial graph has exactly four x-intercepts, one of which appears to touch the axis without crossing. How many total roots does the polynomial have? Describe the possible root structure — what are the real roots' multiplicities, and how many complex roots might there be?
MGU Library Connection
  • Polynomial Factoring and Division Reference — Appendix F of this textbook
  • Complex Number Reference — Appendix E of this textbook
  • Chapter 4: Complex Numbers — review of \(a + bi\), conjugates, and quadratic solutions with complex roots
  • Chapter 5: Polynomial Functions — degree, end behavior, zeros, multiplicity, and synthetic division
  • Chapter 7: Rational Expressions and Functions — polynomial roots now become zeros and asymptotic restrictions
  • MGU Dictionary entries: Fundamental Theorem of Algebra, multiplicity, Rational Root Theorem, irreducible quadratic factor, synthetic division
  • Precalculus / Trigonometry: polynomial root-finding is extended to rational functions, partial fractions, and polynomial approximation methods