Monomials and the exponent rules

When you reach for it

A monomial is a single term: a number, or a number multiplied by one or more variables, where each variable carries a positive whole-number exponent. The number in front is the monomial’s coefficient. So

are all monomials, with coefficients 7, 1, -5 and \displaystyle \frac{2}{3}. When no number is written, the coefficient is 1, as in x, and a lone minus sign means -1, as in -x^2. A variable with no written exponent has exponent 1: x is x^1. By contrast x + 2 is not a monomial (it has two terms), and neither is \displaystyle \frac{3}{x} (the variable sits in a denominator, which no whole-number exponent can express). Later in this lesson \displaystyle \frac{3}{x} will be written 3x^{-1}; that is a fine expression, but its exponent is negative, so it is still not a monomial.

Monomials get multiplied all the time. A rectangle 4a long and 3a wide has area (4a)(3a). A box with edges 2x, 3x and 5x has volume (2x)(3x)(5x). And every product of longer expressions, such as 2x^2(3x^3 - x) or (x + 1)(x^2 - 4), is built out of products of single terms: multiplying a polynomial by a monomial and multiplying polynomials both come down to this move, done once for each pair of terms.

They get divided just as often. If a rectangle has area 36a^5b^3 and one side 4a^2b, the other side is the quotient \displaystyle \frac{36a^5b^3}{4a^2b}. They get raised to powers: a square of side 3x has area (3x)^2, a cube of edge 4p^2 has volume (4p^2)^3. And negative exponents turn up the moment a quotient leaves more factors below the fraction bar than above, or a very small quantity is written in scientific notation, such as 6 \times 10^{-4}.

The goal in each case is to rewrite the expression as one coefficient times one power of each variable. It turns out that this is always possible, and the work is always the same: the numbers combine as numbers, and the exponents of each variable add, subtract or multiply. This lesson proves each of those rules, says exactly what each one needs, and then uses them together.

The procedure

The lesson has five parts. Multiplying monomials comes first, since everything else is built on it. Dividing and raising to a power follow. Negative exponents then remove the last awkward case, and the final part puts all the rules side by side.

Multiplying monomials

Why exponents add. An exponent that is a positive whole number counts factors: x^3 means x \cdot x \cdot x. So

The brackets can be dropped because grouping does not change a product. What remains is a row of x’s, two from the first power and three from the second, and 2 + 3 = 5. Nothing about 2 and 3 was special, which gives the rule, often called the product of powers:

Product of powers. If a is any number and m and n are positive whole numbers, then

The reason is the same count, with letters:

Notice what the rule needs. The two powers must have the same base, because only then is the long row made of one repeated factor. And the exponents must count something, which is why they are positive whole numbers here. The base itself does not change: 2^2 \cdot 2^3 = 2^5, and a quick check agrees, since 4 \cdot 8 = 32 = 2^5.

From powers to monomials. Now take (3x^2)(-4x^3). Each monomial is itself a product, so the whole thing is one product of four factors:

Since order does not change a product, the two numbers can be moved next to each other, and since grouping does not change it either, they can be bracketed together, with the powers of x bracketed together:

The numbers multiply as numbers, signs included. The powers of x multiply by the product of powers. That is the whole move:

• Multiply the coefficients, as ordinary numbers with their signs.
• For each variable, add its exponents. A variable written without an exponent contributes 1. A variable that appears in only one of the factors passes into the answer unchanged, because there is nothing to combine it with.

Both parts are licensed by the same two properties, order and grouping, which hold because every letter stands for a number. Different letters are never combined with each other; they are different bases.

The picture shows the same thing for (3x)(2x). A rectangle 3x wide and 2x tall cuts into 3 \cdot 2 = 6 squares, and each square is x by x, so the area is 6x^2.

A picture like this needs positive lengths and at most three dimensions, so it cannot show a negative coefficient or a product such as x^2 \cdot x^3. The counting argument covers those.

Checking. Because the rewrite is true for every value of the variable, you can test it at one. At x = 2, the left side is (3 \cdot 4)(-4 \cdot 8) = 12 \cdot (-32) = -384, and the right side is -12 \cdot 32 = -384. A mismatch at any value proves an error. A match is good evidence, not a proof.

Dividing monomials

Why exponents subtract. Take x^5 \div x^2, where x \neq 0. Written as a fraction with every factor spelled out,

Each canceled pair is \displaystyle \frac{x}{x} = 1, which is where x \neq 0 is needed: a number divided by itself is 1 only when it is not zero. Two factors cancel from five, and 5 - 2 = 3 are left. When the larger exponent is on the bottom, the leftover factors stay on the bottom instead: \displaystyle \frac{x^2}{x^5} = \frac{1}{x^3}.

Quotient of powers. If a \neq 0 and m and n are positive whole numbers, then

Proof. First, a^n \neq 0, because a product of nonzero numbers is never zero, so every fraction here makes sense. If m > n, then m - n is a positive whole number and m = n + (m - n), so by the product of powers a^m = a^n \cdot a^{m-n}. Then

If m = n, the fraction is a nonzero number divided by itself, which is 1. If m < n, the same splitting works the other way round: a^n = a^m \cdot a^{n-m}, so

That covers every case. \square

From powers to monomials. A quotient of two products splits into a product of quotients:

This is how fractions multiply, read backwards. So a quotient of monomials splits into one fraction for the coefficients and one fraction for each variable:

That is the whole move:

• Divide the coefficients, as a fraction in lowest terms, with its sign.
• For each variable, subtract the smaller exponent from the larger. What is left stays on the side that had more: on top if the top exponent was larger, on the bottom if the bottom one was. A variable with equal exponents cancels to 1. A variable on only one side stays where it is.

Every variable in a denominator is taken to be nonzero, since otherwise the expression has no value. Checking at x = 2, y = 2, the left side is

and the right side is \displaystyle -\frac{5 \cdot 16}{2 \cdot 8} = -5.

Division undoes multiplication, so the picture of (3x)(2x) above can be read backwards. A rectangle of area 6x^2 with one side 2x has other side

which is the 3x across the top of the picture.

A quotient need not be a monomial. The product of two monomials is always a monomial, but a quotient can leave a variable in the denominator: \displaystyle \frac{6x^2}{3x^5} = \frac{2}{x^3}, which is not a monomial. The quotient is a monomial exactly when every variable in the divisor appears in the dividend with at least as large an exponent.

Power of a monomial

Why exponents multiply. A power of a power is a repeated product of the same power:

Three copies of four factors each make 3 \cdot 4 = 12 factors. The rules in this part are treated on their own in Powers of powers, products and quotients.

Power of a power. If m and n are positive whole numbers, then (a^m)^n = a^{mn}.

Proof. By the meaning of the exponent n, (a^m)^n is n factors of a^m. By the product of powers, multiplying them adds the exponent m to itself n times, which gives m \cdot n:

Nothing in this count depended on which positive whole numbers m and n were. \square

Why the exponent reaches every factor. A power of a product is a repeated product, and order and grouping let the factors be sorted:

Power of a product. If n is a positive whole number, then (ab)^n = a^nb^n.

Proof. (ab)^n is n factors of ab, which is n factors of a and n factors of b mixed together. By order and grouping they can be gathered into n factors of a followed by n factors of b, which is a^nb^n. The same argument works for a product of three or more factors. \square

Power of a quotient. If b \neq 0 and n is a positive whole number, then \displaystyle \left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}.

Proof. n factors of \displaystyle \frac{a}{b} multiply tops with tops and bottoms with bottoms, giving n factors of a over n factors of b. \square

From powers to monomials. A monomial is a product, so the power of a product sends the outside exponent to every factor, the coefficient included, and the power of a power then multiplies each variable’s exponent:

That is the whole move:

• Raise the coefficient to the power, sign and all. An even power of a negative number is positive and an odd power is negative: (-2)^4 = 16 but (-2)^3 = -8.
• For each variable, multiply its exponent by the outside exponent. A variable written without an exponent has exponent 1, so it gets the outside exponent itself.

The picture shows why the coefficient must be raised too. A square 3x on each side holds 3 \cdot 3 = 9 small squares of area x^2, so (3x)^2 = 9x^2, not 3x^2.

The exponent belongs only to what it is written against. In (2x)^3 the bracket puts the whole monomial under the exponent, so (2x)^3 = 8x^3; in 2x^3 only the x is cubed. In the same way a minus sign outside the bracket is not raised: -(2x)^2 = -4x^2, while (-2x)^2 = 4x^2.

Negative exponents in expressions

Dividing powers needed three cases, and the third one leaves a fraction. Zero and negative exponents remove the cases. They are definitions, not discoveries: a power with exponent 0 or -3 counts no factors, so it has no meaning until one is chosen. The lesson Zero and negative exponents treats the choice in full; here is the choice and the reason for it.

The zero exponent. When m = n, the quotient of powers gives \displaystyle \frac{a^n}{a^n} = 1. If the rule “subtract the exponents” is to hold here too, it would give a^{n-n} = a^0. So, for a \neq 0,

Negative exponents. When m < n, the quotient of powers gives \displaystyle \frac{a^m}{a^n} = \frac{1}{a^{n-m}}, while subtracting the exponents would give a^{m-n}, which is a^{-(n-m)}. For instance \displaystyle \frac{a^2}{a^5} = \frac{1}{a^3}, and subtracting would give a^{-3}. So, for a \neq 0 and n a positive whole number,

The pattern agrees. Going down a column of powers of 2, each step lowers the exponent by 1 and divides the value by 2, and the definitions are what that pattern gives when it carries on past 2^1:

Both definitions need a \neq 0. The expression 0^{-1} would be \displaystyle \frac{1}{0}, which has no value. And the argument for a^0 = 1 divided a^n by itself, which is impossible when a = 0, since 0^n = 0; so it gives 0^0 no value, and this lesson leaves 0^0 undefined.

Two consequences. For a \neq 0, b \neq 0 and n a positive whole number:

Proof. By the definition, \displaystyle \frac{1}{a^{-n}} = 1 \div \frac{1}{a^n} = a^n, since dividing by a fraction multiplies by its reciprocal. For the second, using the power of a quotient twice,

Both identities hold for every positive whole number n. \square

The move: across the fraction bar. In a fraction whose top and bottom are products, a factor with a negative exponent can cross the fraction bar, and its exponent changes sign. If P and Q stand for the other factors (with Q \neq 0), then

The first is the definition, \displaystyle a^{-n} = \frac{1}{a^n}, put into the fraction. The second is the first consequence, \displaystyle \frac{1}{a^{-n}} = a^n. For example,

Two things limit the move. It is for factors only: a term in a sum cannot cross the bar on its own, because the identities above were about products. And it moves only the base the exponent belongs to:

Every rule now holds for every integer exponent. For nonzero a and b and any integers m and n, positive, negative or zero:

Proof. Throughout, a and b are nonzero, so every power of them is nonzero.

Quotients of positive powers, one case. For positive whole numbers p and q, the three cases of the quotient of powers are now one rule, \displaystyle \frac{a^p}{a^q} = a^{p-q}. For p > q it was already true. For p = q the quotient is 1, which is a^0. For p < q the quotient is \displaystyle \frac{1}{a^{q-p}}, which by definition is a^{-(q-p)} = a^{p-q}.

Products. Any integer is a difference of two positive whole numbers: m = p - q with q = |m| + 1 and p = m + q, which is at least 1. Write m = p - q and n = r - s this way. By the line above, \displaystyle a^m = \frac{a^p}{a^q} and \displaystyle a^n = \frac{a^r}{a^s}. Multiplying the fractions and using the product of powers for positive whole numbers,

Quotients. First, \displaystyle \frac{1}{a^n} = a^{-n} for every integer n: for n > 0 this is the definition; for n = 0 both sides are 1; for n = -k with k > 0 it is the first consequence, \displaystyle \frac{1}{a^{-k}} = a^k. Then \displaystyle \frac{a^m}{a^n} = a^m \cdot \frac{1}{a^n} = a^m \cdot a^{-n} = a^{m-n} by the product rule just proved.

Power of a power. For n a positive whole number, (a^m)^n is n factors of a^m, and the product rule for integers adds m to itself n times, giving a^{mn}. For n = 0 both sides are 1, since a^m \neq 0. For n = -k with k > 0, \displaystyle (a^m)^{-k} = \frac{1}{(a^m)^k} = \frac{1}{a^{mk}} = a^{-mk} = a^{mn}, where the third step is the fact \displaystyle \frac{1}{a^n} = a^{-n} for every integer, since mk may be negative.

Power of a product. For n a positive whole number this is the earlier proof. For n = 0 both sides are 1. For n = -k with k > 0, \displaystyle (ab)^{-k} = \frac{1}{(ab)^k} = \frac{1}{a^kb^k} = \frac{1}{a^k} \cdot \frac{1}{b^k} = a^{-k}b^{-k}.

Power of a quotient. Since \displaystyle \frac{a}{b} = a \cdot b^{-1}, the power of a product and the power of a power give \displaystyle \left(\frac{a}{b}\right)^n = a^n \cdot (b^{-1})^n = a^n \cdot b^{-n} = \frac{a^n}{b^n}. \square

This is what negative exponents buy. Dividing no longer needs cases: subtract the exponents, and a negative result is just another way of saying the factor belongs in the denominator.

Simplified form. An expression with integer exponents is called simplified when each base appears once, no exponent is zero or negative, and the coefficient is a fraction in lowest terms. Negative exponents are a tool for getting there, not usually part of the final answer.

Monomials and exponent rules together

Here are all the rules in one place. In every row m and n are integers, and any base that is divided by, or carries a zero or negative exponent, is nonzero.

An expression that needs several rules follows the order of operations, and a fraction bar groups its top and its bottom: work out powers of brackets first, then the products above and below the bar, then the division. Moving factors with negative exponents across the bar can happen at any point. Any order of correct steps gives the same answer, because each step replaces an expression with one equal to it for every allowed value of the variables. Example 17 does one problem by two routes to show this.

Worked examples

Boundaries

Same base only. The exponents add or subtract because the factors form one row of the same base. In x^3y^2 the bases differ, and the expression is already as simple as it gets. With numbers the same holds: 2^3 \cdot 3^2 = 8 \cdot 9 = 72, and the row of factors mixes 2’s with 3’s, so there is no single base for a sum of exponents to belong to. When different bases share an exponent, as in 2^3 \cdot 5^3, the power of a product applies instead, read backwards: 2^3 \cdot 5^3 = (2 \cdot 5)^3 = 10^3.

Products only. Every rule here is about multiplying, dividing and powering. A sum such as x^2 + x^3 is not a product, and it stays as two terms: x^2 and x^3 are not like terms, so they cannot be combined. Where terms are alike, adding and multiplying do different things to them:

In the same way an exponent does not pass into a sum. (x + 3)^2 is not x^2 + 9: at x = 1 the first is 16 and the second is 10. And (x + y)^{-1} is not x^{-1} + y^{-1}: at x = y = 1 the first is \displaystyle \frac{1}{2} and the second is 2. For the same reason only a factor, never a term of a sum, may cross a fraction bar with its exponent’s sign changed.

Integer exponents, nonzero bases. The rules are proved here for integer exponents. The positive whole-number versions hold for every base, but as soon as a base is divided by, or carries a zero or negative exponent, it must be nonzero: 0^{-1} would be \displaystyle \frac{1}{0}, and 0^0 is left undefined. Fractional exponents such as x^{1/2} are given a meaning in rational exponents, chosen so that the same rules stay true, but only for positive bases there: \bigl((-8)^2\bigr)^{1/2} = 64^{1/2} = 8, while multiplying the exponents would give (-8)^1 = -8.

A quotient may leave the monomials. Products and powers (with positive whole exponents) of monomials are monomials. A quotient is a monomial only when nothing is left in the denominator, so \displaystyle \frac{6x^2}{3x^5} = \frac{2}{x^3} is a simplified quotient but not a monomial.

Where this goes next

Multiplying monomials is the unit step inside all polynomial multiplication. To find 2x^2(3x^3 - x), the distributive property turns it into two monomial products, 6x^5 - 2x^3. Multiplying two polynomials does the same for every pair of terms and then combines like terms. Dividing monomials is the unit step the other way: dividing a polynomial by a monomial divides each term separately, and factoring out a monomial finds the largest monomial that divides every term, which takes the smallest exponent of each variable.

Quotients of monomials are the simplest rational expressions, and multiplying and dividing rational expressions uses exactly the canceling of this lesson, applied to factors that are whole polynomials.

The rules are also how numbers in scientific notation are multiplied and divided: (4 \times 10^3)(2 \times 10^5) has coefficients 4 and 2 and a base of 10, so it is 8 \times 10^8, and negative exponents of 10 are how very small numbers are written.

They explain a fact about degree: since exponents add, the degree of a product of nonzero monomials is the sum of their degrees, the first step toward the same fact for products of polynomials. And the same rules, kept true on purpose, are what give meaning to fractional exponents and then to exponential functions, where the exponent itself is the variable.

Exercises

One variable

Multiply. The same instruction holds through exercise 15.

1. (5x^4)(3x^2)
2. (-6y^3)(4y^7)
3. (-8u^5)(-3u)
4. (t^9)(-t^4)
5. (7k)(2k)
6. (-2n^6)(-n^6)

Several variables

7. (2a^3b)(6a^2b^4)
8. (-5x^2y^3)(4xy)
9. (9m^4n^2)(-m^3n^5)
10. (3p^2q)(8qr^4)
11. \displaystyle \left(\frac{2}{3}x^2y\right)(9xy^6)
12. \displaystyle \left(-\frac{1}{4}c^5d\right)\left(\frac{2}{5}cd^3\right)

Three or more factors

13. (2x)(3x^2)(4x^3)
14. (-a^2b)(5ab^3)(-2a^4)
15. (3y)(-y^2)(-2y^3)(y^4)

Numbers and letters as exponents

16. Write 3^4 \cdot 3^5 as a single power of 3.
17. Multiply (3 \times 10^4)(2 \times 10^6) and write the answer in the form c \times 10^k.
18. (5x^n)(2x^3), where n is a positive whole number.
19. y^{2k} \cdot y^{k+1}, where k is a positive whole number.

Area and volume

20. A rectangle is 7a long and 4ab wide. Write its area as a monomial.
21. A box has edges 2x, 3x and 6x. Write its volume as a monomial, then find the volume when x = 5.

Dividing monomials

Divide. In exercises 22 to 30 every variable is nonzero; write answers with positive exponents.

22. \dfrac{36x^8}{4x^3}
23. \dfrac{-42y^5}{6y}
24. \dfrac{9m^2}{12m^7}
25. \dfrac{-15a^4}{-25a^4}
26. \dfrac{32p^6q^3}{8p^2q^5}
27. \dfrac{45x^3y^9z}{-60x^7y^2z}
28. Write \dfrac{2^{10}}{2^7} as a power of 2, then find its value.
29. \dfrac{x^{n+5}}{x^2}, where n is a positive whole number.
30. A rectangle has area 54u^7v^3 and width 6u^2v. Find its length.

Power of a monomial

Simplify. In exercises 31 to 38 every variable is nonzero.

31. (4x^3)^2
32. (-2a^5)^3
33. (-3m^2n)^4
34. -(3m^2n)^4
35. \displaystyle \left(\frac{1}{2}p^3q^4\right)^3
36. (5y^2)^3(2y)^2
37. \left(\dfrac{x^2y^3}{2z}\right)^4
38. A cube has edges of length 3k^4. Write its volume as a monomial.

Negative exponents in expressions

Write each without zero or negative exponents, and simplify. In exercises 39 to 50 every variable is nonzero.

39. 5^{-2}
40. (-2)^{-3}
41. \displaystyle \left(\frac{3}{4}\right)^{-2}
42. 4^0 + 4^{-1}
43. 7x^{-4}
44. (7x)^{-2}
45. \dfrac{m^{-3}n^5}{p^{-2}}
46. 6a^0b^{-1}
47. x^{-5} \cdot x^9
48. \dfrac{y^{-2}}{y^{-6}}
49. (a^{-3})^2
50. (2x^{-1}y^3)^{-3}

Several rules at once

Simplify, writing answers with positive exponents. In exercises 51 to 53 every variable is nonzero.

51. \dfrac{(3x^2y)^3(2xy^4)}{6x^5y^2}
52. \dfrac{10a^{-3}b^2}{4a^2b^{-5}}
53. \left(\dfrac{(2m^3)^2}{m^{-1}}\right)^{-1}
54. Find (4 \times 10^{-3})(6 \times 10^8) and write it in scientific notation.
55. Find \dfrac{9 \times 10^{-2}}{3 \times 10^{-6}} and write it in scientific notation.

Reasoning

56. Explain, by writing out the factors, why x^4 \cdot x^3 = x^7 and not x^{12}.
57. Explain why 2^3 \cdot 3^4 cannot be written as a single power by adding the exponents, then find its value.
58. The degree of a monomial is the sum of the exponents of its variables. Show that the degree of a product of two nonzero monomials is the sum of their degrees.
59. Explain why the product of two monomials is always a monomial.
60. Give an example showing that the quotient of two monomials need not be a monomial, and say when it is one.
61. Explain why x^0 must equal 1 (for x \neq 0) if subtracting exponents is to work when dividing equal powers, and why the same argument gives no value to 0^0.
62. Show that (2x)^3 and 2x^3 are different expressions by simplifying the first and comparing their values at x = 1.
63. Prove that \displaystyle \left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^n for nonzero a and b and any positive whole number n.
64. Show by trying x = 1 and y = 1 that (x + y)^{-1} is not equal to x^{-1} + y^{-1}, and explain which rule someone would be wrongly using.
65. Using only the definition of a negative exponent and canceling, show that x^5 \cdot x^{-2} = x^3 for x \neq 0.