Extraction of Roots of Algebraic Quantities.

147. Let us first consider the case of monomials, and in order to deduce a rule for extracting the nth root, let us examine the law for the formation of the nth power.

From the definition of a power, it follows that each factor of the root will enter the power, as many times as there are units in the exponent of the power. That is, to form the th power of a monomial,

We form the nth power of the co-efficient for a new co-efficient, and write after this, each letter affected with an exponent equal to n times its primitive exponent.

Conversely, we have for the extraction of the n' root of a monomial, the following

RULE.

Extract the nth root of the numerical co-efficient for a new coefficient, and after this write each letter affected with an exponent

1

equal to th of its exponent in the given monomial; the result

n

will be the required root.

3

Thus, 3/64a9b3c® = 4a3bc2;

and

4/16a8b12c42a2b3c.

From this rule we perceive, that in order that a monomial may be a perfect n' power:

1st. Its co-efficient must be a perfect n' power; and

2d. The exponent of each letter must be divisible by n.

It will be shown, hereafter, how the expression for the root of a quantity, which is not a perfect power, is reduced to its simplest form.

148. Hitherto, ir. finding the power of a monomial, we have paid no attention to the sign with which the monomial may be affected. It has already been shown, that whatever be the sign of a monomial, its square is always positive.

14

Let n be any whole number; then every power of an even degree, as 2n, can be considered as the n1 power of the square; that is, (a2)" = a2n: hence, it follows,

That every power of an even degree, will be essentially posi tive, whether the quantity itself be positive or negative.

Thus,

(±2a2b3c)+=+16a8b12c4.

Again, as every power of an uneven degree, 2n + 1, is but the product of the power of an even degree, 2n, by the first power; it follows that,

Every power of a monomial, of an uneven degrce, has the same sign as the monomial itself.

and (-4a26)3

= 64a6b3.

Hence, (+4a2b)3 = +64a6b3; and

From the preceding reasoning, we conclude,

1st. That when the index of the root of a monomial is uneven, the root will be affected with the same sign as the monomial.

2d. When the index of the root is even, and the monomial a

positive quantity, the root has both

the signs + and

Thus, 4 81a4b12 ± 3ab3; 6

3/64a18 = ± 2a3.

3d. When the index of the root is even, and the monomial negative, the root is impossible;

For, there is no quantity which, being raised to a power of an even degree, will give a negative result. Therefore,

are symbols of operations which it is impossible to execute They are imaginary expressions.

Extraction of the nth Root of Polynomials.

148. Let N denote any polynomial whatever, arranged with reference to a certain letter. Now, the n power of a polynomial is the continued product arising from taking the polynomial n times as a factor: hence, the first term of the product, when arranged with reference to a certain letter, is the n' power of the first term of the polynomial, arranged with reference to the same letter.

Therefore, the n' root of the first term of such a product, will be the first term of the nth root of the product.

Let us denote the first term of the nth root of N by r, and the following terms, arranged with reference to the leading letter of the polynomial, by r', r'', r'"', &c. We shall have,

N = (r+r' + r'' + . . &c.)";

or, if we designate the sum of all the terms after the first

N= (r + s)” = rn + nrn−1s + &c.,→

= r2+nrn−1(r' + p'' + &c. ) + &c.

If now, we subtract r" from N, and designate the remainder by R, we shall have,

which remainder will evidently be arranged with reference to the leading letter of the polynomial; therefore, the first term will contain a higher power of that letter than either of the succeeding terms, and cannot be reduced with any of them. Hence, if we divide the first term of the first remainder, by n times the (n − 1)

power of the first term of the root, the quotient will be the second term of the root.

If now, we place r + r' = u, and denote the sum of the suo ceeding terms of the root by s', we shall have,

N = (u + s')” = u" + nu”−1s' + &c.

un

If now, we subtract us from N, and denote the remainder by

R', we shall have,

R' = N — u* = n(r + r')n−1s' + &c.,

= npm-1(p'' + p''' + &c.) + &c.,

=nrn-1p'' + &c.

In we divide the first term of this remainder by n times the (n-1)th power of the first term of the root, we shall have the third term of the root. If we continue the operation, we shall find that the first term of any new remainder, divided by n times the (n - 1) power of the first term of the root, will give a new term of the root.

It may be remarked, that since the first term of the first remainder is the same as the second term of the given polynomial, we can find the second term of the root, by dividing the second term of the given polynomial by n times the (n-1) power of the first term.

Ilence, for the extraction of the n root of a polynomial, we have the following

RULE.

I. Arrange the given polynomial with reference to one of its letters, and extract the nth root of the first term; this will be the first term of the root.

II. Divide the second term by n times the (n − 1) power of the first term of the root; the quotient will be the second term of the root.

III. Subtract the nth power of the sum of the two terms already found from the given polynomial, and divide the first term of the remainder by n times the (n − 1) power of the first term of the root; the quotient will be the third term of the root.

IV. Continue this operation till a remainder is found equal to 0, or, till one is found whose first term is not divisible by n times the (n-1)th power of the first term of the root: in the former case the root is exact, and the given polynomial a perfect nth power; in the latter case, the polynomial is an imperfect nth power.

149. Let us apply the foregoing rule to the following

EXAMPLES.

1. Extract the cube root of x6—6x5+15xa—20x3+15x2-6x+1.

x6—6x5+15x1—20x3+15x2-6x+1 | x2-2x+1

(x2 —2x)3 —x6—6x5+12x1— 8x3

1st rem.

3x4

3x4-12x3+ &c.

(x2-2x+1)3=x6—6x5+15x1—20x3+15x2-6x+1.

In this example, we first extract the cube root of 26, which gives 22, for the first term of the root. Squaring 2, and multiplying by 3, we obtain the divisor 34: this is contained in the second term - 6x5, -2x times. Then cubing the part of the root found, and subtracting, we find that the first term of the remainder 3x4, contains the divisor once. Cubing the whole root found, we find the cube equal to the given polynomial. Hence, x2 2x+1, is the exact cube root.

4. Find the 4th root of 16a4-96a3x+216a2x2 - 216ax +81x*

16a-96a3x+216a2x2-216ax3+81x4 | 2a-3x

(2α-3x)+= 16aa—96a3x+216a2x2-216ax3+81x4 4 × (2a)3=32u3.

We first extract the 4th root of 16a4, which is 2a. We then raise 2a to the third power, and multiply by 4, the index of the root; this gives the divisor 32a3. This divisor is contained in the second term 96a3x, - 3x times, which is the second term of the root. Raising the whole root found to the 4th power we find the power equal to the given polynomial.

.

5. What is the 4th root of the polynomial,

81ac + 16b1d1 — 96a2cb3d3 · - 216a6c3bd + 216a+c2b2d2.

6. Find the 5th root of

32x5

80x480x3- 40x2+10x — 1.