EQUATIONS.

59. Reducing Equations; Factoring Polynomials.-The process of dividing an equation f(x)=0 by (x−a), a being a root of the equation, is called dividing out or removing the root a; and the equation obtained, one unit lower in degree, and containing one root fewer, is called the reduced equation.

In the example of the preceding article, dividing the root (-2) out of the equation 2x3+x2−x+10=0, we get the reduced equation 2x2-3x+5=0. It is evident that the roots of the reduced equation are roots of the original equation. We are thus enabled in this case to solve the cubic equation completely, its roots being −2, †(3±V−31).

Whenever the left member of an equation can be factored, the solution of the equation can be simplified.

Again, since we know that (3+√−31) and †(3−√ −31) are roots of 2x2-3x+5=0, we know that 2x2-3x+5 can be divided exactly by x−1(3+√ −31) and by x-(3-√−31); hence its factors are [x-1(3+V-31)] [x-‡(3−√−=31)] and some number not involving x, which evidently is 2. Finally we are able to factor the cubic function:

2x3+x2−x+10=2(x+2)[x−4(3+√ −31)]

[x-1(3-V-31)].

Whenever we can solve the equation formed by setting a given polynomial equal to zero, we can factor the polynomial.

60. Factors of a Quadratic.—It is often important to know the nature of the factors of a quadratic expression such as may be represented in general by ax2+bx+c or by ax2+bxy+cy2. Since the factors of ax2+bx+c are found by solving the equation ax2+bx+c=0, they will be real when b2-4ac is positive, imaginary when b2-4ac is negative, and equal when b2-4ac-0. The same discrimination also holds for the form with two vari

ables, as may

be seen from the equation a(z)2 +0 (3)

a()*+b (&)+c=0.

61. Forming Equations. Suppose we wish to form an equation having the roots 1, 2 and 3. Evidently the equation

(x+1)(x−2)(x−3)=0

has these roots, for if we make x=-1, or 2 or 3, one of the factors of the left member becomes zero, and the equation is satisfied. Any other equation having the same roots must be divisible by (x+1), (x-2) and (x-3); if this other equation is only of degree 3, it will be k(x+1)(x−2)(x-3)=0, where k is some constant.

The general principles exemplified here are stated:

An equation having given roots a, b, c, etc., may be formed by setting equal to zero the product of the factors (x−α), (x−b), (x-c), etc.

Two equations of the same degree having the same roots, have also the same coefficients, or can be made to have by dividing one of the equations by a constant factor.

An equation having n roots is of the n-th degree.

62. Imaginary and Irrational Roots.-Given an equation f(x)=0 of which the coefficients are real, and suppose that an imaginary, a+bi (a and b real) is one of its roots; then

so that P=0 and Q=0; consequently

f(a−bi)=P-Qi=0,

so that a-bi is also a root.

Similarly, if the coefficients of f(x)=0 are rational, and a quadratic surd, a+bVc, is one of its roots, a-bvc is also a root.

These facts are generally stated:

Imaginary and quadratic surd roots occur in conjugate pairs. 63. Number of Roots of an Algebraic Equation.-Any algebraic equation of the n-th degree has just n roots; and any polynomial of the n-th degree can be factored (in only one way) into n linear factors.

This follows immediately from the fact that any algebraic equation has at least one root, the proof of which cannot be given here. Assuming it, we see that we could reduce any given equation of the nth degree to one of degree (n-1) by dividing out the assumed root; we could then treat the reduced equation in the same way, obtaining 2 roots in all of the original equation, and a reduced equation of degree n-2. Just n such steps are possible, so that there are n roots. The original equation can be factored in this way into n factors and written in the form

k(x− a)(x—b)(x−c)....=0.

There cannot be any root in addition to what would be found in this way, for if there were more than n roots, the degree of the equation would be higher than the nth. Furthermore, a different set of roots could not be found in some other way, for if we substitute for x a number p different from every one of the roots a, b, c, we get

k(p-a) (p-b) (p—c)....=0,

an untrue relation, since no one of the factors is zero.

Further, we have seen that if the coefficients of an algebraic equation are real, imaginary roots occur in pairs; hence: An algebraic equation of odd degree, if its coefficients are real, has at least one real root.

Examples.

In each of the following, either f(1) or f(−1) is or f(-1) is zero; aided by this information, factor f(x) and give all the roots of f(x)=0.

1. f(x)=3x-2x2+4x-5.

2. f(x) = 2x3-3x2-7x-2.

Factor each of the following:

3. 6x2-7x-3 and x2−x+1. Solve each of the following:

4. (x2-5x+6) (2x2+17x+8)=0. 5. x3-11x2+10x=0.

Form the rational equations of lowest degree having the roots indicated:

6. Roots -,, 1, −2.

7. Roots -1, 1, 4, 0.

8. Roots 100, 1, −1, −2.

9. One root is 3-V2, another -1.

10. Two of the roots are 1-√−3 and 1-√−2.

11. One root is V2-V3.

64. Relations of Roots and Coefficients. From the two articles just preceding, it follows that if an algebraic equation is divided through by the coefficient of its term of highest degree, thus being reduced to the form

-1

x2+P1xn-1+P2xn-2 + P3xn-3 +....+Pn=0,

its left member may be obtained by multiplying together n factors

where a, b, c, etc., are the n roots.

According to the Theorem of the Product of Linear Factors (Art. 53):

P1 = (a+b+c+....) the negative of the sum of the

-

=

roots;

P2=(ab+be+ca+....) the sum of the products of the

P3 (abc+....)

and so on, until finally

Pn=the product of all the roots if n is even, the negative of this product if n is odd.

The chief use that we shall make of this theorem is in checking the results of computations of roots; for this purpose the two results

are used.

P1=(a+b+c+....) and pn=±abc....

64a. Changing the Signs of all the Roots.-If, instead of a, b, c, etc., for the roots of the equation in the preceding article, we had used -a, -b, c, etc., their negatives, the n factors Iwould have been

with the same resulting coefficients except that none of the parentheses would be marked negative. Hence:

The signs of all the roots of an algebraic equation are changed by changing the signs of the coefficients of alternate powers of the variable.

Examples.

1. Write the equation whose roots are the negatives of the roots of 3x3 2x2 + x − 1 = 0; of 4x2 + 3x2+x-5=0; of 4x2-3x+5=0; of 2x3 −2x+1=0.

2. Solve x+4x3 +5x2+2x-2=0, one root being −1+V-1. 3. Solve 6x-13x3-35x2-x+3=0, one root being 2-V3. Solve the following equations:

4. 2x3-19x2+57x-54-0, the roots forming a g. p.

5. 3x-16x2+12x+16=0, one root being double one of the other roots.

6. 6x± — 7x3-16x2+21x-6=0, the product of two of the roots being 3.

7. 12x4x3-54x2+4x+24=0, the sum of two roots being

zero.

8. Show from the Law of Coefficients that if the area of a rectangle is a and its perimeter is p, its dimensions are the roots of s2-2s+a=0.

2