Compound Integral Expressions.

90. Division. Polynomials by Polynomials.

$$\begin{gathered} l*4cr\text{If the divisor (one factor)}\&=\&\&\&a\&+\&b\&+\&c, \\ \text{and the quotient (other factor)}\&=\&\&\&n\&+\&p\&+\&q, \\ \mathrm{\backslash cline}5-9\&\&\&\&an\&+\&bn\&+\&cn \\ \text{then the dividend (product)}\&=\&\{\&+\&ap\&+\&bp\&+\&cp \\ \&\&\&+\&aq\&+\&bq\&+\&cq\text{.} \end{gathered}$$

The first term of the dividend is $an$; that is, the product of $a$, the first term of the divisor, by $n$, the first term of the quotient. The first term $n$ of the quotient is therefore found by dividing $an$, the first term of the dividend, by $a$, the first term of the divisor.

If the partial product formed by multiplying the entire divisor by $n$ be subtracted from the dividend, the first term of the remainder $ap$ is the product of $a$, the first term of the divisor, by $p$, the second term of the quotient; that is, the second term of the quotient is obtained by dividing the first term of the remainder by the first term of the divisor. In like manner, the third term of the quotient is obtained by dividing the first term of the new remainder by the first term of the divisor; and so on.

divide one polynomial by another, therefore,

Arrange both the dividend and divisor in ascending or descending powers of some common letter.

Divide the first term of the dividend by the first term of the divisor.

Write the result as the first term of the quotient.

Multiply all the terms of the divisor by the first term of the quotient.

Subtract the product from the dividend.

If there is a remainder, consider it as a new dividend, and proceed as before.

It is of fundamental importance to arrange the dividend and divisor in the same order with respect to a common letter, and to keep this order throughout the operation.

The beginner should study carefully the processes in the following examples:

1. Divide $x^2+18x+77$ by $x+7$. $$\begin{gathered} r*2cr|l{x}^2\&+\&18x\&+\&77\&x+7 \\ \mathrm{\backslash cline}6-6x^2\&+\&7x\&\&\&x+11 \\ \mathrm{\backslash cline}1-5\&\&11x\&+\&\mathrm{\backslash multicolumn}1@r77 \\ \&\&11x\&+\&\mathrm{\backslash multicolumn}1@r77 \\ \mathrm{\backslash cline}3-5 \end{gathered}$$

The pupil will notice that by this process we have in effect separated the dividend into two parts, $x^2+7x$ and $11x+77$, and divided each part by $x+7$, and that the complete quotient is the sum of the partial quotients $x$ and $11$. Thus,

$$\begin{gathered} x^2+18x+77\&=x^2+7x+11x+77=(x^2+7x)+(11x+77). \\ \therefore \frac{x^2+18x+77}{x+7}\&=\frac{x^2+7x}{x+7}+\frac{11x+77}{x+7}=x+11. \end{gathered}$$

2. Divide $a^2-2ab+b^2$ by $a-b$. $$\begin{gathered} r*2cr|l{a}^2\&-\&2ab\&+\&b^2\&a-b \\ \mathrm{\backslash cline}6-6a^2\&-\&ab\&\&\&a-b \\ \mathrm{\backslash cline}1-5\&-\&ab\&+\&\mathrm{\backslash multicolumn}1@r{b}^2 \\ \&-\&ab\&+\&\mathrm{\backslash multicolumn}1@r{b}^2 \\ \mathrm{\backslash cline}2-5 \end{gathered}$$

3. Divide $a^4-a{b}^3+b^4+2a^2{b}^2-a^{3}b$ by $a^2+b^2$.

Arrange according to the descending powers of $a$. $$\begin{gathered} r*4cr|l{a}^4\&-\&a^{3}b\&+\&2a^2{b}^2\&-\&a{b}^3\&+\&b^4\&a^2+b^2 \\ \mathrm{\backslash cline}10-10a^4\&\&\&+\&a^2{b}^2\&\&\&\&\&a^2-ab+b^2 \\ \mathrm{\backslash cline}1-9\&-\&a^{3}b\&+\&a^2{b}^2\&-\&a{b}^3\&+\&\mathrm{\backslash multicolumn}1@r{b}^4 \\ \&-\&a^{3}b\&\&\&-\&a{b}^3 \\ \mathrm{\backslash cline}2-9\&\&\&+\&a^2{b}^2\&\&\&+\&\mathrm{\backslash multicolumn}1@r{b}^4 \\ \&\&\&+\&a^2{b}^2\&\&\&+\&\mathrm{\backslash multicolumn}1@r{b}^4 \\ \mathrm{\backslash cline}4-9 \end{gathered}$$