Contents / Algebra / Polynomials & Factoring

Chapter 6

Polynomials & Factoring

Polynomial structure, factoring patterns, division, and rational expressions.

Introduction

Polynomial structure, factoring patterns, division, and rational expressions.

A polynomial is an algebraic expression built from variables and constants using only addition, subtraction, and multiplication, where the variable is raised only to non-negative integer powers. Examples include $3 x^{2} + 2 x - 5$ , $x^{7}$ , and constants like 42.

Polynomials are the most important class of functions in algebra and calculus because they are easy to differentiate and integrate, can approximate many other functions, and their equations can be solved in limited cases by radicals. Every polynomial defines a smooth, continuous function whose graph is a curve that eventually behaves like its highest-power term.

6.1Definition, Degree, and Standard Form

Definition 6.1 (Polynomial in one variable). A polynomial in one variable $x$ has the form

a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{1} x + a_{0},

where the $a_{i}$ are constants (the coefficients) and $n$ is a non-negative integer.

Definition 6.2 (Degree, leading coefficient, standard form). The degree of the polynomial is the largest exponent with a nonzero coefficient. The leading coefficient is $a_{n}$ and the constant term is $a_{0}$ . When the terms are written in descending order of degree, the polynomial is in standard form.

A polynomial of degree 0 is a nonzero constant; degree 1 is linear; degree 2 quadratic, etc. (The zero polynomial is a special case: it has no nonzero coefficient at all, so its degree is left undefined — sometimes written as $- \infty$ .)

Intuition. A polynomial is a "sum of scaled powers" of x, like a recipe with only whole-number scoops of x, x squared, x cubed...

The degree tells you how "curvy" or "steep at the ends" the graph will be.

Example 6.3. $p (x) = - 4 x^{3} + 0 \cdot x^{2} + 7 x - 2$ has degree 3, leading coefficient -4, constant term -2. Written in standard form it is already $- 4 x^{3} + 7 x - 2$ (the $x^{2}$ term may be omitted when coefficient is zero).

Exercises for §6.1 →

6.2Terms, Coefficients, and Like Terms

Definition 6.4 (Term and like terms). Each $a_{k} x^{k}$ is a term, and the number of terms is the count of nonzero coefficients. Like terms have the same power of the variable, and can be combined by adding their coefficients.

Writing in standard form makes it easy to read off degree, leading coefficient, and to perform arithmetic.

Intuition. Terms are the separate "chunks" added together. Like terms are chunks that are the same "flavor" (same power) so they merge.

Zero coefficients are invisible; they don't create a term.

Example 6.5. Consider $5 x^{2} + 3 x - 2 x^{2} + 7 - x$ . Combine: $(5 x^{2} - 2 x^{2}) + (3 x - x) + 7 = 3 x^{2} + 2 x + 7$ . Now three terms, degree 2, leading coeff 3, constant 7.

Exercises for §6.2 →

6.3Adding, Subtracting, and Multiplying Polynomials

Addition and subtraction: combine like terms only. Align by powers if helpful.

Multiplication: use the distributive property (every term of first times every term of second) or special patterns (FOIL for binomials, binomial theorem for higher).

Proposition 6.6 (Degree of a product and of a sum). The degree of a product is the sum of the degrees of the factors:

de g (pq) = de g p + de g q .

The degree of a sum or difference is at most the larger of the two degrees, because the leading terms may cancel.

Intuition. Adding polynomials is like combining two shopping lists that use the same units (powers).

Multiplying is like every flavor from one list multiplies with every flavor from the other; collect the results.

Example 6.7.

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

Example 6.8.

(x - 3) (x + 5) = x^{2} + 5 x - 3 x - 15 = x^{2} + 2 x - 15

Exercises for §6.3 →

6.4Graphs and End Behavior of Polynomial Functions

A polynomial function $p (x)$ is defined for all real x. Its graph is a smooth curve with at most (degree - 1) turning points.

Theorem 6.9 (End behaviour of a polynomial). The end behaviour of $p$ — what happens as $x \to + \infty$ and as $x \to - \infty$ — is determined solely by the leading term $a_{n} x^{n}$ :

$n$ even and $a_{n} > 0$ : both ends $\to + \infty$ .
$n$ even and $a_{n} < 0$ : both ends $\to - \infty$ .
$n$ odd and $a_{n} > 0$ : left $\to - \infty$ , right $\to + \infty$ .
$n$ odd and $a_{n} < 0$ : left $\to + \infty$ , right $\to - \infty$ .

The y-intercept is always the constant term p(0).

Intuition. For large |x| the highest power "wins" and dictates whether the graph shoots up or down at each end.

Even powers eventually look like parabolas (U or upside-down U); odd powers look like tilted lines or cubics at the ends.

Example 6.10. For $p (x) = - 2 x^{3} + 4 x^{2} - x + 7$ , degree 3 odd, leading coeff negative: as $x \to + \infty$ , $p (x) \to - \infty$ ; as $x \to - \infty$ , $p (x) \to + \infty$ . $y$ -intercept = 7.

Exercises for §6.4 →

Common Mistakes

Including negative or fractional exponents: $x^{- 1}$ or $x = x^{1/2}$ are not polynomials.
Forgetting to combine like terms before reading degree or number of terms: 3x^2 + 2x^2 is really one term of degree 2.
Thinking the degree of a sum is always the sum of degrees: It is the max (unless leading terms cancel).
Confusing leading coefficient with constant term: Leading is the one with highest power; constant has power zero.
Assuming a polynomial always has as many real roots as its degree: It can have fewer real roots (the rest are complex).
Omitting the sign when the leading coefficient is negative: End behavior flips when $a_{n} < 0$ .
Writing polynomials with terms out of order: Standard descending powers makes comparison and arithmetic easier.

A monomial is the simplest polynomial: an expression consisting of exactly one term. It is a coefficient multiplied by variables raised to non-negative integer powers. Examples: $7$ , $- 3 x$ , $5 x^{2} y^{3}$ , $\frac{1}{2} a b^{2} c^{4}$ , and even $x^{0} = 1$ .

Monomials are the building blocks of all other polynomials. Understanding how to multiply, divide, and find the degree of monomials is the foundation for working with binomials, factoring, and polynomial division later.

6.5Definition and Examples

Definition 6.11 (Monomial). A monomial in several variables is an expression

c \cdot x_{1}^{k_{1}} x_{2}^{k_{2}} \dots x_{m}^{k_{m}},

where $c$ is a real (or complex) coefficient and each $k_{i}$ is a non-negative integer.

If any exponent is negative or fractional, or if there is more than one term (addition/subtraction), it is not a monomial.

Constants are monomials (all exponents zero). The zero monomial is sometimes treated specially.

Intuition. A monomial is one "chunk": a number times some powers of letters glued together by multiplication only.

No plus or minus signs separating different chunks, and no variables in the bottom of a fraction.

Example 6.12. $- 12 x^{4} y z^{2}$ is a monomial: coefficient -12, variables $x$ to power 4, $y$ to 1, $z$ to 2.

Example 6.13. $2 x + 3$ is not; it has two terms added. $4/ x^{2}$ is not (negative exponent).

Exercises for §6.5 →

6.6Degree and Coefficients

Definition 6.14 (Degree of a monomial). The degree of a monomial is the sum of all the exponents appearing on the variables; if there are no variables the degree is $0$ . The coefficient is the numerical factor in front, including its sign.

Two monomials are called like terms precisely when they contain exactly the same variables raised to exactly the same powers; only their coefficients may differ.

Intuition. Degree counts the total "power weight" of all the letters in the chunk.

The coefficient tells you how many copies of that chunk you have (or how much to scale it).

Example 6.15. For $8 a^{3} b^{5}$ , degree = $3 + 5 = 8$ , coefficient = 8.

Solution.

For the constant monomial $- 5$ , degree = 0, coefficient = $- 5$ .

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Exercises for §6.6 →

6.7Operations with Monomials

Method 6.16 (Arithmetic with monomials).

To multiply two monomials: multiply their coefficients and add the exponents of like variables (the product rule of exponents).
To divide monomials (nonzero): divide the coefficients and subtract the exponents of like variables.
To raise a monomial to a power: raise the coefficient to that power and multiply every exponent by the power.

These rules come directly from the laws of exponents and are used constantly when simplifying polynomials.

Intuition. Multiplying monomials is like combining the "recipes": scale the numbers and collect more powers.

Division is the reverse: scale down and peel off powers.

Example 6.17. Multiply: $(6 x^{2} y^{3}) \cdot (- 2 x y^{4}) = 6 \cdot (- 2) \cdot x^{2 + 1} \cdot y^{3 + 4} = - 12 x^{3} y^{7}$ .

Example 6.18. Divide: $(15 a^{4} b^{2} c) / (3 a b^{2}) = (15/3) a^{4 - 1} b^{2 - 2} c^{1 - 0} = 5 a^{3} c$ (assuming $a, b, c$ nonzero).

Exercises for §6.7 →

Common Mistakes

Treating an expression with + or - as a monomial: x^2 + 1 has two monomials added; the whole is a binomial.
Forgetting to add exponents when multiplying like variables: x^2 * x^3 is x^5, not x^6 or x^{2*3} only if power rule confused.
Subtracting exponents on multiplication: That rule is for division.
Including the coefficient when computing degree: Degree ignores the number in front (except for the zero polynomial).
Allowing negative exponents or variables in denominators: Those produce rational expressions, not monomials.
Confusing like terms with same degree only: x^2 y and x y^2 both degree 3 but are not like terms because variable powers differ.
Dividing by zero when simplifying monomials with variables: Always note the variables cannot be zero if they end up in a denominator after simplification.

A binomial is a polynomial with exactly two terms. It is the sum or difference of two monomials: $a + b$ , $3 x^{2} - 5$ , $x y^{2} - 7 z^{3}$ , etc.

Binomials appear constantly in algebra. Their products give rise to important patterns (difference of squares, perfect squares) that allow fast mental math and factoring. Understanding binomials bridges single-term monomials and more complex polynomials.

6.8Definition and Examples

Definition 6.19 (Binomial). A binomial is any expression that can be written as the addition or subtraction of precisely two monomials (after like terms are combined). Its degree is the degree of its highest-degree term.

If the two terms happen to have the same degree, the binomial may look simpler after combining, but normally the terms have different powers.

Intuition. A binomial is two chunks glued with a + or - .

It is one step more complicated than a single monomial but still very manageable.

Example 6.20. $7 x^{4} - 2 x$ is a binomial of degree 4. Terms: $7 x^{4}$ and $- 2 x$ .

Example 6.21. $a^{2} b + \frac{1}{2} ab$ is a binomial (degree 3). Note coefficients can be fractions.

Exercises for §6.8 →

6.9Adding, Subtracting, and Multiplying Binomials

Addition and subtraction of binomials: combine any like terms that appear. The result may remain a binomial or collapse to a monomial.

Multiplication: distribute each term of the first binomial across the second (four multiplications total). The result before combining is four monomials; after combining like terms you usually obtain a trinomial (degree = sum of the two degrees) or sometimes a binomial when middle terms cancel.

Intuition. Adding two binomials is like pouring two small piles of terms into one bucket and merging any that match.

Multiplying is every piece of the first times every piece of the second; then tidy up.

Example 6.22. $(5 x + 3) + (2 x - 8) = 7 x - 5$ (still binomial).

Solution.

$(5 x + 3) - (2 x - 8) = 3 x + 11$ .

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Example 6.23. $(x + 4) (2 x - 1) = x * 2 x + x * (- 1) + 4 * 2 x + 4 * (- 1) = 2 x^{2} - x + 8 x - 4 = 2 x^{2} + 7 x - 4$ (trinomial).

Exercises for §6.9 →

6.10Special Products and Patterns

Three patterns are worth memorizing, because they appear everywhere.

Theorem 6.24 (Special products).

(a + b) (a - b) = a^{2} - b^{2} (difference of squares; the middle terms cancel)

(a + b)^{2} = a^{2} + 2 ab + b^{2}, (a - b)^{2} = a^{2} - 2 ab + b^{2}

(a + b) (c + d) = a c + a d + b c + b d

Recognizing these patterns in reverse is the start of factoring.

Intuition. The difference of squares is a "magic cancellation" that removes the middle.

Squaring a binomial always produces the middle cross term 2ab (twice the product).

Example 6.25. $(x + 7) (x - 7) = x^{2} - 49$ . Fast, no middle term to write.

Example 6.26.

(3 x - 2)^{2} = (3 x)^{2} - 2 * 3 x * 2 + 4 = 9 x^{2} - 12 x + 4

Exercises for §6.10 →

Common Mistakes

Counting a binomial after it collapses: If after adding you get only one term, it is no longer a binomial.
Forgetting the middle term when squaring: (a + b)^2 is never a^2 + b^2.
Sign errors on the middle term of (a - b)^2: It is a^2 - 2ab + b^2 (minus).
Treating (a + b)(a - b) as a^2 + b^2 or a^2 - 2ab + b^2: It is strictly a^2 - b^2.
Not combining like terms after distribution: Leaving four separate terms when two can merge.
Assuming every product of binomials is another binomial: Most expand to trinomials.
Confusing binomial with binomial coefficient (in combinatorics): Here we mean the polynomial with two terms only.

Factorization (or factoring) is the process of rewriting a polynomial as a product of simpler polynomials (or constants). It is the reverse of expanding.

Factoring is one of the most powerful tools in algebra: it lets us solve polynomial equations, simplify rational expressions, find zeros of functions, and recognize patterns that would otherwise be hidden. The goal is always to factor completely over the reals (or as far as possible).

6.11Greatest Common Factor (GCF)

The first step with any polynomial is to factor out the greatest common factor of all its terms, if it exists and is not $1$ .

Method 6.27 (Factoring out the GCF).

Take the GCF of the numerical coefficients.
Take the lowest power of each variable that appears in every term.
Write that product out front, and divide each term by it to get what stays inside.

After pulling out the GCF, the remaining polynomial should have no common factor greater than 1 in all terms.

Intuition. The GCF is the biggest "chunk" that divides every term evenly. Pull it out front like a common multiplier.

What is left inside is the "reduced" version of the expression.

Example 6.28. For $12 x^{4} - 8 x^{3} + 4 x^{2}$ , GCF of coeffs is 4, lowest x power is $x^{2}$ , so $4 x^{2}$ . Inside: $3 x^{2} - 2 x + 1$ . Thus $4 x^{2} (3 x^{2} - 2 x + 1)$ .

Example 6.29.

6 a^{2} b^{3} - 9 a b^{4} + 3 a b^{2} = 3 a b^{2} (2 ab - 3 b^{2} + 1)

Exercises for §6.11 →

6.12Factoring by Grouping

Method 6.30 (Factoring by grouping).

When a polynomial has four or more terms, group them in pairs (or sometimes the first three and the last) so that a common binomial factor appears.
Factor each group.
Pull out the common binomial factor that emerges.

If grouping does not produce a common factor, try a different grouping or conclude it does not factor nicely that way.

Intuition. Grouping is like pairing socks so that each pair has something in common; then the pairs themselves share a bigger pattern.

It turns a messy four-term expression into two factors.

Example 6.31.

x^{3} + x^{2} + 4 x + 4 = (x^{3} + x^{2}) + (4 x + 4) = x^{2} (x + 1) + 4 (x + 1) = (x^{2} + 4) (x + 1)

Example 6.32.

2 x^{3} - 6 x^{2} - 5 x + 15 = (2 x^{3} - 6 x^{2}) + (- 5 x + 15) = 2 x^{2} (x - 3) - 5 (x - 3) = (2 x^{2} - 5) (x - 3)

Exercises for §6.12 →

6.13Factoring Trinomials

Method 6.33 (Factoring a quadratic trinomial). For a quadratic trinomial $a x^{2} + b x + c$ with $a \neq = 0$ :

If $a = 1$ : look for two numbers $m$ , $n$ with $mn = c$ and $m + n = b$ ; then the factorisation is $(x + m) (x + n)$ .
If $a > 1$ : either use trial and error with the factors of $a$ and of $c$ , or use the ac method — find factors of $a c$ that add to $b$ , split the middle term, then group.

Always check by expanding the proposed factors.

Many trinomials do not factor nicely over the integers (prime over the integers).

Intuition. You are looking for two binomials whose "outer" and "inner" cross terms will add up to the middle coefficient.

It's like a number puzzle: find the right pair that multiplies to the constant and adds to the linear coeff.

Example 6.34. Factor $x^{2} - 5 x + 6$ . Numbers that multiply to $6$ and add to $- 5$ : since the product is positive but the sum is negative, both must be negative — try $- 2$ and $- 3$ ( $(- 2) (- 3) = 6$ , $- 2 + (- 3) = - 5$ ). So $(x - 2) (x - 3)$ . Check: $(x - 2) (x - 3) = x^{2} - 3 x - 2 x + 6 = x^{2} - 5 x + 6$ . ✓

Example 6.35. Factor $3 x^{2} + 10 x - 8$ . Here $a = 3, c = - 8$ , so $a c = 3 \cdot (- 8) = - 24$ . Find factors of $- 24$ that add to $b = 10$ : $12$ and $- 2$ ( $12 \cdot (- 2) = - 24$ , $12 + (- 2) = 10$ ). Split the middle term: $3 x^{2} + 12 x - 2 x - 8$ . Group in pairs: $3 x (x + 4) - 2 (x + 4) = (3 x - 2) (x + 4)$ . Check: $(3 x - 2) (x + 4) = 3 x^{2} + 12 x - 2 x - 8 = 3 x^{2} + 10 x - 8$ . ✓

Exercises for §6.13 →

6.14Using Special Identities

Theorem 6.36 (Factoring identities).

a^{2} - b^{2} = (a - b) (a + b) (difference of squares)

a^{2} + 2 ab + b^{2} = (a + b)^{2}, a^{2} - 2 ab + b^{2} = (a - b)^{2}

a^{3} + b^{3} = (a + b) (a^{2} - ab + b^{2}), a^{3} - b^{3} = (a - b) (a^{2} + ab + b^{2})

These allow instant factoring when the polynomial matches the left-hand side exactly.

Intuition. These are the "cheat codes" of factoring. If it looks exactly like one of the patterns, you get the factors for free.

Always check the signs and the middle cross term (or its absence).

Example 6.37.

9 x^{2} - 25 = (3 x)^{2} - 5^{2} = (3 x - 5) (3 x + 5)

Example 6.38. $x^{2} + 6 x + 9 = (x + 3)^{2}$ (perfect square, notice the middle is twice $3 \cdot x$ ).

Exercises for §6.14 →

Common Mistakes

Forgetting to factor out the GCF first: You may miss the complete factorization or make numbers harder.
Wrong sign pairs when factoring trinomials: If product positive but sum negative, both numbers negative.
Stopping too early: "Factor completely" means keep going until no factor can be factored further over the reals.
Assuming every quadratic factors: Many quadratics are irreducible (discriminant negative or not perfect square for integer coeffs).
Mishandling the ac-method split: The two numbers must multiply to a*c and add to b; then attach the correct original a coefficient to the groups.
Confusing sum of squares with difference of squares: x^2 + 4 cannot be factored over the reals (no real b such that b^2 = -4).
Forgetting to check by multiplying back: Always expand your factors to verify they equal the original polynomial.

The difference of two squares is the identity $a^{2} - b^{2} = (a - b) (a + b)$ . It is one of the most frequently used factoring patterns in algebra.

Whenever you see a quantity squared minus another quantity squared, you can factor it instantly into a pair of conjugate binomials. The pattern also helps solve equations and simplify expressions involving squares.

6.15The Algebraic Pattern

Theorem 6.39 (Difference of squares). For any expressions $A$ and $B$ ,

A^{2} - B^{2} = (A - B) (A + B) .

Proof. Expand the right-hand side:

(A - B) (A + B) = A \cdot A + A \cdot B - B \cdot A - B \cdot B = A^{2} + (A B - A B) - B^{2} = A^{2} - B^{2} .

∎

The middle terms always cancel when the binomials are conjugates (same terms, opposite signs).

Intuition. Imagine two squares side by side; their difference can be cut into two rectangular strips whose lengths are (A-B) and widths (A+B) or vice versa.

The cross pieces cancel in the algebra, leaving pure difference.

Example 6.40.

x^{2} - 16 = x^{2} - 4^{2} = (x - 4) (x + 4)

Example 6.41.

49 m^{2} - 25 n^{4} = (7 m)^{2} - (5 n^{2})^{2} = (7 m - 5 n^{2}) (7 m + 5 n^{2})

Exercises for §6.15 →

6.16Recognizing and Applying the Pattern

Method 6.42 (Using the difference of squares).

Confirm the expression is exactly "something squared" minus "something else squared".
Identify $A$ and $B$ (they may themselves be binomials or carry coefficients).
Write the factors $(A - B) (A + B)$ .
Check whether either factor can be factored further (often the sum factor cannot, over the reals).

Higher even powers can be treated as squares of lower powers: x^4 - 1 = (x^2)^2 - 1^2 , then factor more.

Intuition. Look for the two square signs and the minus between them.

Pull the square roots out as the A and B; the signs in the factors are always opposite.

Example 6.43.

4 x^{2} - 9 y^{2} = (2 x)^{2} - (3 y)^{2} = (2 x - 3 y) (2 x + 3 y)

Example 6.44. $x^{4} - 81 = (x^{2})^{2} - 9^{2} = (x^{2} - 9) (x^{2} + 9)$ . Then $x^{2} - 9$ factors again as difference of squares: $(x - 3) (x + 3)$ . Overall $(x - 3) (x + 3) (x^{2} + 9)$ .

Exercises for §6.16 →

6.17Solving Equations Using the Pattern

Set a difference of squares equal to zero and factor: (A - B)(A + B) = 0.

Then A - B = 0 or A + B = 0.

This usually produces two solutions that are opposites when the constant is a perfect square and there is no linear term.

Be careful: after factoring, discard any solutions that make original denominators zero if rational expressions are involved, though pure polynomial equations have no such restriction.

Intuition. Factoring turns the equation into "one of these two pieces must be zero".

Each piece gives a simple linear equation.

Example 6.45. Solve $x^{2} - 121 = 0$ .

Solution.

$(x - 11) (x + 11) = 0 \Rightarrow x = 11$ or $x = - 11$ .

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Example 6.46. Solve $16 x^{2} - 25 = 0$ .

Solution.

$(4 x - 5) (4 x + 5) = 0 \Rightarrow 4 x = 5$ or $4 x = - 5 \Rightarrow x = 5/4$ or $x = - 5/4$ .

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Exercises for §6.17 →

Common Mistakes

Writing both factors with the same sign: (a + b)(a + b) would give a^2 + 2ab + b^2, not the difference.
Forgetting that the sum factor (A + B) usually does not factor further over the reals: A^2 + B^2 has no real roots.
Treating x^2 - 7 as difference of squares over integers: 7 is not a perfect square, so it stays (x - sqrt(7))(x + sqrt(7)) or remains unfactored in integer coeffs.
Missing higher-power cases: x^4 - 16 is (x^2 - 4)(x^2 + 4) and then x^2-4 factors again.
Sign error when A or B has a minus already: e.g. x^2 - (y - 1)^2 requires care with the inner parentheses.
Assuming the pattern works for sum of squares: x^2 + 4 never factors nicely over reals.
Not checking the original expression really is squares: x^2 - 2x + 1 is (x-1)^2, a square, not a difference.

The sum and difference of cubes identities are:

a^{3} + b^{3} = (a + b) (a^{2} - ab + b^{2})

a^{3} - b^{3} = (a - b) (a^{2} + ab + b^{2})

These allow us to factor cubic expressions that match the pattern exactly. Unlike difference of squares, the quadratic factor that results is always irreducible over the real numbers.

They appear when solving cubic equations and when simplifying expressions with cubes.

6.18Sum of Cubes

Theorem 6.47 (Sum of cubes). Whenever two cubed terms are added, they factor as the sum of the two roots times a quadratic:

a^{3} + b^{3} = (a + b) (a^{2} - ab + b^{2}) .

Proposition 6.48 (The quadratic factor is irreducible). For $b \neq = 0$ , the quadratic factor $a^{2} - ab + b^{2}$ never breaks into real linear factors.

Proof. Treated as a quadratic in $a$ , the factor $a^{2} - ab + b^{2}$ has discriminant

(- b)^{2} - 4 b^{2} = - 3 b^{2},

which is negative for every $b \neq = 0$ .∎

Intuition. Sum of cubes is like the volume of two cubes combined can be expressed using the side lengths in a factored way.

The quadratic "leftover" is always "bulky" and doesn't break into real linears.

Example 6.49.

x^{3} + 8 = x^{3} + 2^{3} = (x + 2) (x^{2} - 2 x + 4)

Example 6.50.

27 + y^{3} = 3^{3} + y^{3} = (3 + y) (9 - 3 y + y^{2})

Exercises for §6.18 →

6.19Difference of Cubes

Theorem 6.51 (Difference of cubes). For a subtraction of two cubes the sign in the quadratic flips to a plus for the middle term:

a^{3} - b^{3} = (a - b) (a^{2} + ab + b^{2}) .

Again the quadratic factor $a^{2} + ab + b^{2}$ has discriminant $- 3 b^{2} < 0$ and stays prime over the reals.

Intuition. Difference of cubes works the same as sum of cubes, but the sign inside the quadratic flips to a plus: $a^{3} - b^{3} = (a - b) (a^{2} + ab + b^{2})$ .

Memory hook: the sign of the linear factor $(a - b)$ copies the sign between the cubes (minus here), and the middle term of the quadratic always takes the opposite sign (plus here). "SOAP": Same, Opposite, Always Positive.

Example 6.52.

x^{3} - 1 = (x - 1) (x^{2} + x + 1)

Example 6.53.

64 m^{3} - 27 n^{3} = (4 m - 3 n) ((4 m)^{2} + (4 m) (3 n) + (3 n)^{2}) = (4 m - 3 n) (16 m^{2} + 12 mn + 9 n^{2})

Exercises for §6.19 →

6.20Recognizing the Patterns and Solving Equations

Method 6.54 (Applying the cube identities).

Confirm the expression is a pure cube plus or minus a pure cube, with no other terms.
Write it as $A^{3} \pm B^{3}$ , where $A$ and $B$ may contain coefficients or other variables.
Apply the formula.
Factor any further difference of squares that appears in the factors (as in $x^{6} - 64$ ).

When solving A^3 - B^3 = 0 or similar set to zero, the real solution is usually just A = B (the quadratic contributes none).

Intuition. Spot the two cubed pieces and the + or - between them.

Write the sum/diff of roots times the "guard" quadratic.

Example 6.55. Solve $x^{3} + 27 = 0$ .

Solution.

$(x + 3) (x^{2} - 3 x + 9) = 0$ .
$x + 3 = 0 \Rightarrow x = - 3$ (the only real solution).

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Example 6.56. Factor $x^{6} - 64$ completely over the integers.

Solution.

$x^{6} - 64 = (x^{2})^{3} - 4^{3} = (x^{2} - 4) (x^{4} + 4 x^{2} + 16) = (x - 2) (x + 2) (x^{4} + 4 x^{2} + 16)$ .

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Exercises for §6.20 →

Common Mistakes

Mixing up the signs in the quadratic: For sum of cubes the middle is minus; for difference it is plus.
Thinking the quadratic always factors: It never does over the reals for nonzero b.
Forgetting that A or B can be more than a single variable: (2x)^3 + 1 still works with A=2x, B=1.
Treating x^3 + y^3 as (x + y)^3: That expands to three terms plus 3xy terms; not the same.
Stopping after the first factor when further factoring is possible: x^6 - 64 can go to linear factors for the diff-of-squares part.
Trying to factor sum of squares with these formulas: They apply only to cubes.
Sign error on the linear factor: For a^3 - b^3 the (a - b) has the minus; for sum it is plus.

Polynomial long division is the process of dividing one polynomial (the dividend) by another (the divisor) to produce a quotient polynomial and a remainder polynomial whose degree is strictly less than the divisor's degree.

It is the direct analogue of the long division algorithm you learned for numbers and is the general method when the divisor is not linear (synthetic division is a shortcut only for linear divisors).

The fundamental identity is:

dividend = (divisor) × (quotient) + remainder

6.21The Long Division Algorithm

Method 6.57 (Polynomial long division). For a divisor of any degree:

Arrange both polynomials in standard descending powers, inserting $0$ coefficients for missing terms.
Divide the leading term of the current dividend by the leading term of the divisor; this is the next term of the quotient.
Multiply the entire divisor by this quotient term.
Subtract the result from the current dividend, aligning like terms.
Bring down the next term (or the remainder so far becomes the new dividend).
Repeat until the degree of the remaining polynomial is less than the degree of the divisor.

The final remaining polynomial is the remainder.

Intuition. It is exactly like dividing big numbers: guess the next digit (here the next term), multiply, subtract, bring down.

You keep going until what's left is "smaller" than the divisor (lower degree).

Example 6.58. Divide $x^{2} + 5 x + 6$ by $x + 2$ .

Solution.

$x^{2} / x = x$ (first quotient term).
$x * (x + 2) = x^{2} + 2 x$ .
Subtract from $x^{2} + 5 x + 6$ : $(5 x - 2 x) + 6 = 3 x + 6$ .
$3 x / x = 3$ .
$3 * (x + 2) = 3 x + 6$ .
Subtract: 0.
Quotient $x + 3$ , remainder 0.

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Exercises for §6.21 →

6.22Quotient and Remainder

Theorem 6.59 (Division algorithm for polynomials). Given a dividend $f (x)$ and a nonzero divisor $d (x)$ , there are a unique quotient $Q (x)$ and a unique remainder $R (x)$ with $de g R < de g d$ such that

f (x) = d (x) Q (x) + R (x) .

If $R = 0$ , the divisor is an exact factor of the dividend.

The quotient has degree = deg(dividend) - deg(divisor) (when exact or before remainder).

This identity is the basis for polynomial rings, partial fraction decomposition, and many algebraic manipulations.

Intuition. The remainder is what is "left over" that you cannot divide any more.

The equation is like "total = groups × size + leftovers".

Example 6.60. Divide $2 x^{3} + 3 x^{2} - x + 5$ by $(x + 1)$ .

Solution.

$2 x^{3} / x = 2 x^{2}$ (first quotient term). $2 x^{2} (x + 1) = 2 x^{3} + 2 x^{2}$ . Subtract: $(3 x^{2} - 2 x^{2}) - x + 5 = x^{2} - x + 5$ .
$x^{2} / x = x$ . $x (x + 1) = x^{2} + x$ . Subtract: $(- x - x) + 5 = - 2 x + 5$ .
$- 2 x / x = - 2$ . $- 2 (x + 1) = - 2 x - 2$ . Subtract: $5 - (- 2) = 7$ . Degree $0 <$ degree of divisor, so we stop.
Quotient $2 x^{2} + x - 2$ , remainder $7$ .
Check: $(x + 1) (2 x^{2} + x - 2) + 7 = 2 x^{3} + x^{2} - 2 x + 2 x^{2} + x - 2 + 7 = 2 x^{3} + 3 x^{2} - x + 5$ . Matches the dividend. ✓

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Exercises for §6.22 →

6.23Examples and Applications

Long division is used to:

Factor polynomials when a factor is suspected (if remainder 0).
Rewrite improper rational expressions (deg num >= deg den) as polynomial + proper fraction.
Prove the factor theorem or remainder theorem (when divisor is linear, remainder = f(c) for root c).

Always write missing powers with zero coefficients to keep columns aligned.

Intuition. When the "fraction" of polynomials is top-heavy, division gives the integer part (the quotient polynomial) plus a leftover fraction.

It turns hard division into addition and multiplication checks.

Example 6.61. Divide $x^{3} - 8$ by $(x - 2)$ .

Solution.

$x^{3} / x = x^{2}$ . $x^{2} (x - 2) = x^{3} - 2 x^{2}$ . Subtract: $2 x^{2}$ .
$2 x^{2} / x = 2 x$ . $2 x (x - 2) = 2 x^{2} - 4 x$ . Subtract: $4 x$ .
$4 x / x = 4$ . $4 (x - 2) = 4 x - 8$ . Subtract: $0$ .
Quotient $x^{2} + 2 x + 4$ , remainder 0. (Indeed $x^{3} - 8 = (x - 2) (x^{2} + 2 x + 4)$ .)

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Exercises for §6.23 →

Common Mistakes

Forgetting zero coefficients for missing powers: Columns misalign and you drop terms or get wrong signs.
Subtracting incorrectly (sign errors): Remember "subtract" means add the opposite; distribute the minus carefully.
Stopping when remainder degree equals divisor degree: Must continue until strictly lower degree.
Assuming remainder is always zero: Most divisions leave a nonzero remainder.
Confusing with synthetic division: Synthetic is faster only for linear divisors; long division works generally and shows the full process.
Writing the quotient with wrong degree: It should be exactly deg(dividend) - deg(divisor) before considering remainder.
Not verifying the final identity: Always multiply back (divisor * quotient + remainder) and confirm it equals the dividend.

Synthetic division is a compact, efficient method for dividing a polynomial by a linear binomial of the form (x − c). It uses only the coefficients of the dividend and the number c, avoiding the repeated writing of the divisor and variables.

It is especially useful for testing possible rational roots (by the rational root theorem) and for evaluating polynomials via the remainder theorem.

6.24When and Why to Use Synthetic Division

Use synthetic division when the divisor is exactly (x − c) for some number c (c can be negative or fractional).

It cannot be used directly for quadratic or higher-degree divisors; fall back to long division in those cases.

The method produces the same quotient and remainder as long division but in far less space and time.

Intuition. Long division writes the whole (x - c) over and over. Synthetic keeps only the "c" and the numbers, doing the multiply-and-add dance in a single row.

It is the "fast lane" for linear divisors.

Example 6.62. To divide by $(x - 2)$ you only need the number 2 and the coefficients of the dividend. No need to write " $x - 2$ " repeatedly.

Exercises for §6.24 →

6.25The Synthetic Division Procedure

Method 6.63 (Synthetic division). Set up:

Write the coefficients of the dividend in a row, including $0$ for any missing powers.
Write the root $c$ to the left (or the right — conventions vary).
Bring the first coefficient straight down.

Then for each subsequent column:

Multiply the number just brought down (or previously computed) by $c$ .
Add that product to the next coefficient in the dividend row.
Write the sum below.

The numbers in the bottom row (all but the very last) are the coefficients of the quotient, in descending order. The final number is the remainder.

For negative c the signs take care of themselves through the arithmetic.

Intuition. Bring down, multiply by c, add to next, repeat.

It's like a chain of "scale by c and accumulate".

Example 6.64. Divide $x^{3} - 6 x^{2} + 11 x - 6$ by $(x - 1)$ , $c = 1$ . Coefficients: 1 -6 11 -6

  1 | 1   -6   11   -6
        |     1    -5    6
        ---------------
          1   -5    6    0

Quotient $x^{2} - 5 x + 6$ , remainder 0.

Example 6.65. Divide $3 x^{3} + 0 x^{2} - 5 x + 7$ by $(x + 2)$ , so $c = - 2$ (because $x + 2 = x - (- 2)$ ). Coefficients: $3, 0, - 5, 7$ .

  -2 | 3    0    -5     7
     |     -6    12   -14
     ------------------------
       3   -6     7    -7

Bring down $3$ . $3 \cdot (- 2) = - 6$ ; add to $0$ to get $- 6$ . $- 6 \cdot (- 2) = 12$ ; add to $- 5$ to get $7$ . $7 \cdot (- 2) = - 14$ ; add to $7$ to get $- 7$ . Reading the bottom row: quotient $3 x^{2} - 6 x + 7$ , remainder $- 7$ . Check: $(x + 2) (3 x^{2} - 6 x + 7) - 7 = 3 x^{3} - 6 x^{2} + 7 x + 6 x^{2} - 12 x + 14 - 7 = 3 x^{3} - 5 x + 7$ . Matches. ✓

Exercises for §6.25 →

6.26Interpreting Results and the Remainder Theorem

The bottom row gives the coefficients of $Q (x)$ , followed by $R$ .

Theorem 6.66 (Remainder theorem). When a polynomial $f$ is divided by $x - c$ , the remainder is

R = f (c) .

So synthetic division instantly evaluates $f$ at $c$ .

Theorem 6.67 (Factor theorem). If $R = 0$ then $x - c$ is an exact factor of $f$ , and $c$ is a root of $f$ .

The quotient row can be used to factor f(x) = (x - c) * Q(x) when R=0, then repeat on Q if desired.

Intuition. The last number is "what you get when you plug c into the polynomial" without doing the full substitution.

Zero at the end means you found a root and can peel off that linear factor.

Example 6.68. For $f (x) = x^{3} + 4 x^{2} - x + 7$ , run synthetic division with $c = - 2$ . Coefficients: $1, 4, - 1, 7$ .

Solution.

Bring down $1$ . Then $1 \cdot (- 2) = - 2$ , and $4 + (- 2) = 2$ .
$2 \cdot (- 2) = - 4$ , and $- 1 + (- 4) = - 5$ .
$- 5 \cdot (- 2) = 10$ , and $7 + 10 = 17$ .
Bottom row is $1, 2, - 5, 17$ : quotient $x^{2} + 2 x - 5$ , remainder $17$ .
By the remainder theorem, $f (- 2) = 17$ . Sanity check by direct substitution: $(- 2)^{3} + 4 (- 2)^{2} - (- 2) + 7 = - 8 + 16 + 2 + 7 = 17$ . ✓

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Exercises for §6.26 →

Common Mistakes

Forgetting to insert zero coefficients for missing powers: The columns shift and all subsequent numbers become wrong.
Using the wrong c (sign error): For divisor (x + 3) you must use c = -3, not +3.
Misreading which numbers are the quotient: All bottom numbers except the very last (remainder) form the quotient.
Thinking synthetic works for non-linear divisors: It does not; the multiply-by-single-c assumption fails.
Arithmetic slips on negative c: The signs propagate correctly if you multiply and add carefully each step.
Stopping before writing the remainder: The process always produces one more bottom number than the number of quotient coefficients.
Confusing the bottom row with the original coefficients: They are the derived successive sums/products.

A rational expression is a quotient of two polynomials p(x)/q(x), where q(x) is not the zero polynomial. When viewed as a function it is called a rational function.

Rational expressions are the algebraic analogue of numeric fractions. They can be simplified, multiplied, divided, added, and subtracted using the same rules, provided we carefully track the values that make any denominator zero (the excluded values from the domain).

6.27Definition and Domain

Definition 6.69 (Rational expression). A rational expression is a quotient

\frac{p ( x )}{q ( x )}

of two polynomials with $q$ not the zero polynomial. Its domain is all real numbers except the roots of $q (x)$ .

If $de g p < de g q$ the expression is proper; otherwise it is improper, and division rewrites it as a polynomial plus a proper rational expression.

The expression is undefined wherever the denominator vanishes, even if the numerator also vanishes there (a hole may appear after simplification).

Intuition. Think of it as a fraction where the top and bottom are built from powers of x instead of plain numbers.

You can never let the bottom become zero; those x values are forbidden from the start.

Example 6.70. For (x^2 + 1)/(x - 5) the domain is all reals except x=5.

Example 6.71. (x^2 - 1)/(x - 1) has domain all reals except x=1, even though it simplifies to x+1 (with a hole at x=1).

Exercises for §6.27 →

6.28Simplifying Rational Expressions

Method 6.72 (Simplifying a rational expression).

Factor numerator and denominator completely.
Cancel any common factors, provided they are not zero at the points of interest.
State the excluded values that were cancelled — the simplified expression equals the original only where the cancelled factors were nonzero.

Never cancel before factoring, and never cancel constants that are not factors in the same way.

Intuition. Canceling is removing a matching "brick" from top and bottom, but only if that brick isn't zero (because zero bricks would have made the whole thing undefined).

The simplified version is valid everywhere except the removed points.

Example 6.73. (x^2 - 4x + 3)/(x - 1) = (x-1)(x-3)/(x-1) = x - 3 , for x ≠ 1.

Example 6.74. (2x^2 + 6x)/(2x^2 + 4x) = 2x(x + 3) / 2x (x + 2) = (x + 3)/(x + 2) , for x ≠ 0 and x ≠ -2.

Exercises for §6.28 →

6.29Multiplication and Division

To multiply: multiply numerators and multiply denominators, then simplify by canceling common factors before or after.

To divide: multiply by the reciprocal of the second expression, then simplify.

After any operation, state the domain of the result (union of restrictions).

Intuition. Multiplying fractions: tops together, bottoms together, then cancel what you can.

Dividing fractions: flip the second and multiply.

Example 6.75. (x + 3)/(x - 2) * (x^2 - 4)/(x + 1) = (x + 3)/(x - 2) * (x - 2)(x + 2)/(x + 1) = (x + 3)(x + 2)/(x + 1) , for x ≠ 2, x ≠ -1.

Example 6.76. Divide $\frac{x - 1}{x + 4} \div \frac{x ^{2} - 1}{2 x + 8}$ .

Solution.

Flip the second fraction and multiply: $\frac{x - 1}{x + 4} \cdot \frac{2 x + 8}{x ^{2} - 1}$ .
Factor everything: $2 x + 8 = 2 (x + 4)$ and $x^{2} - 1 = (x - 1) (x + 1)$ . So the product is $\frac{( x - 1 ) \cdot 2 ( x + 4 )}{( x + 4 ) ( x - 1 ) ( x + 1 )}$ .
Cancel the common $(x + 4)$ and $(x - 1)$ : result $\frac{2}{x + 1}$ .
Domain: the excluded values come from every denominator seen along the way — the original $x + 4 \neq = 0$ , and dividing by $\frac{x ^{2} - 1}{2 x + 8}$ requires $x^{2} - 1 \neq = 0$ and $2 x + 8 \neq = 0$ . So $x \neq = - 4, x \neq = 1, x \neq = - 1$ .

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Exercises for §6.29 →

6.30Adding and Subtracting

Method 6.77 (Adding and subtracting rational expressions).

Find a common denominator; the LCD is often the product of the distinct factors, or the LCM of the factored denominators.
Rewrite each fraction with that common bottom.
Add or subtract the numerators.
Simplify the resulting single fraction if possible.

Intuition. To add "apple slices over different plates" you need the same plate size (common denominator), then add the apple counts on top.

The final plate size may be reducible.

Example 6.78. 1/x + 2/(x + 1) = (x + 1 + 2x ) / [x(x + 1)] = (3x + 1)/(x^2 + x) , for x ≠ 0, x ≠ -1.

Example 6.79. (x)/(x^2 - 9) - 1/(x - 3) = x/((x-3)(x+3)) - (x+3)/((x-3)(x+3)) = [x - (x + 3)] / (x^2 - 9) = -3 / (x^2 - 9) , for x ≠ ±3.

Exercises for §6.30 →

Common Mistakes

Canceling before factoring or canceling non-common factors: (x+1)/(x+2) cannot cancel the +1 and +2.
Forgetting to exclude canceled values from the domain: The simplified expression is not equal to the original at those points (holes).
Using the wrong common denominator when adding: The LCD must be a multiple of each original denominator.
Distributing the minus sign incorrectly when subtracting: (a/b - c/d) requires minus on the entire second numerator after common den.
Assuming the domain is only the final denominator after simplifying: Original restrictions from any intermediate denominator carry forward if not canceled properly.
Treating rational expressions like polynomials (adding denominators etc.): They follow fraction rules, not polynomial rules.
Ignoring that an improper rational expression can first be divided to a polynomial + proper fraction: Sometimes that is the simplest form.

About the practice questions. They check that you can carry out this chapter's computations correctly, and each one is graded on a single answer. They are not proof exercises: working through them confirms the mechanics, not that you could prove the results yourself. For that, re-read the statements above and try to reconstruct their proofs with the page closed.

Practice this chapter