10. Find the square root of 25 a1b2 — 40a3b2c + 76a2b3c2 -48ab2c3+36b2c* — 30 a⭑bc + 24a3bc2 — 36 a2bc3+9a1c2. Ans. 5ab3ac-4abc+6bc2. 150. We will conclude this subject with the following remarks: (1) A binomial can never be a perfect square, since we know that the square of the most simple polynomial (viz., a binomial) contains three distinct parts, which cannot experience any reduction amongst themselves. Thus, the expression a2+b2 is not a perfect square: it wants the term ± 2ab, in order that it should be the square of a ±b. (2) In order that a trinomial, when arranged, may be a perfect square, its two extreme terms must be squares, and the middle term must be the double product of the square roots of the two others. Therefore, to obtain the square root of a trinomial when it is a perfect square, Extract the roots of the two extreme terms, and give these roots the same or contrary signs, according as the middle term is positive or negative. To verify it, see if the double product of the two roots is the same as the middle term of the trinomial. Thus, 9 a® — 48 a1b2 + 64 a2b1 is a perfect square, since and also √9a3a and √64ab8ab2; 2 x 3 a3 X 8ab2 48 a1b2 the middle term. = = But 4a2+14 ab +962 is not a perfect square: for although 4a2 and +962 are the squares of 2a and 3b, yet 2×2a ×3b is not equal to 14 ab. (3) In the series of operations required by the general rule, when the first term of one of the remainders is not exactly divisible by twice the first term of the root, we may conclude that the proposed polynomial is not a perfect square. This is an evident consequence of the course of reasoning by which we have arrived at the general rule for extracting the square root. (4) When the polynomial is not a perfect square, it may sometimes be simplified (see § 139). Take, for example, the expression √a3b+4a2b2+4ab3. The quantity under the radical sign is not a perfect square; but it can be put under the form ab(a2 + 4 ab+462). Now, the factor within the parenthesis is evidently the square of a+2b, whence we may conclude that √a3b+4a2b2+4ab3 = (a + 2b)√ab. Take also the expression √2ab — 4 ab2 + 2b3. √2a2b-4ab2+2b3 = (a - b)√2b. CUBE ROOT OF POLYNOMIALS. 150 a. From the rules for forming the powers of binomials, we have (a+b)3 = a3 +3a2b+3ab2 + b3. Hence the cube of a binomial is arranged with reference to the descending powers of its first term, and is equal to the cube of the first term plus three times the square of the first term multiplied by the second, plus other terms. In reversing the process, to find the cube root of a polynomial containing not more than four terms, we must arrange it with reference to one of its letters; then find the cube root of the first term for the first term of the root, and subtract its cube from the polynomial. The first term of the remainder will be three times the square of the first term of the root multiplied by the second; and therefore, to find the second term of the root, divide the first term of the remainder by three times the square of the first term of the root. As an example, extract the cube root of a3+3a2b+3ab2+b3. (a + b)3 = a3 +3a2b+3ab2+b3 0 The root is exact. Suppose it be required to raise x+y+c to the third power. We may write it in the form of a binomial, thus: (x+y)+c, in which the sum of x and y is taken for the first term. Then, as before, we have [(x + y) + c]3 == (x + y)3 + 3 (x + y)2 c + 3 (x + y) c2 +c3. Now, the cube root of the second member has three terms. The first two terms, x and y, are obtained just as in the preceding example. When the cube of (x+y), or (x+y)3, is subtracted, there will be a second remainder consisting of 3(x+y)2c, plus other terms. The first term of this remainder is 3c; that is, it is three times the square of the first term of the root multiplied by the third term of the root; and hence, to find the third term of the root, divide the first term of the second remainder by three times the square of the first term of the root. To illustrate by an example, extract the cube root of 266x15x20x15x2-6x+1. x-6x+15x-20x+15x2-6x+1|x2-2x+1 (x2 — 2x)3 = xˆ — 6x+12x1- 8.2-3 3 x*- etc. (x2-2x+1)3=x-6x+15x-20x3+15x2-6x+1 | 3x1, Divisor. In this example, we first extract the cube root of 2o, which gives x2, for the first term of the root. Squaring x2, and multiplying by 3, we obtain the divisor 32: this is contained in the second term - 2x times. Then cubing the sum of these two terms, and subtracting, we find that the first term of the remainder, 3x, contains the divisor once. Cubing the sum of the three terms and subtracting, we find a remainder 0. Hence x2 - 2x + 1 is the exact cube root. Hence we have the following rule: Arrange the given polynomial with reference to one of its letters, and extract the cube root of the first term. This will be the first term of the root. Divide the second term of the polynomial by three times the square of the first term of the root. The quotient will be the second term of the root. Subtract the cube of the sum of the two terms of the root from the given polynomial, and divide the first term of the remainder by three times the square of the first term of the root. quotient will be the third term of the root. The Continue this operation till a remainder 0 is found, or until one is found whose first term is not divisible by three times the square of the first term of the root. In the former case the root is exact: in the latter case the polynomial is an imperfect third power. Exercises. Find the cube roots of the following polynomials: 1. 8x-12x2+6x-1. 2. 15x+75x+125. 3. x+6x-40296x64. Ans. 2x-1. Ans. x + 5. Ans. x+2x-4. 4. 8x-12x+30 x 25 x3 +30x2-12x+8. Ans. 2x2 - x + 2. 5. 64a - 288 a3 + 1080 a3 — 1458 a — 729. Ans. 4a6a-9. 6. 1-6x+21x3 — 44x3 + 63 xa − 54x2 + 27 xo. Ans. 1-2x+3x2. By extracting the required root of the first and the last terms, two terms of the root may in general be found, from which the remaining ones may often be determined by inspection. The whole root may then be verified as above. |