Operations with rational expressions | Lesson (article) | Khan Academy (2024)

What are rational expressions?

Rational expressions look like fractions that have variables in their denominators (and often numerators too). For example, x2x+3 is a rational expression. Just as we can add, subtract, multiply, and divide fractions, we can perform the four operations on rational expressions.

Rational expressions also represent the division of one polynomial expression by another. For example, x2x+3 represents the division of x2 by x+3, for which we can find a quotient and a remainder.

In this lesson, we'll learn to:

  1. Simplify rational expressions
  2. Add, subtract, multiply, and divide rational expressions
  3. Rewrite rational expressions in the form of quotients and remainders

This lesson builds upon the following skills:

  • Factoring quadratic and polynomial expressions
  • Operations with polynomials

You can learn anything. Let's do this!

How do I simplify rational expressions?

Intro to rational expression simplification

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Reducing rational expressions to lowest terms

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Just like fractions, but with polynomial factoring

You're probably familiar with simplifying fractions like 510; we can factor out a 5 from both the numerator and the denominator and cancel them, leaving us with 12.

510=1525=12

With rational expressions, we can also cancel out factors that appear in both the numerator and the denominator. These factors can be polynomials!

For example, we can simplify the rational expression x+12x+2 by factoring out x+1 from both the numerator and the denominator and canceling them, leaving us with 12.

x+12x+2=1(x+1)2(x+1)=12

On the SAT, the numerators and denominators of rational expressions can also be quadratic expressions and higher order polynomials, so the ability to factor these expressions fluently is key to your success.

Try it!

try: find the simplified rational expression

x25x+6x2+x6=(x2)(x3)(x2)(x+3)

The rational expression x25x+6x2+x6 is factored as shown above. Since both the numerator and the denominator contain

as a factor, we can cancel the identical factors to simplify the expression.

x25x+6x2+x6 is equivalent to

.

How do I multiply and divide rational expressions?

Multiplying & dividing rational expressions: monomials

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Multiplying & dividing rational expressions: monomials

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Multiplying and dividing rational expressions

The same rules for multiplying and dividing fractions apply to multiplying and dividing rational expressions.

When multiplying two rational expressions:

abcd=acbd

When dividing two expressions, recall that dividing by a fraction is equivalent to multiplying by that fraction's reciprocal:

(ab)(cd)=abdc=adbc

However, to avoid lengthy polynomial operations, it's recommended that you factor and cancel any cancellable factors before you write out the final expression.

To multiply two rational expressions:

  1. Factor any factorable polynomial expressions in the numerators and the denominators.
  2. Cancel any identical factors that appear in both the numerators and the denominators of the expressions.
  3. Multiply the remaining numerators and multiply the remaining denominators.

Dividing two rational expressions is similar to multiplying; just remember that dividing by an expression is equivalent to multiplying by the reciprocal of the same expression.

Let's look at some examples!

What is the product of x2x+3 and x+3x ?

x2x+3x+3x=xxx+3x+3x=x

x is the product of x2x+3 and x+3x.

If f(x)=x2+2x+1x+3 and g(x)=xx2+4x+3, what is f(x)g(x) ?

Both x2+2x+1 and x2+4x+3 can be factored:

  • x2+2x+1=(x+1)(x+1)
  • x2+4x+3=(x+1)(x+3)

Dividing by a rational expression is equivalent to multiplying by its reciprocal:

f(x)g(x)=(x2+2x+1x+3)(xx2+4x+3)=x2+2x+1x+3x2+4x+3x=(x+1)(x+1)x+3(x+1)(x+3)x=(x+1)(x+1)x+3(x+1)(x+3)x=(x+1)(x+1)(x+1)x=(x+1)3x

f(x)g(x) is (x+1)3x.

Try it!

try: identify factorable numerators and denominators

x2+1x+1x2+3x+23x2+3

Before multiplying the two rational expressions shown above, Raj wants to factor the numerators and denominators. Which of the following can be factored?

Choose all answers that apply:

Choose all answers that apply:

  • x2+1

  • x+1

  • x2+3x+2

  • 3x2+3

How do I add and subtract rational expressions?

Adding rational expressions with different denominators

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Adding rational expression: unlike denominators

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Adding and subtracting rational expressions

The same rules for adding and subtracting fractions apply to adding and subtracting rational expressions.

When adding or subtracting two rational expressions with unlike denominators:

ab±cd=adbd±bcbd=ad±bcbd

Remember that you can only add and subtract the numerators of the rational expressions if the expressions have a common denominator! In most cases, the easiest way to find a common denominator is to multiply the two unlike denominators, b and d.

To add and subtract two rational expressions:

  1. Find a common denominator for the two expressions. In most cases, the product of the two denominators would work.
  2. Rewrite the equivalent form of each rational expression using the common denominator.
  3. Add or subtract the numerators of the expressions while retaining the common denominator.
  4. Combine like terms and write the result.
  5. Factor and/or cancel as needed.

Let's look at some examples!

What is the sum of x2x+3 and 4x+3x+3 ?

Since the two rational expressions have the same denominator, we can add the numerators and keep the denominator the same.

x2x+3+4x+3x+3=x2+4x+3x+3=(x+1)(x+3)x+3=(x+1)(x+3)x+3=x+1

x+1 is the sum of x2x+3 and 4x+3x+3.

What is the difference 2xx+33x+1 ?

We need to find a common denominator for 2xx+3 and 3x+1. One common denominator is simply the product of the denominators, (x+3)(x+1).

To rewrite both rational expressions with the common denominator, we multiply the numerator and denominator of each rational expression by the denominator of the other expression:

2xx+33x+1=2xx+3x+1x+13x+1x+3x+3=2x(x+1)3(x+3)(x+3)(x+1)=(2x)(x)+(2x)(1)+(3)(x)+(3)(3)(x)(x)+(x)(1)+(3)(x)+(3)(1)=2x2+2x3x9x2+x+3x+3=2x2x9x2+4x+3

2x2x9x2+4x+3 is the difference 2xx+33x+1.

Try it!

try: rewrite expressions with common denominators

Celeste wants to find the sum of 3x and x2x3. In order to add the two rational expressions, she must first find a common denominator by

the two denominators, x and 2x3.

Next, she needs to rewrite each rational expression using the common denominator. The first term can be rewritten as 3(2x3x(2x3), and the second term can be rewritten as

.

How do I rewrite a rational expression as a quotient and a remainder?

Dividing polynomials by linear expressions

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Dividing polynomials by linear expressions

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Polynomial long division

We can represent any rational expression as a(x)b(x), where a and b are polynomial expressions in terms of x.

For example, for a(x)=x2+2x+4 and b(x)=x+3, a(x)b(x)=x2+2x+4x+3.

When dividing a and b, we can find quotient polynomial q and remainder polynomial r such that:

a(x)b(x)=q(x)+r(x)b(x)

Where the degree of r is less than the degree of b. Since b is usually a first degree polynomial (ax+b) on the SAT, r is usually a constant.

When dividing two polynomials using long division, we focus on the highest degree terms in the numerator and the denominator first. For example, for x2+2x+4x+3, the highest degree term in the numerator is x2, and the highest degree term in the denominator is x. The first question we ask is "what is x2 divided by x ?"

x2x=x, so we write x as the first term of the quotient q, find the product of x and the divisor x+3, then subtract the product from a. This eliminates the x2 term in the dividend.

xx+3|x2+2x+4(x2+3x)x2x+4

Next, we do the same to what's left of the dividend, x+4. We ask "what is x divided by x ?"

xx=1, so we write 1 as the second term of the quotient q, find the product of 1 and the divisor x+3, then subtract the product from x+4. This eliminates the x term in the dividend.

x1x+3|x2+2x+4(x2+3x)x2x+4x2(x3)x2+2x+7

This leaves us with the constant 7. Since the degree of 7 is lower than the degree of x+3, we can stop dividing here and write our quotient and remainder.

  • q(x)=x1
  • r(x)=7

Therefore, x2+2x+4x+3=x1+7x+3.

Another strategy to find the quotient and the remainder is to group the numerator, which requires us to split the numerator of a rational expression into a polynomial divisible by the denominator and the remainder.

You don't need to know how to group the numerator, but it may save you time on the test.

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Dividing quadratics by linear expressions with remainders

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Dividing quadratics by linear expressions with remainders: missing x-term

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To divide polynomial expressions a(x) and b(x) using long division:

  1. Divide the highest degree term of a by the highest degree term of b. This gives you a term of the quotient.
  2. Multiply the result of Step 1 by b.
  3. Subtract the result of Step 2 from a. Be careful when subtracting negatives!
  4. Repeat the divide-multiply-subtract steps using what's left of the dividend until the result is of a lower degree than b.
  5. The terms calculated in the "divide" steps form the quotient q. The leftover polynomial with a lower degree than b is the remainder r.
  6. Write the result as q(x)+r(x)b(x).

Example: For f(x)=x2 and g(x)=x+3, rewrite f(x)g(x) in the form q(x)+r(x)g(x).

Let's divide f(x) by g(x) to find the quotient and remainder.

First, we can divide x2 by x to get x.

xx+3|x2+0x(x2+3x)x23x

Keep in mind that multiplying x by x+3 creates an x-term, 3x. Since x2 has no x-term, we're effectively subtracting 3x from 0x.

Next, we can divide 3x by x to get 3:

x3x+3|x2+0x(x2+3x)x23x+0.(3x9)x2+3x(.)9

Similarly, multiplying 3 by x+3 creates a constant term, 9. Since 3x has no constant term, we're effectively subtracting 9 from (or adding 9 to) 0.

Since our divisor is a first degree term and only a constant term remains, r is of a lower degree than g, and we can stop dividing. The quotient is x3, and the remainder is 9. We can write the expression as x3+9x+3.

f(x)g(x)=x3+9x+3

Try it!

Try: take the first steps of polynomial long division

6x3+3x214x22x+1

In the rational expression above, the numerator is a

polynomial and the denominator is a first degree polynomial. Therefore, the quotient must be a

polynomial.

To find the first term of the quotient, we must divide the highest degree term in the numerator,

, by the highest degree term in the denominator, 2x.

The first term of the quotient is

.

Your turn!

Practice: multiply two rational expressions

Which of the following is equivalent to 2x+1x+1x+2 ?

Choose 1 answer:

Choose 1 answer:

  • 2x+1

  • 2x+2

  • x+32x+3

  • 2x+22x+3

Practice: multiply two rational expressions

x2+5x+6x+4x2+4xx+3

Which of the following is equivalent to the expression above?

Choose 1 answer:

Choose 1 answer:

  • xx+2

  • x+2x

  • x+2

  • x2+2x

practice: subtract two rational expressions

4x+132x3

Which of the following is equivalent to the expression above?

Choose 1 answer:

Choose 1 answer:

  • 14x

  • 12x2x3

  • 5x42x2x3

  • 5(x3)2x2x3

practice: rewrite a rational expression as quotient and remainder

Which of the following is equivalent to x2+5x+2x+5 ?

Choose 1 answer:

Choose 1 answer:

  • x+1

  • x+2

  • x+2x+5

  • x+14x+5

Operations with rational expressions | Lesson (article) | Khan Academy (2024)

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