Algebraic Expressions: Terms, Simplifying, Expanding and Factoring

What is an algebraic expression?

An algebraic expression is a mathematical phrase made of numbers, variables (letters standing for numbers) and operations such as +, −, ×, ÷ and powers. Examples are 2x+3, a^2-2ab+b^2 and (x+1)/(x). An expression is a phrase, not a sentence: it has no equals sign and nothing to solve, but it has a value once the letters are given values.

Expressions are built from terms, the parts joined by + or −. Each term has a coefficient (the number multiplying the letters) and a variable part (the letters with their exponents). A term with no letter is a constant.

Three habits make this easier. The sign in front of a term belongs to the term, so 3x^2-5x+7 has the terms 3x^2, -5x and 7. A letter on its own has coefficient 1, so x=1x. And the degree of a term is the sum of the exponents of its letters, so 4xy has degree 2; the degree of an expression is the largest degree among its terms.

The usual names are by number of terms: a monomial has one term, a binomial two and a trinomial three. A sum of terms whose letters have whole-number exponents is a polynomial; its degree gives it a name, linear (1), quadratic (2) or cubic (3). Not every expression is a polynomial: \dfrac1x and √x are algebraic expressions but not polynomials. A linear expression such as 4n+1 is also how the n-th term of an arithmetic sequence is written.

Expression or equation? An equation has an equals sign and says two expressions are equal, as in 2x+3=11; you solve it. An expression such as 2x+3 you simplify, evaluate or factor. Type any expression below to see it taken apart.

How do you combine like terms?

Like terms have exactly the same variable part, the same letters with the same exponents, and you combine them by adding their coefficients and keeping the variable part: 3x+5x=8x. Only the coefficients may differ.

Why it works. It is the distributive law read backward: 3x+5x=(3+5)x=8x, as three of something plus five of the same thing is eight of it. That is also why unlike terms cannot be merged: 3x+5x^2 is three of one thing and five of another, and there is no common thing to count.

Worked example: simplify 5x^2+3x-2x^2+7x-4.

1. Group the like terms: (5x^2-2x^2)+(3x+7x)-4.
2. Combine each group: 3x^2+10x-4.

Nothing else can be combined, so 3x^2+10x-4 is the simplified form. It is conventional to write the terms from the highest degree down. Use the tool below to check any expression you simplify by hand.

How do you evaluate an algebraic expression?

To evaluate an expression, replace every letter with its value, writing each value in brackets, and then calculate with the usual order of operations: brackets, then powers, then × and ÷, then + and −. The brackets are what keep negative values safe, and the order of operations with negative numbers has traps of its own.

Evaluate 2x^2-3x at x=-3:

1. Substitute in brackets: 2(-3)^2-3(-3).
2. Power first: 2·9-3(-3).
3. Multiply: 18+9.
4. Add: 27.

The classic slip is skipping the brackets: 2·-3^2 would give -18, because the power would apply to the 3 alone. A second slip is squaring the coefficient too: in 2x^2 only x is squared, so at x=-3 it is 2·9=18, not (2·(-3))^2=36.

A table of values evaluates one expression at many inputs. For 2x^2-3x:

Substitution is also how you test. Two expressions are equivalent when they have the same value for every input. One input where they differ proves they are not equivalent: (x+3)^2 and x^2+9 both give 9 at x=0, but at x=1 they give 16 and 10. The reverse does not hold. Agreeing at a few inputs proves nothing, since x^2 and 2x agree at both x=0 and x=2 yet are different expressions.

How do you expand brackets?

To expand brackets, multiply every term inside by whatever is outside, using the distributive law a(b+c)=ab+ac; for two brackets, multiply every term of the first by every term of the second.

For a number outside: 3(x+2)=3x+6. A minus sign in front of a bracket multiplies by −1 and flips every sign inside: -(x-4)=-x+4.

For two binomials the rule is (a+b)(c+d)=ac+ad+bc+bd, four products. Many classes call this FOIL (first, outer, inner, last), which is the same four products in a fixed order. Example:

(x+2)(x-3)=x· x+x·(-3)+2· x+2·(-3)=x^2-3x+2x-6=x^2-x-6.

The area model shows why. A rectangle of width a+b and height c+d has area (a+b)(c+d), the rectangle formula from the area of a quadrilateral. Cut it into four pieces and the areas are ac, bc, ad and bd; the pieces make up the whole, so the four products add up to the product of the sums. Move the sliders to see it with numbers.

Three products come up so often that they are worth recognizing:

In the difference of squares the two middle terms -ab and +ab cancel, which is why no x term is left. A square is never just the squares of the parts: (a+b)^2 is a product of two brackets, (a+b)(a+b), and the cross terms ab appear twice.

How do you factor an algebraic expression?

To factor an expression is to write it as a product, the reverse of expanding: x^2-x-6 becomes (x+2)(x-3). Work through the same checklist every time, in this order, and check the answer by expanding it again.

1. Take out the common factor. 6x^2+9x=3x(2x+3), since 3 and x divide both terms. Do this first, because it makes everything after it smaller.
2. Two terms: a difference of squares? a^2-b^2=(a-b)(a+b), so x^2-25=(x-5)(x+5) and 4x^2-9=(2x-3)(2x+3). A sum of squares such as x^2+9 does not factor over the real numbers.
3. Three terms: a perfect square? a^2±2ab+b^2=(a± b)^2, so x^2+6x+9=(x+3)^2. The test is that the middle term is twice the product of the square roots of the outer terms.
4. Three terms, leading coefficient 1. For x^2+bx+c find two numbers whose product is c and whose sum is b: they are the numbers in (x+p)(x+q). For x^2-5x+6 the numbers are -2 and -3 (product 6, sum −5), so it is (x-2)(x-3).
5. Three terms, leading coefficient not 1. For ax^2+bx+c find two numbers whose product is ac and whose sum is b, split the middle term with them, and group. For 2x^2+7x+3 the product is 6 and the sum 7, so the numbers are 1 and 6:

2x^2+x+6x+3=x(2x+1)+3(2x+1)=(2x+1)(x+3).

Reading the signs for x^2+bx+c. If c is positive the two numbers have the same sign, the sign of b: x^2+7x+12=(x+3)(x+4) and x^2-7x+12=(x-3)(x-4). If c is negative they have opposite signs, and the larger one takes the sign of b: x^2+2x-15=(x+5)(x-3).

Not everything factors. An expression is fully factored when no factor can be broken further. The trinomial x^2+x+1 has no whole-number factors at all, and x^2-2 has none with whole numbers, although with radicals it is (x-√2)(x+√2). The tool below follows the checklist and says which step it used.

How do you simplify algebraic fractions?

To simplify an algebraic fraction, factor the top and the bottom, then cancel the factors they share; you may cancel factors, never terms. This is where factoring pays off.

(x^2-9)/(x+3)=((x-3)(x+3))/(x+3)=x-3.

(x^2-1)/(x^2-x)=((x-1)(x+1))/(x(x-1))=(x+1)/(x).

Restrictions. A fraction is undefined where its bottom is zero, and canceling does not remove that. The first fraction above is not defined at x=-3, although x-3 is. So the honest statement is (x^2-9)/(x+3)=x-3 for x≠-3. Try x=-3 in code below: the original gives an error or NaN.

Terms cannot be canceled. In (x+3)/(3) the 3 on top is part of a sum, not a factor, so it does not cancel. Test with x=3: the fraction is \dfrac63=2, while x+1=4. What is true is (x+3)/(3)=\dfrac x3+1.

Multiplying and dividing work as with numbers: multiply tops and bottoms, \dfrac ab·\dfrac cd=(ac)/(bd), and to divide, multiply by the reciprocal, \dfrac ab÷\dfrac cd=\dfrac ab·\dfrac dc. Factor first and cancel before multiplying out.

Adding and subtracting need a common denominator, just as fractions of numbers such as \frac12+\frac13 do:

\dfrac1x+\dfrac1x+1=(x+1)/(x(x+1))+(x)/(x(x+1))=(2x+1)/(x(x+1)).

The common denominator is the product of the different factors, here x(x+1), with each fraction rewritten over it.

How do you turn words into algebra?

To turn words into algebra, name the unknown quantity with a letter and translate each phrase into an operation, taking care over the order of "less than" and "from". Algebra is a short way of writing a sentence.

"Less than" and "from" reverse the order: "7 less than n" starts from n, so it is n-7, not 7-n.

A worked example. A notebook costs 5 dollars and a pen costs 3 dollars. For x notebooks and y pens the total is 5x+3y dollars. With 3 notebooks and 2 pens, substitute: 5·3+3·2=21 dollars. The expression is a recipe that works for any quantities, and that is the point of using letters.

How do you work with algebraic expressions in Python and JavaScript?

In code an expression you want to evaluate becomes a function, and an expression you want to manipulate symbolically needs a library such as SymPy in Python. Plain Python and JavaScript only calculate with numbers; they do not know that x^2-9=(x-3)(x+3).

>>> from sympy import symbols, expand, factor, simplify
>>> x = symbols('x')
>>> expand((x + 2) * (x - 3))
x**2 - x - 6
>>> factor(x**2 - 5*x + 6)
(x - 3)*(x - 2)
>>> factor(x**2 + 1)          # no real factors, so it is returned unchanged
x**2 + 1
>>> simplify((x**2 - 9) / (x + 3))
x - 3
>>> (3*x**2 - 5*x + 1).subs(x, 2)
3

const f = (x) => 3 * x ** 2 - 5 * x + 1
f(2)     // 3
f(-2)    // 23

const g = (x) => (x ** 2 - 9) / (x + 3)
g(2)     // -1
g(3)     // 0, the same as 3 - 3
g(-3)    // NaN, because the original is 0 / 0 there

#include <iostream>
using namespace std;

double f(double x) { return 3 * x * x - 5 * x + 1; }
double g(double x) { return (x * x - 9) / (x + 3); }

int main() {
  cout << f(2) << " " << f(-2) << endl;          // 3 23
  cout << g(2) << " " << g(3) << endl;           // -1 0
  // A program cannot expand or factor for you, but it can test an identity at many points
  int bad = 0;
  for (int x = -10; x <= 10; x++) {
    if ((x + 2) * (x - 3) != x * x - x - 6) bad++;        // expand
    if (x * x - 5 * x + 6 != (x - 2) * (x - 3)) bad++;    // factor
  }
  cout << "failures: " << bad << endl;
  return 0;
}

/* output:
3 23
-1 0
failures: 0
*/

The pitfalls, in order of how often they bite:

A numeric test such as f(2) can catch a wrong simplification but cannot prove a right one: it checks one input, exactly the limit described in the section on evaluating.

What are the common mistakes in algebraic expressions?

The most common mistakes with algebraic expressions are the nine below, each with the correct statement and a number that shows why.

Practice: test your understanding

These questions mix everything above. A wrong answer costs nothing here: read the hint and try again.

Summary: algebraic expressions at a glance

• Parts: an expression is terms joined by + and −; a term has a coefficient and a variable part, and a term with no letter is a constant.
• Like terms: same letters, same exponents; add coefficients, keep the variable part.
• Evaluate: put each value in brackets, then follow the order of operations. A few matching inputs never prove two expressions equal.
• Expand: a(b+c)=ab+ac and (a+b)(c+d)=ac+ad+bc+bd; (a± b)^2=a^2±2ab+b^2; (a+b)(a-b)=a^2-b^2.
• Factor: common factor first, then a difference of squares, a perfect square, or two numbers with product c and sum b; check by expanding.
• Fractions: factor, then cancel factors, never terms; note where the bottom is zero.
• Words: name the unknown, and watch the order in "less than".
• Code: write 3 * x, not 3x; use ** for powers; use SymPy for symbolic work.

Glossary: key terms

Algebraic expression

A combination of numbers, variables and operations with no equals sign, such as 3x squared minus 5x plus 7.

Variable

A letter that stands for a number that can change or is not yet known.

Constant

A term with no variable, whose value never changes, such as the 7 in 3x plus 7.

Term

One part of an expression, separated from the others by a plus or minus sign, which carries its own sign.

Coefficient

The number that multiplies the variables in a term, as the minus 5 in minus 5x.

Like terms

Terms with exactly the same variables raised to the same exponents, which differ at most in their coefficients.

Polynomial

A sum of terms in which each variable has a whole-number exponent, such as x squared minus x minus 6.

Degree

The largest sum of exponents of the variables in any single term of a polynomial.

Binomial

A polynomial with exactly two terms, such as x plus 3.

Distributive law

The rule that a times the sum of b and c equals a times b plus a times c.

Expanding

Removing brackets by multiplying out a product, so that (x plus 2)(x minus 3) becomes x squared minus x minus 6.

Factoring

Writing an expression as a product of simpler expressions, the reverse of expanding.

Difference of squares

An expression of the form a squared minus b squared, which factors as a minus b times a plus b.

Algebraic fraction

A fraction whose top and bottom are algebraic expressions, defined only where the bottom is not zero.

Frequently asked questions

References

1. James Stewart, Lothar Redlin and Saleem Watson (2016). Precalculus: Mathematics for Calculus (7th ed.). Cengage Learning.
2. Muhammad ibn Musa al-Khwarizmi (820). The Compendious Book on Calculation by Completion and Balancing (Al-kitab al-mukhtasar fi hisab al-jabr wa-l-muqabala).
3. François Viète (1591). In artem analyticem isagoge.
4. René Descartes (1637). La Géométrie.
5. Florian Cajori (1928). A History of Mathematical Notations. Open Court Publishing Company.
6. SymPy Development Team. SymPy tutorial: simplification. docs.sympy.org. https://docs.sympy.org/latest/tutorials/intro-tutorial/simplification.html.

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