Multiplying and dividing rational expressions
Two fractions with polynomials in them are multiplied or divided. How do you get one fraction in simplest form, and which values of the variable are still off limits?
Where you are 3 in Simplifying rational expressions
When you reach for it
A rational expression is a fraction whose numerator and denominator are polynomials, such as \displaystyle \frac{x^2 - 16}{2x + 6}. You reach for this skill when two of them are multiplied or divided and you want the result as one fraction, as simple as it can be, together with the values of the variable where the original was never defined.
It comes up whenever rates and ratios carry a variable: a speed times a time, a count per sheet times sheets per hour, the quotient of two functions. It is also the first half of fraction arithmetic with polynomials; adding and subtracting, which need a common denominator, come next.
The procedure
Everything here is the arithmetic of multiplying fractions and dividing fractions, applied to polynomials. That works because at any allowed value of the variable each polynomial is just a number, and number fractions obey these rules.
Multiplying
For polynomials p, q, r, s with q \ne 0 and s \ne 0,
Numerators multiply, denominators multiply. Then the product is simplified by canceling common factors, using \displaystyle \frac{ak}{bk} = \frac{a}{b} for k \ne 0, the same move that simplifying a rational expression rests on.
Factor before you multiply. Only a factor of the whole numerator and the whole denominator can cancel. Multiplying out first buries those factors inside a large polynomial that then has to be factored again. So factor every numerator and denominator completely first, watching for the special patterns.
Cancel across the two fractions. A factor on top of the first fraction may cancel with a factor underneath the second. Once multiplied, all the top factors sit in one numerator and all the bottom factors in one denominator, and the commutative and associative properties of multiplication let you reorder and regroup them so that the matching pair stands together.
Dividing
For polynomials p, q, r, s with q \ne 0, r \ne 0 and s \ne 0,
Dividing by a fraction is multiplying by its reciprocal. The reason is the meaning of division: \displaystyle \frac{p}{q} \div \frac{r}{s} is the expression that, times \displaystyle \frac{r}{s}, gives back \displaystyle \frac{p}{q}, and
After the flip, the division is a multiplication and everything above applies. A divisor that is a polynomial, such as x^2 - 16, is the fraction \displaystyle \frac{x^2 - 16}{1}, with reciprocal \displaystyle \frac{1}{x^2 - 16}.
The excluded values
The product \displaystyle \frac{p}{q} \cdot \frac{r}{s} is undefined wherever q = 0 or s = 0. A quotient \displaystyle \frac{p}{q} \div \frac{r}{s} is undefined in those places and also where r = 0: there the divisor equals 0, and nothing can be divided by zero. After the flip, r is a denominator, which is where that restriction becomes visible.
Read every excluded value from the factored form before canceling. Canceling removes a factor from sight but not from the original problem, so a value that made the original undefined stays excluded from the answer. The answer is the simplified fraction together with the list, for example \displaystyle \frac{x + 4}{2}, x \ne -3, 0, 4.
Worked examples
Factor, cancel, multiply
Multiply, and state the excluded values.
Factor each part: x^2 - 16 = (x - 4)(x + 4) is a difference of squares, 2x + 6 = 2(x + 3), x^2 + 3x = x(x + 3) and x^2 - 4x = x(x - 4).
The denominators are zero at x = -3, x = 0 and x = 4, so those are excluded. Now cancel x - 4, x + 3 and x, each of which appears once on top and once below:
The answer has no variable in its denominator, yet the restrictions stay: at x = 4 the original product divides by zero. Check at x = 2: the original is \displaystyle \frac{-12}{10} \cdot \frac{10}{-4} = 3, and \displaystyle \frac{2 + 4}{2} = 3.
Dividing adds a restriction
Divide, and state the excluded values.
Factor completely, taking out the common factor first: 3x^2 - 12 = 3(x^2 - 4) = 3(x - 2)(x + 2), and x^2 + x = x(x + 1).
The denominators give x \ne 0 and x \ne -1. The divisor’s numerator x - 2 must not be zero either, since at x = 2 the divisor is 0, so x \ne 2. Now flip the divisor and multiply:
So the quotient is \displaystyle \frac{15(x + 2)}{x + 1}, x \ne -1, 0, 2. The value 2 is easy to lose: it never appears in a denominator of the problem as written.
A pair of opposite factors
Multiply, and state the excluded values.
Factor: x^2 + 2x = x(x + 2), x^2 - 4 = (x - 2)(x + 2) and x^2 - 7x + 10 = (x - 2)(x - 5). The denominators exclude x = 0, -2, 2, 5.
The factors 5 - x and x - 5 are opposites: 5 - x = -(x - 5), so their quotient is -1 wherever x \ne 5. Canceling them leaves that -1 behind:
Restrictions that are fractions
Divide, and state the excluded values.
Factor: 2x^2 + 7x + 3 = (2x + 1)(x + 3), x^2 - 9 = (x - 3)(x + 3), 4x^2 - 1 = (2x - 1)(2x + 1) and x^2 - 6x + 9 = (x - 3)^2.
The denominators exclude x = 3 and x = -3. The divisor’s numerator, (2x - 1)(2x + 1), is zero at \displaystyle x = \frac{1}{2} and \displaystyle x = -\frac{1}{2}, so those go too. Flip and cancel:
The quotient is \displaystyle \frac{x - 3}{2x - 1}, \displaystyle x \ne -3, -\frac{1}{2}, \frac{1}{2}, 3. Of the two factors of (x - 3)^2 on top, one cancels the x - 3 below and one remains.
A factor that is never zero
Divide, and state the excluded values.
The numerator is a difference of cubes, a^3 - 8 = (a - 2)(a^2 + 2a + 4). Also a^2 - 4 = (a - 2)(a + 2) and 3a + 6 = 3(a + 2).
The denominators exclude a = 2 and a = -2. The divisor’s numerator a^2 + 2a + 4 adds nothing: completing the square gives (a + 1)^2 + 3, which is at least 3 for every real a. Flip and cancel:
Every factor cancels except the 3, so the quotient is the constant 3, for a \ne -2, 2.
Units cancel like factors
Yevgeny’s printing shop cuts each sheet of card into business cards, x + 2 across and x - 2 down. Cards are packed in boxes of 5x + 10 (five rows of x + 2). The press prints 300 sheets an hour. How many boxes does it fill per minute? Evaluate for x = 8.
A sheet holds (x + 2)(x - 2) = x^2 - 4 cards. Chain the rates so that each unit appears once on top and once below, as with conversion factors:
Cards, sheets and hours cancel, leaving boxes per minute. The numbers and polynomials cancel by the same rule:
The press fills x - 2 boxes a minute. For x = 8 that is 6 boxes a minute. Check directly: a sheet holds 10 \cdot 6 = 60 cards, five sheets a minute give 300 cards, and a box holds 50. Here x > 2, since a sheet must have at least one row.
Boundaries
Only factors cancel. In \displaystyle \frac{x + 4}{x + 7} nothing cancels: the x is part of a sum, not a factor of the whole numerator. This is why factoring comes first.
Restrictions come from the original problem. List them before canceling, from every denominator and, in a division, from the divisor’s numerator. A simplified answer without its list is a different expression, defined at points where the original is not.
Only real zeros restrict. A factor such as a^2 + 2a + 4 or x^2 + 1 has no real zero, so it adds no excluded value.
Division is not commutative. \displaystyle \frac{p}{q} \div \frac{r}{s} flips the divisor, never the first fraction. In a chain such as A \div B \cdot C, work from left to right: only B is flipped.
In context, the variable has its own limits. In Example 6 the algebra allows any x \ne -2, but the sheet needs x > 2.
Where this goes next
Adding and subtracting rational expressions needs one more idea, the common denominator, and uses this lesson’s factoring and canceling throughout. A fraction whose numerator and denominator are themselves fractions is a division in disguise, so it is simplified with the flip learned here. When the expressions become functions, the quotient of two functions is undefined where the divisor is zero, the same extra restriction as in Example 2. Solving a rational equation then relies on knowing which values were excluded, and rational functions show those values as holes and asymptotes on a graph.
Exercises
Multiply
Simplify, and state the excluded values.
- \displaystyle \frac{x^2 - 36}{4x + 8} \cdot \frac{x^2 + 2x}{x^2 - 6x}
- \displaystyle \frac{6y^2}{5y - 10} \cdot \frac{y^2 - 4}{9y}
- \displaystyle \frac{x^2 + 7x + 12}{x^2 - x - 6} \cdot \frac{x^2 - 9}{x^2 + 8x + 16}
- \displaystyle \frac{7 - x}{x + 5} \cdot \frac{x^2 + 5x}{x^2 - 49}
- \displaystyle \frac{2m - 2}{m^2 + m} \cdot \frac{m^2 + 2m + 1}{m^2 - 1}
Divide
Simplify, and state the excluded values.
- \displaystyle \frac{x^2 - 25}{3x} \div \frac{x + 5}{6x^2}
- \displaystyle \frac{4a + 12}{a^2 - 2a} \div \frac{a + 3}{a - 2}
- \displaystyle \frac{x^2 - 3x - 10}{x^2 - 4} \div \frac{x^2 - 10x + 25}{2x - 4}
- \displaystyle \frac{x^2 + 4x}{x - 1} \div (x^2 - 16)
- \displaystyle \frac{3n^2 + 5n - 2}{n^2 - 9} \div \frac{9n^2 - 1}{n^2 + 3n}
Cubes, opposites and two variables
- \displaystyle \frac{x^3 + 27}{x^2 - 9} \cdot \frac{x - 3}{x^2 - 3x + 9}, with excluded values.
- \displaystyle \frac{2 - y}{y^2 + y} \div \frac{y^2 - 4}{y^2 - y}, with excluded values.
- \displaystyle \frac{a^2 - b^2}{4ab} \div \frac{a - b}{8b^2}, with the conditions on a and b.
Restrictions
- List every excluded value of \displaystyle \frac{x + 1}{x - 4} \div \frac{x + 6}{x^2 - 1}, then simplify it.
- Theophil writes \displaystyle \frac{x^2 - 1}{x} \div \frac{x - 1}{x} = x + 1 for every x. Correct his statement.
Printing shop
- Aldona’s press prints 720 sheets an hour. Each sheet holds x^2 + 3x stickers (x rows of x + 3), and stickers are sold in packs of 4x + 12. Write the number of packs filled per minute as a simplified expression, and evaluate it for x = 6.
- Stanislava’s shop prints a run of n^2 + 10n leaflets at 5n + 50 leaflets a minute. How many minutes does the run take? Evaluate for n = 40.
Working backwards
- Find a rational expression R with \displaystyle \frac{x^2 - 9}{x + 2} \cdot R = x - 3, and state the values the product must exclude.
Reasoning
- In \displaystyle \frac{x + 3}{x} \div \frac{x - 2}{x + 1} no denominator contains x - 2. Explain why x = 2 is still excluded.
- Explain why, in a product, a factor in the numerator of the first fraction may cancel a factor in the denominator of the second.
- Explain why \displaystyle \frac{x}{x - 1} \div \frac{x^2 + 1}{x} excludes x = 0 and x = 1 but no value from x^2 + 1.
Answers
- \displaystyle \frac{x + 6}{4}, x \ne -2, 0, 6. Factored: \displaystyle \frac{(x - 6)(x + 6)}{4(x + 2)} \cdot \frac{x(x + 2)}{x(x - 6)}.
- \displaystyle \frac{2y(y + 2)}{15}, y \ne 0, 2. Factored: \displaystyle \frac{6y^2 (y - 2)(y + 2)}{5(y - 2) \cdot 9y}, and \displaystyle \frac{6y}{45} = \frac{2y}{15}.
- \displaystyle \frac{(x + 3)^2}{(x + 2)(x + 4)}, x \ne -4, -2, 3. Factored: \displaystyle \frac{(x + 3)(x + 4)}{(x - 3)(x + 2)} \cdot \frac{(x - 3)(x + 3)}{(x + 4)^2}.
- \displaystyle -\frac{x}{x + 7}, x \ne -7, -5, 7. Factored: \displaystyle \frac{7 - x}{x + 5} \cdot \frac{x(x + 5)}{(x - 7)(x + 7)}, and \displaystyle \frac{7 - x}{x - 7} = -1.
- \displaystyle \frac{2}{m}, m \ne -1, 0, 1. Factored: \displaystyle \frac{2(m - 1)}{m(m + 1)} \cdot \frac{(m + 1)^2}{(m - 1)(m + 1)}.
- 2x(x - 5), x \ne -5, 0. Flip: \displaystyle \frac{(x - 5)(x + 5)}{3x} \cdot \frac{6x^2}{x + 5}; the divisor’s numerator excludes -5.
- \displaystyle \frac{4}{a}, a \ne -3, 0, 2. Flip: \displaystyle \frac{4(a + 3)}{a(a - 2)} \cdot \frac{a - 2}{a + 3}.
- \displaystyle \frac{2}{x - 5}, x \ne -2, 2, 5. Flip: \displaystyle \frac{(x - 5)(x + 2)}{(x - 2)(x + 2)} \cdot \frac{2(x - 2)}{(x - 5)^2}.
- \displaystyle \frac{x}{(x - 1)(x - 4)}, x \ne -4, 1, 4. The divisor is \displaystyle \frac{x^2 - 16}{1}, so multiply by \displaystyle \frac{1}{(x - 4)(x + 4)} and cancel x + 4.
- \displaystyle \frac{n(n + 2)}{(n - 3)(3n + 1)}, \displaystyle n \ne -3, -\frac{1}{3}, 0, \frac{1}{3}, 3. Flip: \displaystyle \frac{(3n - 1)(n + 2)}{(n - 3)(n + 3)} \cdot \frac{n(n + 3)}{(3n - 1)(3n + 1)}.
- 1, x \ne -3, 3. x^3 + 27 = (x + 3)(x^2 - 3x + 9), and x^2 - 3x + 9 has no real zero.
- \displaystyle \frac{1 - y}{(y + 1)(y + 2)}, y \ne -2, -1, 0, 1, 2. Flip: \displaystyle \frac{2 - y}{y(y + 1)} \cdot \frac{y(y - 1)}{(y - 2)(y + 2)}, and \displaystyle \frac{2 - y}{y - 2} = -1.
- \displaystyle \frac{2b(a + b)}{a}, with a \ne 0, b \ne 0, a \ne b. Flip: \displaystyle \frac{(a - b)(a + b)}{4ab} \cdot \frac{8b^2}{a - b}.
- x \ne -6, -1, 1, 4, and the quotient is \displaystyle \frac{(x + 1)^2(x - 1)}{(x - 4)(x + 6)}. The denominators give 4, 1, -1; the divisor’s numerator gives -6.
- \displaystyle \frac{x^2 - 1}{x} \div \frac{x - 1}{x} = x + 1 only for x \ne 0, 1. At x = 0 both fractions are undefined; at x = 1 the divisor is 0.
- 3x packs a minute, so 18 when x = 6. \displaystyle \frac{x(x + 3)}{4(x + 3)} \cdot \frac{720}{60} = \frac{x}{4} \cdot 12. The algebra excludes x = -3; here x counts rows, so x > 0 and that value never arises.
- \displaystyle \frac{n}{5} minutes, so 8 minutes when n = 40. Time is leaflets divided by rate: \displaystyle \frac{n(n + 10)}{5(n + 10)}. The algebra excludes n = -10; here the run and the rate are both positive, which forces n > 0, so that value never arises.
- \displaystyle R = \frac{x + 2}{x + 3}, with x \ne -3, -2. The product \displaystyle \frac{(x - 3)(x + 3)}{x + 2} \cdot \frac{x + 2}{x + 3} cancels to x - 3.
- At x = 2 the divisor is \displaystyle \frac{0}{3} = 0, and dividing by zero is undefined. After the flip the problem reads \displaystyle \frac{x + 3}{x} \cdot \frac{x + 1}{x - 2}, and x - 2 is a denominator.
- The product equals one fraction, the product of the numerators over the product of the denominators. By the commutative and associative properties the factors in each can be reordered and regrouped, so the two matching factors can stand over each other, where \displaystyle \frac{ak}{bk} = \frac{a}{b} for k \ne 0 removes them.
- x = 1 and x = 0 make a denominator zero. x^2 + 1 is the divisor’s numerator and could exclude a value only by being zero, but x^2 \ge 0 gives x^2 + 1 \ge 1 for every real x.