8.2 Division of polynomials (Page 2/2)

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The process was to divide, multiply, and subtract.

Review of subtraction of polynomials

A very important step in the process of dividing one polynomial by another is subtraction of polynomials. Let’s review the process of subtraction by observing a few examples.

1. Subtract $x - 2$ from $x - 5;$ that is, find $(x - 5) - (x - 2) .$

Since $x - 2$ is preceded by a minus sign, remove the parentheses, change the sign of each term, then add.

$\frac{\begin{array}{l} x - 5 \\ - (x - 2) \end{array}}{} = \frac{\begin{array}{l} x - 5 \\ - x + 2 \end{array}}{- 3}$

The result is $- 3.$

2. Subtract $x^{3} + 3 x^{2}$ from $x^{3} + 4 x^{2} + x - 1.$

Since $x^{3} + 3 x^{2}$ is preceded by a minus sign, remove the parentheses, change the sign of each term, then add.

$\frac{\begin{array}{l} x^{3} + 4 x^{2} + x - 1 \\ - (x^{3} + 3 x^{2}) \end{array}}{} = \frac{\begin{array}{l} x^{3} + 4 x^{2} + x - 1 \\ - x^{3} - 3 x^{2} \end{array}}{x^{2} + x - 1}$

The result is $x^{2} + x - 1.$

3. Subtract $x^{2} + 3 x$ from $x^{2} + 1.$

We can write $x^{2} + 1$ as $x^{2} + 0 x + 1.$

$\frac{\begin{array}{l} x^{2} + 1 \\ - (x^{2} + 3 x) \end{array}}{} = \frac{\begin{array}{l} x^{2} + 0 x + 1 \\ - (x^{2} + 3 x) \end{array}}{} = \frac{\begin{array}{l} x^{2} + 0 x + 1 \\ - x^{2} - 3 x \end{array}}{- 3 x + 1}$

Dividing a polynomial by a polynomial

Now we’ll observe some examples of dividing one polynomial by another. The process is the same as the process used with whole numbers: divide, multiply, subtract, divide, multiply, subtract,....

The division, multiplication, and subtraction take place one term at a time. The process is concluded when the polynomial remainder is of lesser degree than the polynomial divisor.

Sample set b

Perform the division.

$\begin{array}{l} \frac{x - 5}{x - 2} . & We are to divide x - 5 by x - 2. \end{array}$

Long division showing x minus two dividing x minus five with the comment 'Divide x into x' on the right side. This division is not performed completely. See the longdesc for a full description.

$\begin{array}{l} 1 - \frac{3}{x - 2} \\ Thus, \\ \frac{x - 5}{x - 2} = 1 - \frac{3}{x - 2} \end{array}$

$\begin{array}{l} \frac{x^{3} + 4 x^{2} + x - 1}{x + 3} . & We are to divide x^{3} + 4 x^{2} + x - 1 by x + 3. \end{array}$

Long division showing x plus three dividing x cube plus four x square plus x minus one with the comment 'Divide x into x cube' on the right side. This division is not performed completely. See the longdesc for a full description

$\begin{array}{l} x^{2} + x - 2 + \frac{5}{x + 3} \\ Thus, \\ \frac{x^{3} + 4 x^{2} + x - 1}{x + 3} = x^{2} + x - 2 + \frac{5}{x + 3} \end{array}$

Practice set b

Perform the following divisions.

$\frac{x + 6}{x - 1}$

$1 + \frac{7}{x - 1}$

$\frac{x^{2} + 2 x + 5}{x + 3}$

$x - 1 + \frac{8}{x + 3}$

$\frac{x^{3} + x^{2} - x - 2}{x + 8}$

$x^{2} - 7 x + 55 - \frac{442}{x + 8}$

$\frac{x^{3} + x^{2} - 3 x + 1}{x^{2} + 4 x - 5}$

$x - 3 + \frac{14 x - 14}{x^{2} + 4 x - 5} = x - 3 + \frac{14}{x + 5}$

Sample set c

$\begin{array}{l} Divide 2 x^{3} - 4 x + 1 by x + 6. \\ \begin{array}{l} \frac{2 x^{3} - 4 x + 1}{x + 6} & \begin{array}{l} Notice that the x^{2} term in the numerator is missing . \\ We can avoid any confusion by writing \end{array} \\ \frac{2 x^{3} + 0 x^{2} - 4 x + 1}{x + 6} & Divide, multiply, and subtract . \end{array} \end{array}$

Steps of long division showing the quantity x plus six dividing the quantity two x cubed plus zero x squared minus four x minus plus one. See the longdesc for a full description

$\frac{2 x^{3} - 4 x + 1}{x + 6} = 2 x^{3} - 12 x + 68 - \frac{407}{x + 6}$

Practice set c

Perform the following divisions.

$\frac{x^{2} - 3}{x + 2}$

$x - 2 + \frac{1}{x + 2}$

$\frac{4 x^{2} - 1}{x - 3}$

$4 x + 12 + \frac{35}{x - 3}$

$\frac{x^{3} + 2 x + 2}{x - 2}$

$x^{2} + 2 x + 6 + \frac{14}{x - 2}$

$\frac{6 x^{3} + 5 x^{2} - 1}{2 x + 3}$

$3 x^{2} - 2 x + 3 - \frac{10}{2 x + 3}$

Exercises

For the following problems, perform the divisions.

$\frac{6 a + 12}{2}$

$3 a + 6$

$\frac{12 b - 6}{3}$

$\frac{8 y - 4}{- 4}$

$- 2 y + 1$

$\frac{21 a - 9}{- 3}$

$\frac{3 x^{2} - 6 x}{- 3}$

$- x (x - 2)$

$\frac{4 y^{2} - 2 y}{2 y}$

$\frac{9 a^{2} + 3 a}{3 a}$

$3 a + 1$

$\frac{20 x^{2} + 10 x}{5 x}$

$\frac{6 x^{3} + 2 x^{2} + 8 x}{2 x}$

$3 x^{2} + x + 4$

$\frac{26 y^{3} + 13 y^{2} + 39 y}{13 y}$

$\frac{a^{2} b^{2} + 4 a^{2} b + 6 a b^{2} - 10 a b}{a b}$

$a b + 4 a + 6 b - 10$

$\frac{7 x^{3} y + 8 x^{2} y^{3} + 3 x y^{4} - 4 x y}{x y}$

$\frac{5 x^{3} y^{3} - 15 x^{2} y^{2} + 20 x y}{- 5 x y}$

$- x^{2} y^{2} + 3 x y - 4$

$\frac{4 a^{2} b^{3} - 8 a b^{4} + 12 a b^{2}}{- 2 a b^{2}}$

$\frac{6 a^{2} y^{2} + 12 a^{2} y + 18 a^{2}}{24 a^{2}}$

$\frac{1}{4} y^{2} + \frac{1}{2} y + \frac{3}{4}$

$\frac{3 c^{3} y^{3} + 99 c^{3} y^{4} - 12 c^{3} y^{5}}{3 c^{3} y^{3}}$

$\frac{16 a x^{2} - 20 a x^{3} + 24 a x^{4}}{6 a^{4}}$

$\frac{8 x^{2} - 10 x^{3} + 12 x^{4}}{3 a^{3}} or \frac{12 x^{4} - 10 x^{3} + 8 x^{2}}{3 a^{3}}$

$\frac{21 a y^{3} - 18 a y^{2} - 15 a y}{6 a y^{2}}$

$\frac{- 14 b^{2} c^{2} + 21 b^{3} c^{3} - 28 c^{3}}{- 7 a^{2} c^{3}}$

$\frac{2 b^{2} - 3 b^{3} c + 4 c}{a^{2} c}$

$\frac{- 30 a^{2} b^{4} - 35 a^{2} b^{3} - 25 a^{2}}{- 5 b^{3}}$

$\frac{x + 6}{x - 2}$

$1 + \frac{8}{x - 2}$

$\frac{y + 7}{y + 1}$

$\frac{x^{2} - x + 4}{x + 2}$

$x - 3 + \frac{10}{x + 2}$

$\frac{x^{2} + 2 x - 1}{x + 1}$

$\frac{x^{2} - x + 3}{x + 1}$

$x - 2 + \frac{5}{x + 1}$

$\frac{x^{2} + 5 x + 5}{x + 5}$

$\frac{x^{2} - 2}{x + 1}$

$x - 1 - \frac{1}{x + 1}$

$\frac{a^{2} - 6}{a + 2}$

$\frac{y^{2} + 4}{y + 2}$

$y - 2 + \frac{8}{y + 2}$

$\frac{x^{2} + 36}{x + 6}$

$\frac{x^{3} - 1}{x + 1}$

$x^{2} - x + 1 - \frac{2}{x + 1}$

$\frac{a^{3} - 8}{a + 2}$

$\frac{x^{3} - 1}{x - 1}$

$x^{2} + x + 1$

$\frac{a^{3} - 8}{a - 2}$

$\frac{x^{3} + 3 x^{2} + x - 2}{x - 2}$

$x^{2} + 5 x + 11 + \frac{20}{x - 2}$

$\frac{a^{3} + 2 a^{2} - a + 1}{a - 3}$

$\frac{a^{3} + a + 6}{a - 1}$

$a^{2} + a + 2 + \frac{8}{a - 1}$

$\frac{x^{3} + 2 x + 1}{x - 3}$

$\frac{y^{3} + 3 y^{2} + 4}{y + 2}$

$y^{2} + y - 2 + \frac{8}{y + 2}$

$\frac{y^{3} + 5 y^{2} - 3}{y - 1}$

$\frac{x^{3} + 3 x^{2}}{x + 3}$

$x^{2}$

$\frac{a^{2} + 2 a}{a + 2}$

$\frac{x^{2} - x - 6}{x^{2} - 2 x - 3}$

$1 + \frac{1}{x + 1}$

$\frac{a^{2} + 5 a + 4}{a^{2} - a - 2}$

$\frac{2 y^{2} + 5 y + 3}{y^{2} - 3 y - 4}$

$2 + \frac{11}{y - 4}$

$\frac{3 a^{2} + 4 a - 4}{a^{2} + 3 a + 3}$

$\frac{2 x^{2} - x + 4}{2 x - 1}$

$x + \frac{4}{2 x - 1}$

$\frac{3 a^{2} + 4 a + 2}{3 a + 4}$

$\frac{6 x^{2} + 8 x - 1}{3 x + 4}$

$2 x - \frac{1}{3 x + 4}$

$\frac{20 y^{2} + 15 y - 4}{4 y + 3}$

$\frac{4 x^{3} + 4 x^{2} - 3 x - 2}{2 x - 1}$

$2 x^{2} + 3 x - \frac{2}{2 x - 1}$

$\frac{9 a^{3} - 18 a^{2} + 8 a - 1}{3 a - 2}$

$\frac{4 x^{4} - 4 x^{3} + 2 x^{2} - 2 x - 1}{x - 1}$

$4 x^{3} + 2 x - \frac{1}{x - 1}$

$\frac{3 y^{4} + 9 y^{3} - 2 y^{2} - 6 y + 4}{y + 3}$

$\frac{3 y^{2} + 3 y + 5}{y^{2} + y + 1}$

$3 + \frac{2}{y^{2} + y + 1}$

$\frac{2 a^{2} + 4 a + 1}{a^{2} + 2 a + 3}$

$\frac{8 z^{6} - 4 z^{5} - 8 z^{4} + 8 z^{3} + 3 z^{2} - 14 z}{2 z - 3}$

$4 z^{5} + 4 z^{4} + 2 z^{3} + 7 z^{2} + 12 z + 11 + \frac{33}{2 z - 3}$

$\frac{9 a^{7} + 15 a^{6} + 4 a^{5} - 3 a^{4} - a^{3} + 12 a^{2} + a - 5}{3 a + 1}$

$(2 x^{5} + 5 x^{4} - 1) \div (2 x + 5)$

$x^{4} - \frac{1}{2 x + 5}$

$(6 a^{4} - 2 a^{3} - 3 a^{2} + a + 4) \div (3 a - 1)$

Exercises for review

( [link] ) Find the product. $\frac{x^{2} + 2 x - 8}{x^{2} - 9} \cdot \frac{2 x + 6}{4 x - 8} .$

$\frac{x + 4}{2 (x - 3)}$

( [link] ) Find the sum. $\frac{x - 7}{x + 5} + \frac{x + 4}{x - 2} .$

( [link] ) Solve the equation $\frac{1}{x + 3} + \frac{1}{x - 3} = \frac{1}{x^{2} - 9} .$

$x = \frac{1}{2}$

( [link] ) When the same number is subtracted from both the numerator and denominator of $\frac{3}{10}$ , the result is $\frac{1}{8}$ . What is the number that is subtracted?

( [link] ) Simplify $\frac{\frac{1}{x + 5}}{\frac{4}{x^{2} - 25}} .$

$\frac{x - 5}{4}$

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Source: OpenStax, Algebra i for the community college. OpenStax CNX. Dec 19, 2014 Download for free at http://legacy.cnx.org/content/col11598/1.3

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