Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

74. To Divide a Polynomial by a Polynomial. EXAMPLE 1. Divide 3 x2 +17 x+20 by x+4. SOLUTION. Dividend, 3 x2+17 x+20 | x+4 Divisor. 3 times (x+4) gives 3 x2+12 x 3 x+5 Quotient. Ans. Subtracting gives

5 x+20 5 times (x+4) gives

5 x+20 Subtracting gives

EXPLANATION. The solution above consists of several steps taken in a definite order as follows.

(1) The terms of both dividend and divisor are arranged according to the descending powers of x. (See § 33.)

(2) The first term in the dividend is divided by the first term of the divisor, giving the first term in the quotient. That is, the 3 x2 is divided by the x, giving 3 x.

(3) The 3 x thus obtained is now multiplied by the divisor and the result subtracted from the dividend, giving 5 x+20. This 5 x+20 is therefore what is left of the dividend after the terms in it that come from multiplying the divisor by the first term of the quotient have been subtracted.

(4) 5 x+20 is now treated as a new dividend and the process carried on as before, this time to obtain the second term of the quotient. That is, the 5 x is divided by the x of the divisor, giving 5 as the second term of the quotient.

(5) The 5 thus obtained is now multiplied by the divisor and subtracted to form a new dividend as before. This, however, turns out in the present instance to give zero (all terms canceling) so no further terms are left in the dividend to be considered. Hence the complete quotient is 3 x+5.

CHECK. Multiplying the divisor, x+4, by the quotient, 3 x+5, in the way shown in § 53, gives the dividend, 3 x2+17 x+20.

EXAMPLE 2. Divide 6 x3 – 7 x2 +1 by 2 x– 1.

SOLUTION. Observe that the dividend contains an ~3 term and an a2 term but no x term; that is, the first power of x is missing. In such cases it is best in arranging the work to leave a space for the

missing term, after which the process is the same as in Example 1. The process is given in full below.

6 x3— 7 x2+( )+1/2 x-1 Divisor.
6 23—3 x2 3x2–2 2–1 Quotient. Ans.

-4 x2+( )+1
-4 x2+2 x

- 2 x+1
-2x+1

[ocr errors]

In case the result of the last subtraction is not zero (note that it was zero in Examples 1 and 2) then, as in arithmetic, the final result of the division is a quotient and a remainder. This is illustrated in the following example.

EXAMPLE 3. Divide 5 x4 — 23+x+1 by x2 – 1.
SOLUTION. 5 24— 23+( )+x+1|22–1 : Divisor.

5 24 -5 x 5 x2– +5 Quotient. Ans.

- 23+5 x2+x
– x3 + x

5 x2 +1
5 x2 -5

+6 Remainder. CHECK. As in arithmetic, when the remainder is added to the product of the divisor and quotient, the result should be the dividend. In other words, (x2-1)(5 x2— x+5)+6 should equal 5 24— 23+x+1, and we readily find this to be so. The work is left to the pupil.

WRITTEN EXERCISES Carry out each of the following indicated divisions, and check each answer. 1. (3 x2 – 2 x-1)=(x-1). 2. (3 x3 – 4 x2+2x– 1) = (x-1). 3. (15 x2+x-2) =(3 x— 1). 4. (4 y3+2 y2-1)=(2 y–1). [Hint. The first power of y is missing in the dividend. See Example 2 in § 74.]

5. (a3 — 3 a2+2 a-6) = (a?+2). 6. (m5+m4 — m3 – 1) = (m2 – 1). 7. (23+x2– x+2) (22 — 3+1).

SOLUTION. Here, the divisor is a trinomial (instead of a binomial as in preceding examples). The division, however, is carried out as before. Thus,

x3+x2– 2+2 32- x+1 Divisor.
23— x2+x x+2 Quotient. Ans.

2 x2—2 x+2
2 x2—2 x+2

0
8. (6 y3 – 5 y2 +7 y-2) = (2 yay+2).
9. (a4+16+4 a?) --(2 ata2+4).

(Hint. First arrange in descending powers of a.] 10. (25 — 61 x—60) = (x2 – 2 x—3). 11. (24 — 1) = (x+1). 13. (23+1) = (2x2 — x+1). 12. (25+1) = (x+1).

14. (x4 — 1) = (x3 — x2+x— 1). 15. Find the quotient and remainder in each of the following indicated divisions. (See Example 3 in § 74.)

(a) (2 x2 +3 x+1) = (x+2). (6) (6 x3 – 5 x2 +7 x+1)=(3 x— 1). (c) (24 – 3 x3+x2+2 x – 1) = (x2 — 3 —2). 16. Divide 2 x2+xy-y2 by x+y. [HINT. Here two letters, x and y, occur, but the process is the same as before. The quotient is 2 x- y.)

17. Divide a3 — 2 a26+2 ab? b3 by a-b.
18. Divide mon +2 mn3 — m2 — 2 n2 by m2 +2 na.
[Hint. Arrange in descending powers of m.]
19. Divide 24 y4 by x-y.
20. Divide x4 + y4 by x+y.
21. Divide x3+y3+23 — 3 xyz by x+y+z.
For further exercises on this topic, see Appendix, p. 301.

75. * Special Cases of Division.t We found in § 65 that a2–62 was the product of (ab) and (a+b); that is, Q2—62=(ab) (a+b). This is the same as saying that (a2—62)+(a,b)=a+b, or that (a?— 12) = (a+b)=Q-6. Whence, we may say that ar— b2 is divisible by either a-b or a+b without a remainder; or, as we often say, the division is exact. This also appears by actual division as follows.

a +( )-b2|a-b
a2- ab a+b

tab-b2
tab-62

a2+( )62|a+b.
q2+ab a-6

- ab-62
- ab-62

Suppose now we consider the difference of two cubes: q3— 53. Is this likewise divisible by either a-b or a+b without a remainder; that is, is it “exactly divisible” by a,b and a+b? To answer, let us divide out and see.

[ocr errors][merged small][merged small]

Thus we see that a3— 53 is exactly divisible by a-6, the quotient being a2+ab+b2, and we see also that a3-03 is not exactly divisible by a+b, since we then have the remainder -2 63.

Similarly, if we consider the sum (instead of the difference) of two cubes; that is, if we consider a:+b3, we find (see Ex. 1 below) that it is exactly divisible by a+b, the quotient being q2- ab+62, and we find also that a3+63 is not exactly divisible by a-b, the remainder being 2 63.

+ This may be omitted at the discretion of the teacher from the first year course. The same is true of the other articles and exercises that are starred (*) hereafter.

These results give us the following formulas.
* Formula IX a. 23— 13=(a,b)(a2+ab+62).
* Formula IX b. q3+b3=(a+b)(az-ab+62).

Note. These Formulas give us the following useful rule: The difference of the cubes of two numbers is divisible by the difference of the numbers, while the sum of the cubes of two numbers is divisible by the sum of the numbers.

EXERCISES 1. Show that (as stated above) the quotient of a3+63 divided by a+b is q?ab+62.

2. Show that a3+63 is not exactly divisible by a,b, but that there is a remainder of 2 63.

3. Show that a4–64 is exactly divisible by either a+b or a-b.

4. Show that a2+62 is not exactly divisible by either a-b or a+b. Find the remainder in each case.

5. Show that a++b4 is not exactly divisible by either a-b or a+b.

76. * Factoring the Sum or Difference of Two Cubes. Formulas IX a and IX b show that the sum of two cubes, or the difference of two cubes, may always be expressed as the product of two factors.

EXAMPLE 1. Factor 8 x3—27 .
SOLUTION. 8 23—27 ys= (2 x)3– (3 y)3

= (2 x—3 y)[(2 x)2+(2 x)(3 y)+(3 y)2] (Formula IX a)

= (2 2–3 y) (4 x2+6 xy+9 y). Ans. EXAMPLE 2. Factor mon3+64 p3. Solution. m3n3+64 p3= (mn)3+(4 p)3

= (mn+4 p)[(mn)2(mn)(4 p)+(4 p)2). (Formula IX b) = (mn+4 p) (m2n24 mnp+16 pa). Ans.

EXERCISES Factor each of the following expressions. 1. Q3–8 63. 5. 8 x3+125. 2. 23+8 63. 6. 1+m3. 3. 23–1.

7. a3b3-c3ds. 4. 23+1.

8. (2-y)3+(x+y)8.

9. 8(m+n)3+125 ns. 10. 1-(a-6)3. 11. (x−y)3–8. 12. x3y29—216.

« ΠροηγούμενηΣυνέχεια »