Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

When the division can not be exactly performed, the quotient may be expressed in the form of a fraction, and this fraction may be reduced to its lowest terms according to a method to be explained in Art. 79.

SIGNS IN DIVISION.

(73.) The proper sign to be prefixed to a quotient may be deduced from the principles already established for multiplication, since the product of the divisor and quotient must be equal to the dividend. Hence, because +ax+b=+ab

+ab÷+b=+a

[blocks in formation]

When both the dividend and divisor have the same sign, the quotient will have the sign +; when they have different signs, the quotient will have the sign --.

Examples.

1. Divide -15y' by 3y.

Here it is plain that the answer must be -5y; for 3yx-5y=-15y'.

2. Divide 40abd by -5a3b.

3. Divide -58x'y' by 2x1y3.

4. Divide -18a'c'd'x' by -3a2c3x'.

Ans. -Sa'd.

Ans. -29x1y3.

Ans. +6a'cd'x.

QUEST.-When the division can not be exactly performed, what is to be done? How do we determine the proper sign to be prefixed to the quotient? Give the rule for the signs.

[blocks in formation]

9. Divide -72a'c'd'x' by -6a2cx3. 10. Divide 84b'c'y" by -12b3c3y3.

Ans.

Ans.

CASE II.

(74.) When the divisor is a monomial and the div idend a polynomial.

We have seen, Art. 57, that when a single term is multiplied into a polynomial, the former enters into svery term of the latter.

Thus, if we multiply a by a+b, we obtain for a product a'+ab.

Hence, if we divide a2+ab by a, the quotient must be a+b.

Therefore, when the divisor is a monomial and the dividend a polynomial, we have the following

RULE.

Divide each term of the dividend by the divisor, as in the former Case.

Examples.

1. Divide 3x+6x+3ax-15x by 3x.

Ans. x2+2x+a-5.

Ans. c+4x-3a.

2. Divide 3abc+12abx-9ab by 3ab.

QUEST.-What is the second case? Give the rule when the divi

dend is a polynomial.

3. Divide 40a'b3+60a3b1-17ab by -ab.

Ans. -40a'b3-60ab+17.

4. Divide 15abc-10a cx2+5a'c'd' by -5a2c. Ans. -3b+2x2-acď.

5. Divide 25ab2-15ab'c-10ab by -5ab.

Ans.

6. Divide 6a2x1y°—12a3x'y'+15a*x'y' by 3a2x'y'.

Ans.

7. Divide 27a3x2+24a1x3-9a3x* by 3ax2.

Ans.

8. Divide 12a'y°-16a3y'+20a°y'-28a'y' by -4a'y3.

CASE III.

Ans.

(75.) When the divisor and dividend are both polynomials.

Let it be required to divide a2+2ab+b2 by a+b.

The object of this operation is to find a third polynomial which, multiplied by the second, will reproduce the first.

It is evident that the dividend is composed of all the partial products arising from the multiplication of each term of the divisor by each term of the quotient, these products being added together and reduced. Hence, if we divide the first term a' of the dividend by the first term a of the divisor, we shall obtain a term of the quotient, which is a. Multiplying each term of the divisor by a, and subtracting the product a'+ab from the proposed dividend, the remainder may be regarded as the product of the divisor by the remaining terms of the quotient. We shall then obtain another

QUEST.-What is the third case? How do we obtain the first term of the quotient?

4

term of the quotient by dividing the first term of the remainder ab by the first term of the divisor a, which gives b. Multiplying the divisor by b, and subtracting as before, we find nothing remains. Hence a+b is the exact quotient.

The operation may be exhibited as follows:

The dividend is a2+2ab+b2 | a+b is the divisor.

a2+ ab a+b is the quotient.

ab+b is the first remainder.

ab+b2

0

It is generally convenient in algebra to place the divisor on the right of the dividend, and the quotient directly under the divisor.

In this example we have arranged the terms of the divisor and dividend in the order of the powers of the letter a. When the terms are thus arranged, we shall always obtain a term of the quotient by dividing the first term on the left of the dividend by the first term on the left of the divisor. Therefore, before commencing the division, the terms should be arranged in the order of the powers of one of the letters.

deduce the following

Hence we

(76.) RULE FOR THE DIVISION OF POLYNOMIALS. 1. Arrange the terms of the dividend and divisor in the order of the powers of one of the letters.

2. Divide the first term on the left of the dividend by the first term on the left of the divisor; the result will be the first term of the quotient.

QUEST.--In what order must we arrange the terms? Give the rule for the division of polynomials.

3. Multiply the divisor by this term, and subtract the product from the dividend.

4. Divide the first term of the remainder by the first term of the divisor; the result will be the second term of the quotient.

5. Multiply the divisor by this term, and subtract the product from the last remainder. Continue the same operation till all the terms of the dividend are exhausted.

If the divisor is not exactly contained in the dividend, the quantity which remains after the division is finished must be placed over the divisor in the form of a fraction, and annexed to the quotient.

Ex. 2. Divide x3-a3+3a2x-3ax2 by x-a.

We here arrange the letters in the order of the powers of x.

[blocks in formation]

Ex. 3. Divide a'+2ab+2ab'+b' by a'+ab+b2.

Ans. a+b.

Ex. 4. Divide 24a3-42a'x+19ax2-15x3 by 2a-3x.
Ans. 12a-3ax+5x2.

Ex. 5. Divide 6a2+17a'b+24ab2+16b' by 3a+4b.

Ans. 2a+3ab+4b2.

QUEST-If there is a remainder, what must be done with it?

« ΠροηγούμενηΣυνέχεια »