| Primary source | OpenStax College Algebra 2e, Section 1.4: "Polynomials" |
| Direct link | https://openstax.org/books/college-algebra-2e/pages/1-4-polynomials |
| Supplementary | OpenStax College Algebra 2e, Section 1.6: "Rational Expressions" |
| Supplementary link | https://openstax.org/books/college-algebra-2e/pages/1-6-rational-expressions |
College Algebra 2e Section 1.4 begins on page 58 of the PDF.
Two algebraic expressions that look different can be the same function.
The expression $\dfrac{x^2 - 4}{x - 2}$ and the expression $x + 2$ have the same value for every $x \neq 2$. They are algebraically equivalent after canceling the common factor $(x - 2)$. Simplification is the process of finding the form that reveals structure most clearly.
"Simpler" has a technical meaning here: a form is simpler when it has no compound fractions, no unfactored common factors, no irrational expressions in a denominator, and no like terms left uncombined. Each of these is a specific goal, not a vague aesthetic preference.
The point of simplification is not to make expressions shorter for their own sake. It is to put expressions in a form where further work -- solving, evaluating, differentiating, integrating -- is possible or easier.
Combining like terms: Add or subtract coefficients of terms with identical variable parts. \[ 5x^2 - 3x + 2x^2 + 7x = 7x^2 + 4x \]
Distributive property: $a(b + c) = ab + ac$ (works in both directions).
Fraction arithmetic with variables (LCD method): \[ \frac{a}{b} + \frac{c}{d} = \frac{ad + bc}{bd} \] (Use the LCD, not the product $bd$, when simplification is needed.)
Canceling in rational expressions: Only factors can be canceled, not terms. \[ \frac{(x+2)(x-3)}{(x+2)(x+5)} = \frac{x-3}{x+5} \quad (x \neq -2) \]
Rationalizing the denominator: Multiply numerator and denominator by the conjugate of the radical expression in the denominator.
Two terms are like terms if they have exactly the same variable part (same variables, same exponents). Only like terms can be combined by addition or subtraction.
| Expression | Like terms? | Result |
|---|---|---|
| $3x^2$ and $5x^2$ | Yes | $8x^2$ |
| $4xy$ and $7xy$ | Yes | $11xy$ |
| $3x^2$ and $5x$ | No | Cannot combine |
| $4xy$ and $4x^2y$ | No | Cannot combine |
Example 1. Simplify $7x^3 - 2x + 5x^3 + 9 - 4x + 1$.
Group like terms: \[ (7x^3 + 5x^3) + (-2x - 4x) + (9 + 1) = 12x^3 - 6x + 10 \]
Organizing tip. When an expression is long, rearrange the terms so like terms are adjacent before combining. Skipping this step and trying to add non-adjacent like terms in your head is a reliable source of sign errors.
Distributing over a sum: \[ a(b_1 + b_2 + \ldots + b_n) = ab_1 + ab_2 + \ldots + ab_n \]
Example 2. Expand and simplify $3(2x - 5) - 4(x + 1)$.
Distribute: \[ 6x - 15 - 4x - 4 \]
Combine like terms: \[ (6x - 4x) + (-15 - 4) = 2x - 19 \]
Common Error -- Sign Error with Negative Leading Factor. In $-4(x + 1)$, the $-4$ multiplies both $x$ AND $1$. The result is $-4x - 4$, not $-4x + 1$. Read the factor as $(-4) \cdot x + (-4) \cdot 1$.
FOIL (multiplying two binomials). FOIL stands for First, Outer, Inner, Last -- a checklist for applying the distributive property twice.
\[ (a + b)(c + d) = \underbrace{ac}_{\text{First}} + \underbrace{ad}_{\text{Outer}} + \underbrace{bc}_{\text{Inner}} + \underbrace{bd}_{\text{Last}} \]
Example 3. Expand $(3x - 2)(x + 5)$.
\[ 3x \cdot x + 3x \cdot 5 + (-2) \cdot x + (-2) \cdot 5 = 3x^2 + 15x - 2x - 10 = 3x^2 + 13x - 10 \]
FOIL is just the distributive property applied systematically. For three or more factors, distribute one factor at a time.
Example 4. Expand $(x + 1)(x - 2)(x + 3)$.
First: $(x+1)(x-2) = x^2 - x - 2$.
Then multiply by $(x + 3)$: \[ (x^2 - x - 2)(x + 3) = x^3 + 3x^2 - x^2 - 3x - 2x - 6 = x^3 + 2x^2 - 5x - 6 \]
A rational expression is a ratio of two polynomials: $\dfrac{P(x)}{Q(x)}$.
To simplify: Factor numerator and denominator completely, then cancel common factors.
Example 5. Simplify $\dfrac{x^2 - 9}{x^2 + x - 6}$.
Factor numerator: $x^2 - 9 = (x-3)(x+3)$.
Factor denominator: $x^2 + x - 6 = (x-3)(x+2)$.
Cancel the common factor $(x - 3)$: \[ \frac{(x-3)(x+3)}{(x-3)(x+2)} = \frac{x+3}{x+2} \quad (x \neq 3) \]
You must state the excluded value. Canceling $(x-3)$ is only valid when $x \neq 3$. The original expression is undefined at $x = 3$ (denominator is zero). The simplified expression $\dfrac{x+3}{x+2}$ is defined at $x = 3$, so without the restriction, they are not equal. This restriction matters in limits (MATH161): the limit as $x \to 3$ of the original expression is $\dfrac{6}{5}$, and the excluded value is exactly the point being approached.
Only factors can be canceled, not terms.
Critical Error. $\dfrac{x + 5}{x + 3} \neq \dfrac{5}{3}$. You cannot cancel the $x$'s because they are terms being added, not factors being multiplied. Canceling is only legal when the entire numerator has a factor in common with the entire denominator.
Test: let $x = 1$. Left side: $\dfrac{6}{4} = \dfrac{3}{2}$. Right side: $\dfrac{5}{3}$. These are not equal, confirming the cancellation was illegal.
To add or subtract rational expressions:
Example 6. Simplify $\dfrac{3}{x-1} + \dfrac{5}{x+2}$.
Denominators: $(x-1)$ and $(x+2)$. LCD $= (x-1)(x+2)$.
\[ \frac{3(x+2)}{(x-1)(x+2)} + \frac{5(x-1)}{(x-1)(x+2)} = \frac{3x + 6 + 5x - 5}{(x-1)(x+2)} = \frac{8x + 1}{(x-1)(x+2)} \]
Check: no common factors between $8x + 1$ and $(x-1)(x+2)$; the expression is fully simplified.
Example 7. Simplify $\dfrac{2}{x^2 - 4} - \dfrac{1}{x + 2}$.
Factor: $x^2 - 4 = (x-2)(x+2)$.
LCD $= (x-2)(x+2)$.
\[ \frac{2}{(x-2)(x+2)} - \frac{x-2}{(x-2)(x+2)} = \frac{2 - (x-2)}{(x-2)(x+2)} = \frac{2 - x + 2}{(x-2)(x+2)} = \frac{4 - x}{(x-2)(x+2)} \]
Distribute the minus sign. When subtracting a fraction, the minus applies to every term in the numerator of the second fraction. Writing $2 - x - 2$ instead of $2 - (x-2) = 2 - x + 2$ is a common and costly error.
Multiplication: Multiply numerators, multiply denominators. Factor first and cancel before multiplying to keep numbers small.
Example 8. Simplify $\dfrac{x^2 - 1}{x + 3} \cdot \dfrac{x^2 + 5x + 6}{x - 1}$.
Factor first: \[ \frac{(x-1)(x+1)}{x+3} \cdot \frac{(x+2)(x+3)}{x-1} \]
Cancel $(x - 1)$ and $(x + 3)$: \[ \frac{(x+1)(x+2)}{1} = (x+1)(x+2) \quad (x \neq 1, \, x \neq -3) \]
Division: Multiply by the reciprocal.
Example 9. Simplify $\dfrac{x^2 - 4}{x + 1} \div \dfrac{x - 2}{x^2 - 1}$.
Invert the second fraction and multiply: \[ \frac{x^2 - 4}{x + 1} \cdot \frac{x^2 - 1}{x - 2} = \frac{(x-2)(x+2)}{x+1} \cdot \frac{(x-1)(x+1)}{x-2} \]
Cancel $(x-2)$ and $(x+1)$: \[ (x+2)(x-1) \quad (x \neq 2, \, x \neq -1, \, x \neq 1) \]
A complex fraction (or compound fraction) has a fraction in the numerator, the denominator, or both. There are two methods.
Method 1 (multiply through by the LCD). Find the LCD of all fractions inside the expression, then multiply every term in the numerator and denominator by the LCD.
Example 10. Simplify $\dfrac{\dfrac{1}{x} + \dfrac{1}{y}}{\dfrac{1}{x} - \dfrac{1}{y}}$.
LCD of all inner fractions: $xy$.
Multiply numerator and denominator by $xy$: \[ \frac{xy \cdot \dfrac{1}{x} + xy \cdot \dfrac{1}{y}}{xy \cdot \dfrac{1}{x} - xy \cdot \dfrac{1}{y}} = \frac{y + x}{y - x} \]
A denominator is rationalized when it contains no radicals. This is a conventional form that makes further computation (especially differentiation) easier.
Case 1: Monomial radical denominator. Multiply numerator and denominator by the radical.
Example 11. Rationalize $\dfrac{3}{\sqrt{5}}$.
\[ \frac{3}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{3\sqrt{5}}{5} \]
Case 2: Binomial radical denominator. Multiply by the conjugate. The conjugate of $a + \sqrt{b}$ is $a - \sqrt{b}$, and their product is $a^2 - b$ (no radical).
Example 12. Rationalize $\dfrac{4}{3 - \sqrt{2}}$.
Multiply by the conjugate $3 + \sqrt{2}$: \[ \frac{4}{3 - \sqrt{2}} \cdot \frac{3 + \sqrt{2}}{3 + \sqrt{2}} = \frac{4(3 + \sqrt{2})}{9 - 2} = \frac{4(3 + \sqrt{2})}{7} = \frac{12 + 4\sqrt{2}}{7} \]
Why rationalize? In calculus, many differentiation rules produce expressions with $\sqrt{x}$ in the denominator. Rationalizing allows you to compute limits and derivatives by direct substitution instead of requiring special techniques.
| Error | Example | Correction |
|---|---|---|
| Canceling terms (not factors) | $\dfrac{x+5}{x+3} = \dfrac{5}{3}$ | Cannot cancel; factor first |
| Dropping a sign on distribution | $-4(x+1) = -4x + 1$ | Both terms: $-4x - 4$ |
| Not distributing the minus sign | $\dfrac{3}{x} - \dfrac{x+2}{x} = \dfrac{3 - x + 2}{x}$ | Must be $\dfrac{3 - (x+2)}{x} = \dfrac{1-x}{x}$ |
| Wrong LCD | Adding $\dfrac{1}{4}$ and $\dfrac{1}{6}$ using LCD $= 24$ | LCD $= 12$; smaller LCD is better |
| Forgetting excluded values | $\dfrac{(x-3)(x+2)}{x-3} = x+2$ with no restriction | State $x \neq 3$ |
Problem 1. Simplify: $4x^2 - 7x + 3 + 2x^2 + 5x - 1$.
$(4x^2 + 2x^2) + (-7x + 5x) + (3 - 1) = 6x^2 - 2x + 2$
Problem 2. Expand and simplify: $2(x + 3) - 5(2x - 1)$.
$2x + 6 - 10x + 5 = -8x + 11$
Problem 3. Simplify: $\dfrac{x^2 - 16}{x^2 - 2x - 8}$.
Numerator: $(x-4)(x+4)$.
Denominator: $(x-4)(x+2)$.
Cancel $(x-4)$: $\dfrac{x+4}{x+2}$, $x \neq 4$.
Problem 4. Simplify: $\dfrac{2}{x} + \dfrac{3}{x^2}$.
LCD $= x^2$.
$\dfrac{2x}{x^2} + \dfrac{3}{x^2} = \dfrac{2x + 3}{x^2}$
No common factors to cancel. Done.
Problem 5. Simplify: $\dfrac{x}{x+1} - \dfrac{2}{x-1}$.
LCD $= (x+1)(x-1)$.
$\dfrac{x(x-1)}{(x+1)(x-1)} - \dfrac{2(x+1)}{(x+1)(x-1)} = \dfrac{x^2 - x - 2x - 2}{(x+1)(x-1)} = \dfrac{x^2 - 3x - 2}{(x+1)(x-1)}$
Check numerator for factors: discriminant $= 9 + 8 = 17$, not a perfect square. No further cancellation.
Problem 6. Rationalize the denominator: $\dfrac{\sqrt{3}}{\sqrt{6} - \sqrt{2}}$.
Conjugate: $\sqrt{6} + \sqrt{2}$.
$\dfrac{\sqrt{3}(\sqrt{6}+\sqrt{2})}{(\sqrt{6})^2 - (\sqrt{2})^2} = \dfrac{\sqrt{18} + \sqrt{6}}{6 - 2} = \dfrac{3\sqrt{2} + \sqrt{6}}{4}$
Problem 7. Simplify: $\dfrac{\dfrac{x}{y} - \dfrac{y}{x}}{\dfrac{1}{x} + \dfrac{1}{y}}$.
LCD of inner fractions: $xy$.
Numerator: $\dfrac{x}{y} - \dfrac{y}{x} = \dfrac{x^2 - y^2}{xy}$.
Denominator: $\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{y + x}{xy}$.
Divide: $\dfrac{x^2 - y^2}{xy} \div \dfrac{x + y}{xy} = \dfrac{x^2 - y^2}{xy} \cdot \dfrac{xy}{x+y} = \dfrac{(x-y)(x+y)}{x+y} = x - y$.
$(x \neq 0, \, y \neq 0, \, x \neq -y)$
Problem 8. Simplify: $\dfrac{x+h}{x(x+h)} - \dfrac{1}{x+h} + \dfrac{1}{x}$.
The first fraction simplifies: $\dfrac{x+h}{x(x+h)} = \dfrac{1}{x}$.
So the expression becomes $\dfrac{1}{x} - \dfrac{1}{x+h} + \dfrac{1}{x} = \dfrac{2}{x} - \dfrac{1}{x+h}$.
LCD $= x(x+h)$:
$\dfrac{2(x+h)}{x(x+h)} - \dfrac{x}{x(x+h)} = \dfrac{2x + 2h - x}{x(x+h)} = \dfrac{x + 2h}{x(x+h)}$
Problem 9. In calculus, the difference quotient of $f(x) = \dfrac{1}{x}$ is $\dfrac{f(x+h) - f(x)}{h}$. Simplify it completely.
$f(x+h) = \dfrac{1}{x+h}$ and $f(x) = \dfrac{1}{x}$.
$\dfrac{f(x+h)-f(x)}{h} = \dfrac{\dfrac{1}{x+h} - \dfrac{1}{x}}{h}$
Numerator: $\dfrac{x - (x+h)}{x(x+h)} = \dfrac{-h}{x(x+h)}$.
Divide by $h$:
$\dfrac{-h}{x(x+h)} \cdot \dfrac{1}{h} = \dfrac{-1}{x(x+h)}$
As $h \to 0$, this approaches $\dfrac{-1}{x^2}$, which is the derivative of $\dfrac{1}{x}$ (covered in MATH161 Chapter 2).
Think of an algebraic expression as a sentence. Simplification is editing.
A sentence like "The person who is tall and who also is named Alex, Alex walked quickly to the store" can be rewritten as "The tall person named Alex walked quickly to the store." No meaning is lost; the structure is clearer.
A rational expression like $\dfrac{(x-3)(x+2)}{(x-3)(x+5)}$ contains a repeated factor $(x - 3)$ that adds no information for $x \neq 3$ -- it appears in both the numerator and denominator, so its net effect is multiplication by $1$. After canceling: $\dfrac{x+2}{x+5}$.
The key question in simplification is always: "What can I cancel or combine without changing the meaning?" For polynomials, "meaning" means the set of output values at each input. Combining like terms changes the form, not the function. Canceling a common factor changes the form and removes a hole (excluded point) -- which is why you must name that hole.
For students who find math intimidating: The most important single skill in this lesson is the rule that only factors can be canceled, not terms. If you learn one thing and forget everything else, learn this: test any proposed cancellation by trying a specific value of $x$. If both sides give different numbers, the cancellation was wrong.
For students interested in proof: The statement "only factors can be canceled" follows from the definition of multiplication. $\dfrac{ab}{ac} = \dfrac{b}{c}$ because $\dfrac{ab}{ac} = a \cdot \dfrac{1}{a} \cdot \dfrac{b}{c} = 1 \cdot \dfrac{b}{c}$. The key step requires $a$ to be a MULTIPLIED factor, not an added term. If $a$ is added, no such factoring of $1/a$ exists.
For students interested in careers: Simplifying rational expressions corresponds to reducing a transfer function in electrical engineering, simplifying a conditional probability formula in statistics, or canceling units in dimensional analysis. The algebraic structure is identical across all these applications.