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divisible by a-b; also a®-, and so on for any positive value of n.
The quotients obtained by dividing the difference of the same powers of two quantities by the difference of the quan. tities follow a simple law. Thus,
etc. (an_6")=(a–5)=an-stan-26+an–382 + .... ta?bn–3 +abn-3
+nThe exponents of a decrease by unity, while those of 6 in. crease by unity.
84. It may also be proved that the difference of like even powers of any two quantities is always divisible by the sum of the quantities. Thus, (a2—62)=(a+b)=a_b.
etc. Also, the sum of like odd powers of any two quantities is always divisible by the sum of the quantities. Thus, (a +53)--(a+b)=a? - ab +62
(a +65)=(a+b)=a4-a3b+ a2b2-ab3 +64.
etc. The exponents of a and 6 follow the same law as in Art. 83, but the signs of the terms are alternately plus and minus.
85. When exact division is impossible.—One polynomial can not be divided by another polynomial containing a letter which is not found in the dividend ; for it is impossible that one quantity multiplied by another which contains a certain letter should give a product not containing that letter.
A monomial is never divisible by a polynomial, because every polynomial multiplied by another quantity gives a product containing at least two terms not susceptible of reduction.
Yet a binomial may be divided by a polynomial containing any number of terms.
Thus, a_64 is divisible by a +a?b+ab2 +63, and gives for & quotient a-b.
To resolve a Polynomial into Factors. 86. When a polynomial is capable of being resolved into factors, the factors can generally be discovered by inspection, or from the law of formation.
If all the terms of a polynomial have a common factor, that factor is a factor of the polynomial; and the other factor may be found by dividing the polynomial by the common factor.
EXAMPLES. 1. Resolve 3a2b2 + 3ab3 + 3ab2c into factors.
Ans. 3ab2(a+b+c). 2. Resolve 5a'b2_102373-5a-74-5a2b2 into factors.
Ans. 5a2b2(a2-2ab-62-1). 3. Resolve 6a2b2c3—12ab2c3mx— 18ab-coy into factors.
Ans. 6ab2c3(a—2mx— 3y). 4. Resolve 7ab2_7a263_7a2b2c into factors. 5. Resolve 8a2bc+12ab2c-16abc into factors. 6. Resolve 10ab?cmx-5abcy +5abc into factors.
87. When two terms of a trinomial are perfect squares, and the third term is twice the product of their square roots, the trinomial will be the square of the sum or difference of these roots, Arts. 66 and 67, and may be resolved into factors accordingly.
EXAMPLES. 1. Resolve aa-2ab +62 into factors. Ans. (a-6)(a−b). 2. Resolve a2 +4ab+462 into factors.
Ans. (a +26)(a +26). 3. Resolve a’ - 6ab+962 into factors.
4. Resolve 9a2_24ab+1662 into factors. 5. Resolve 25a- 60a2b3 + 3666 into factors. 6. Resolve 4m?n? - 4mn+1 into factors. 7. Resolve 49a"64-168a3b3 +144a2b2 into factors. 8. Resolve n3 +2n+n into three factors. 9. Resolve 16a+62-24a-bmw +9m222 into factors. 10. Resolve mana + 2moni + mane into three factors.
88. If a binomial consists of two squares connected by the minus sign, it must be equal to the product of the sum and difference of the square roots of the two terms, Art. 69, and may
be resolved into factors accordingly.
1. Resolve 4a2-962 into factors. Ans. (2a+36) (20—36). 2. Resolve 9a2b2_16a-c2 into factors. 3. Resolve a x-9ax into three factors. 4. Resolve a' _64 into three factors. 5. Resolve a6-66 into its factors. 6. Resolve a8-18 into four factors. 7. Resolve 1-2 into two factors. 8. Resolve 4-5 into two factors.
89. If the two terms of a binomial are both powers of the same degree, it may generally be resolved into factors accord. ing to the principles of Arts. 82-84.
EXAMPLES 1. Resolve ai_73 into its factors. Ans. (a? + ab +62) a–). 2. Resolve a' +63 into its factors. 3. Resolve a6_76 into four factors. 4. Resolve a3_873 into its factors. 5. Resolve 8a-1 into its factors. 6. Resolve 8q3_863 into three factors. 7. Resolve 1+2763 into its factors. 8. Resolve 8a3+2763 into its factors. 9. Resolve al6_716 into five factors.
GREATEST COMMON DIVISOR. — LEAST COMMON MULTIPLE.
90. A common divisor of two quantities is a quantity which will divide them both without a remainder. Thus 2ab is a common divisor of 6a2b2x and 10a%b3y.
91. A prime factor is one that can not be resolved into any other factors. It is, therefore, divisible only by itself and unity. Thus the quantity 2a2—2ab is the product of the three prime factors 2, a, and a–6.
92. The greatest common divisor of two quantities is the greatest quantity which will divide each of them without a remainder. It is the continued product of all the prime factors which are common to both. The term greatest here refers to the degree of a quantity, or of its leading term, and not to its arithmetical value.
93. When both quantities can be resolved into prime factors by methods already explained, the greatest common divisor may be found by the following
RULE. Resolve both quantities into their prime factors. The continued product of all those factors which are common to both, will be the greatest common divisor required.
EXAMPLES. 1. Find the greatest common divisor of 4a2bx and 6ab2ac?. Resolving into factors, we have
4a2bx=2a x 2a xbxx.
6ab22a x 36 x x x x x x x. The common factors are 2a, b, and x. Hence the greatest common divisor is 2abx.
2. Find the greatest common divisor of 4am+4bmand 3an +3bn. Resolving into factors, we have
4am+4bm-=2m x 2m(a+b).
3an+3bn=3n(a+b). Hence a+b is the greatest common divisor. 3. Find the greatest common divisor of 23— 43 and 2—yo.
203 — Yo=(x-4) (22 +ay+y?).
22-y2=(x,y) (x+y). Hence s-y is the greatest common divisor.
4. Find the greatest common divisor of 35a-bmx? and 42am.
5. Find the greatest common divisor of 3aac-6abæ+36200 and 4aly-46.
6. Find the greatest common divisor of 9mx2 - 6x + m and 9nx-n.
7. Find the greatest common divisor of 12a2—36ab+2762 and 8a2_1862.
94. When the given quantities can not be resolved into prime factors by inspection, the greatest common divisor may be found by applying the following principle:
The greatest common divisor of two quantities is the same with the greatest common divisor of the least quantity, and their remainder after division.
To prove this principle, let the greatest of the two quantities be represented by A, and the least by B. Divide A by B; let the entire part of the quotient be represented by Q, and the remainder by R. Then, since the dividend must be equal to the product of the divisor by the quotient, plus the remainder, we shall have A=QB+R.
Now every number which will divide B will divide QB; and every number which will divide R and QB will divide R+QB, or A. That is, every number which is a common divisor of B and R is a common divisor of A and B.
Again: every number which will divide A and B will di.