squares and higher even powers of i, and hence P and Q will not be changed by changing the sign of i. Therefore when a Bi is substituted for x in the given expression the result will be P - Qi.

If now the given expression vanishes when a + Bi is substituted for x, we have P + Qi = 0.

Hence, as P and Q are real, we must have both P = 0 and Q = 0, and therefore P - Qi = 0.

Hence if the given expression vanishes when a + Bi is substituted for x, it will also vanish when a - Bi is substituted for x.

Therefore [Art. 88] if x a ẞi is a factor of the given expression, x-a + Bi will also be a factor.

Thus, if any expression rational and integral in x, and with all its coefficients real, be divisible by either of two conjugate complex expressions it will also be divisible by the other.

CHAPTER XV.

SQUARE AND CUBE ROOTS.

193. WE have already shewn how to find the square of a given algebraical expression; and we have now to shew how to perform the inverse operation, namely that of finding an expression whose square will be identically equal to a given algebraical expression. It will be seen that our knowledge of the mode of formation of squares will enable us in many cases to write down by inspection the square root of a given expression.

194. From the identity

a2+2ab+b2= (a + b)2,

we see that when a trinomial expression consists of the sum of the squares of any two quantities plus (or minus) twice their product, it is equal to the square of their sum (or difference).

Hence, to write down the square root of a trinomial expression which is a perfect square, arrange the expression according to descending powers of some letter; the square root of the whole expression will then be found by taking the square roots of the extreme terms with the same or with different signs according as the sign of the middle term is positive or negative.

Thus, to find the square root of

4a-12a+b3 + 966.

The square roots of the extreme terms are ± 2a* and +363. Hence, the middle term being negative, the required square root is ± (2a1 — 3b3).

Note. In future only one of the two square roots of an expression will be given, namely that one for which the sign of the first term is positive: to find the other root all the signs must be changed.

195. When an expression which contains only two different powers of a particular letter is arranged according to ascending or descending powers of that letter, it will only consist of three terms. For example, the expression a2 + b2 + c2+2bc + 2ca + 2ab when arranged according to powers of a is the trinomial

a2 + 2a (b+c) + (b2 + c2 + 2bc).

It follows therefore from the preceding article that however many terms there may be in an expression which is a perfect square, the square root can be written down by inspection, provided that the expression contains only two different powers of some particular letter.

Ex. 1. To find the square root of

a2+b2+c2+2bc + 2ca+2ab.

Arranged according to powers of a, we have
a2+2a (b+c) + (b+c)2, that is {a+(b+c)}2.
Hence the required square root is a+b+c.

Ex. 2. To find the square root of

4x2+9y+162*+ 12x2y2 – 16x2z2 – 24y2x2.

The given expression is

that is,

which is

4x4 + 4x2 (3y2 - 4z2) +9y1 — 24y2x2 + 16x4,

(2x2)2+2 (2x2) (3y2 — 4z2) + (3y2 — 4z2)2,

{2x2+(3y2-422)}2.

Hence the required square root is 2x2+3y2–422.

Ex. 3. To find the square root of

a2+2abx+ (b2+2ac) x2+2bcx3 +c2x2.

Arrange according to powers of a; we then have
a2+2a (bx + cx2)+b2x2+2bcx3 + c2x+,

that is,

a2+2a (bx + cx2)+(bx+cx2)2.

Hence the required square root is a + bx + cx2.

Ex. 4. To find the square root of

x6 - 2x+3x+2x3 (y − 1) + x2 (1 − 2y) +2xy + y2.

The expression only contains y2 and y; we therefore arrange it according to powers of y, and have

y2+2y (x3- x2+x)+x6 − 2x5 + 3x1 − 2x3 + x2.

Now, if the expression is a complete square at all, the last of the three terms must be the square of half the coefficient of y; and it is easy to verify that

(x3 − x2+x)2 = x6 – 2x5 + 3x1 – 2x3 +x2.

Hence the required square root is y+x3- x2+x.

196. To find the square root of any algebraical expression.

Suppose that we have to find the square root of (A + B)2, where A stands for any number of terms of the root, and B for the rest; the terms in A and B being arranged according to descending (or ascending) powers of some letter, so that every term in A is of higher (or lower) degree in that letter than any term of B.

Also suppose that the terms in A are known, and that we have to find the terms in B.

Subtracting A from (A + B), we have the remainder (2A + B) B.

Now from the mode of arrangement it follows that the term of the highest (or lowest) degree in the remainder is twice the product of the first term in A and the first term in B.

Hence, to obtain the next term of the required root, that

is, to obtain the highest (or lowest) term of B, we subtract from the whole expression the square of that part of the root which is already found, and divide the highest (or lowest) term of the remainder by twice the first term of the root.

The first term of the root is clearly the square root of the first term of the given expression; and, when we have found the first term of the root, the second and other terms of the root can be obtained in succession by the above process.

For example, to find the square root of

x6-4x56x1 - 8x3+9x2 - 4x +4.

The process is written as follows:

(x3- 2x2+x-2)2 = x6 − 4x5 + 4x4 – 8x3 +9x2 - 4x +4

We first take the square root of the first term of the given expression, which must be arranged according to ascending or descending powers of some letter: we thus obtain x3, the first term of the required root.

Now subtract the square of x3 from the given expression, and divide the first term of the remainder, namely -4x5, by 2x3: we thus obtain - 2x2, the second term of the root.

Now subtract the square of x3 – 2x2 from the given expression, and divide the first term of the remainder, namely 2x4, by 2x3: we thus obtain x, the third term of the root.

Now subtract the square of x3- 2x2+x from the given expression, and divide the first term of the remainder, namely - 4x3, by 2x3: we thus obtain 2, the fourth term of the root.

Subtract the square of x3- 2x2+x-2 from the given expression and there is no remainder.

Hence x3- 2x2+x-2 is the required square root.

The squares of x3, x3 - 2x2, &c. are placed under the given expression, like terms being placed in the same column, so that in every case the first term of the remainder is obvious.

197. The square root of an algebraical expression may also be obtained by means of the theorem of Art. 91.

Take for example the case just considered.