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12.3: Division of Polynomials

Difficulty Level: At Grade Created by: CK-12

Learning Objectives

  • Divide a polynomial by a monomial.
  • Divide a polynomial by a binomial.
  • Rewrite and graph rational functions.

Introduction

A rational expression is formed by taking the quotient of two polynomials.

Some examples of rational expressions are

2xx214x23x+42x9x2+4x5x2+5x12x32x+3

Just as with rational numbers, the expression on the top is called the numerator and the expression on the bottom is called the denominator. In special cases we can simplify a rational expression by dividing the numerator by the denominator.

Divide a Polynomial by a Monomial

We’ll start by dividing a polynomial by a monomial. To do this, we divide each term of the polynomial by the monomial. When the numerator has more than one term, the monomial on the bottom of the fraction serves as the common denominator to all the terms in the numerator.

Example 1

Divide.

a) 8x24x+162

b) 3x2+6x1x

c) 3x218x+69x

Solution

a) 8x24x+162=8x224x2+162=4x22x+8

b) 3x3+6x1x=3x3x+6xx1x=3x2+61x

c) 3x218x+69x=3x29x18x9x+69x=x32+23x

A common error is to cancel the denominator with just one term in the numerator.

Consider the quotient 3x+44.

Remember that the denominator of 4 is common to both the terms in the numerator. In other words we are dividing both of the terms in the numerator by the number 4.

The correct way to simplify is:

3x+44=3x4+44=3x4+1

A common mistake is to cross out the number 4 from the numerator and the denominator, leaving just 3x. This is incorrect, because the entire numerator needs to be divided by 4.

Example 2

Divide 5x310x2+x255x2.

Solution

5x310x2+x255x2=5x35x210x25x2+x5x2255x2

The negative sign in the denominator changes all the signs of the fractions:

5x35x2+10x25x2x5x2+255x2=x+215x+5x2

Divide a Polynomial by a Binomial

We divide polynomials using a method that’s a lot like long division with numbers. We’ll explain the method by doing an example.

Example 3

Divide x2+4x+5x+3.

Solution

When we perform division, the expression in the numerator is called the dividend and the expression in the denominator is called the divisor.

To start the division we rewrite the problem in the following form:

x+3)x2+4x+5¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

We start by dividing the first term in the dividend by the first term in the divisor: x2x=x.

We place the answer on the line above the x term:

x+3)x2+4x+5¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯x

Next, we multiply the x term in the answer by the divisor, x+3, and place the result under the dividend, matching like terms. x times (x+3) is x2+3x, so we put that under the divisor:

x+3)x2+4x+5¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯x  x2+3x

Now we subtract x2+3x from x2+4x+5. It is useful to change the signs of the terms of x2+3x to x23x and add like terms vertically:

x+3)x2+4x+5¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯xx23x x

Now, we bring down the 5, the next term in the dividend.

x+3)x2+4x+5¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯xx23x x+5

And now we go through that procedure once more. First we divide the first term of x+5 by the first term of the divisor. x divided by x is 1, so we place this answer on the line above the constant term of the dividend:

x+3)x2+4x+5¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯x + 1x23x x+5

Multiply 1 by the divisor, x+3, and write the answer below x+5, matching like terms.

x+3)x2+4x+5¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯x + 1x23x x+5 x+3

Subtract x+3 from x+5 by changing the signs of x+3 to x3 and adding like terms:

x+3)x2+4x+5¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯x + 1x23x x+5  x32

Since there are no more terms from the dividend to bring down, we are done. The quotient is x+1 and the remainder is 2.

Remember that for a division with a remainder the answer is quotient+remainderdivisor. So the answer to this division problem is x2+4x+5x+3=x+1+2x+3.

Check

To check the answer to a long division problem we use the fact that

(divisor×quotient)+remainder=dividend

For the problem above, here’s how we apply that fact to check our solution:

(x+3)(x+1)+2=x2+4x+3+2=x2+4x+5

The answer checks out.

To check your answers to long division problems involving polynomials, try the solver at http://calc101.com/webMathematica/long-divide.jsp. It shows the long division steps so you can tell where you may have made a mistake.

Rewrite and Graph Rational Functions

In the last section we saw how to find vertical and horizontal asymptotes. Remember, the horizontal asymptote shows the value of y that the function approaches for large values of x. Let’s review the method for finding horizontal asymptotes and see how it’s related to polynomial division.

When it comes to finding asymptotes, there are basically four different types of rational functions.

Case 1: The polynomial in the numerator has a lower degree than the polynomial in the denominator.

Take, for example, y=2x1. We can’t reduce this fraction, and as x gets larger the denominator of the fraction gets much bigger than the numerator, so the whole fraction approaches zero.

The horizontal asymptote is y=0.

Case 2: The polynomial in the numerator has the same degree as the polynomial in the denominator.

Take, for example, y=3x+2x1. In this case we can divide the two polynomials:

x1)3x+2¯¯¯¯¯¯¯¯¯¯¯ 33x+35

So the expression can be written as y=3+5x1.

Because the denominator of the remainder is bigger than the numerator of the remainder, the remainder will approach zero for large values of x. Adding the 3 to that 0 means the whole expression will approach 3.

The horizontal asymptote is y=3.

Case 3: The polynomial in the numerator has a degree that is one more than the polynomial in the denominator.

Take, for example, y=4x2+3x+2x1. We can do long division once again and rewrite the expression as y=4x+7+9x1. The fraction here approaches zero for large values of x, so the whole expression approaches 4x+7.

When the rational function approaches a straight line for large values of x, we say that the rational function has an oblique asymptote. In this case, then, the oblique asymptote is y=4x+7.

Case 4: The polynomial in the numerator has a degree that in two or more than the degree in the denominator. For example: y=x3x1.

This is actually the simplest case of all: the polynomial has no horizontal or oblique asymptotes.

Example 5

Find the horizontal or oblique asymptotes of the following rational functions.

a) y=3x2x2+4

b) y=x13x26

c) y=x4+1x5

d) y=x33x2+4x1x22

Solution

a) When we simplify the function, we get y=312x2+4. There is a horizontal asymptote at y=3.

b) We cannot divide the two polynomials. There is a horizontal asymptote at y=0.

c) The power of the numerator is 3 more than the power of the denominator. There are no horizontal or oblique asymptotes.

d) When we simplify the function, we get y=x3+6x7x22. There is an oblique asymptote at y=x3.

Notice that a rational function will either have a horizontal asymptote, an oblique asymptote or neither kind. In other words, a function can’t have both; in fact, it can’t have more than one of either kind. On the other hand, a rational function can have any number of vertical asymptotes at the same time that it has horizontal or oblique asymptotes.

Review Questions

Divide the following polynomials:

  1. 2x+42
  2. x4x
  3. 5x355x
  4. x2+2x5x
  5. 4x2+12x364x
  6. 2x2+10x+72x2
  7. x3x2x2
  8. 5x493x
  9. x312x2+3x412x2
  10. 36x+x39x3
  11. x2+3x+6x+1
  12. x29x+6x1
  13. x2+5x+4x+4
  14. x210x+25x5
  15. x220x+12x3
  16. 3x2x+5x2
  17. 9x2+2x8x+4
  18. 3x243x+1
  19. 5x2+2x92x1
  20. x26x125x4

Find all asymptotes of the following rational functions:

  1. x2x2
  2. 1x+4
  3. x21x2+1
  4. x4x29
  5. x2+2x+14x1
  6. x3+14x1
  7. xx3x26x7
  8. x42x8x+24

Graph the following rational functions. Indicate all asymptotes on the graph:

  1. x2x+2
  2. x31x24
  3. x2+12x4
  4. xx23x+2

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