Consequently the roots of the cubic are x=3, x=

(2) Given x1+4x3-3x2-8x+4=0, to find x. Here the factors are

x2±√(—4x3 +3x2+8x—4);

or x2+2x±√(7x2+8x—4);
or x2+2x-2±√(3x2);

Therefore the component quadratics are

3+/21

2

x2+(2+3)x-2=0, and x2+(2/3)x-2=0; Solving these by the second rule, we have from the first 2x+2+3=±√ (8+(2+√3)}

−2−√3±√{8+(2+√3)3},

and from the second x=

2

̧−2+√3±√ {8+(2—√3)"} ̧

2

which are the four values of a sought, or the roots of the biquadratic.

3. Given x1-8x3+15x2+4x-8=0, to find x.

From the first, 2x-4=√(2—2√33 +16)=

√{2(9-√33)}.. x=2± √ {2(9—√33)}

And from the second, x=2±√{2(9+√33)} |

2

(4) Given 5x+4x3-3x2-4x+2=0.

2

1

Multiplying by 5, in order to make the first term a square,

we have

25x+20x+15x-20x+10=0;

The factors of the first member are

5x+(-20x3-15x+20x-10);

or 5x+2x±√(-11x+20x-10).

Here the quadratic expression under the radical is -(11x220x+10); and 4x11x10>202.. whatever real value be put for x, the expression 11a-20x+ 10 is always positive: and, consequently, the expression under the radical is always negative, whatever real value be put for x (see pp. 180, 182); hence no real value of x can ever make either of the factors zero; consequently the roots of the proposed equation are all imaginary.

OC

Rejecting the value x=0 given by the first of these, as belonging to the biquadratic and not to the proposed cubic, we

have x=

a

2

for one root; the others may be readily found by

solving the quadratic, as sufficiently explained.

(7) Given the equation x-4x+7x2-4x+1=0, to find x.

x2±√(4x3-7x2+4x-1);

or x2-2x+√(-3x2+4x-1);
or x2-2x+1±√x2;

..the component equations are x-(2-√-1)x+1=0, and x2-(2+√—1)x+1=0. From the first, 2x-(2—√—1)=±√{−4+(2~√—1)}

EXAMPLES FOR EXERCISE,

(1) x3-2x2+3x-2=0.
(2) x1-2x3-3x2-2x+k=0.
(3) x1+4x3+4x2+k=0.
(4) x1—4x3+ 2x2+4x+k=0.
(5) x1+6x3+x2—24x+k=0.
(6) x1-6x3+5x+12x+k=0.
(7) 8x1-16x3+24x2—2x+}=0.
(8) x1+8x3-5x2-24x+9=0.
(9) x3-3x2+7x2-6x+4=0.

3

(10) x-2x+5x3-3x+7=0.

2

(11) 5x-36x+67·2x2-8.64x+288=0.

NOTE.-Many examples might be proposed which the preceding method would not enable us to resolve into rational factors; and it might appear to the learner that when the method is applied to such examples, the operation is fruitless, and only so much work thrown away. It would be of comparatively little value if such were the case; but the process here explained is of considerable assistance in advancing the analysis and solution of the equation under all circumstances, as is sufficiently shown in the work referred to at page 190. Its leading us at once to the complete solution in such particular cases as those selected above, is only a col. lateral advantage of it, which has been exclusively dwelt upon here because it may be illustrated and applied independently of all knowledge beyond quadratic equations, and because, moreover, the practice of the method in particular examples will be a good introduction to its use in the doctrine of equations in general.

87. It may be here remarked that there always exists a value for the symbol u (page 186) that will cause the quantity under the radical to become a complete square: it is not suitable to this place to show how it may be determined generally, that is, for every particular example. In the cases above, it has been found to be such as to render the quantity under the radical monomial; but when this cannot be done, there are cases of the cubic in which the quantity under the radical may be made a complete trinomial square by the same sort of inspection, whenever the equation has a rational root, as the following examples, all except the last, from other books, will sufficiently show. (1) Let the equation be x-6x+4=0, or x2-6x+4x=0,

x2±√(6x2—4x);

or x2-1±√(4x2-4x+1)= x2-1±(2x-1); ..the component equations are x2—2x=0, x2+2x-2=0; so that the roots of the cubic are x=2, and x=-1±√3. (2) Let the equation be x3-3x2+4=0, or x1—3x3 +4x=0 x2±√(3x3-4x);

or x2

2

x2 — 3 x + 2 ± √ ( 25 x 2 — 10x+4) = x2—3x+2±

(5x-2).. the component equations are a3+x=0, aa—4x

+4=0, so that the roots of the cubic are 1, 2, and 2.

(3) Let the equation be 2+3x2-6x-8-0, or x+3x -6x-8x=0,

..the component equations are x2+4x=0, x--x-2=0; so that the roots of the cubic are -4, -1, and 2.

(4) To resolve the equation +(a+b)x2 +(ab+c)x+ac =0, which may be any cubic whatever, into rational factors. Multiplying by x, the factors of the resulting biquadratic are x2±√{~(a+b)x3—(ab+c)x3—acx},

a+b

C

x+ +

2

2

2

-b

4

+마을 (0)

hence the rational factors of the biquadratic are

x+ax, and x2+bx+c;

+(')'} =

and therefore those of the cubic, x+a, and x2+ bx+c.

This general investigation confirms the remark made above, namely, that after removing the term containing a3 from under the radical, such a value of u always exists as to render the expression under the radical in the next step a complete square; the proper value of u above is The trinomial square be

с

2

2

comes reduced to the monomial () when ab, in which case

-c is the coefficient of x2 in the second step, so that

at once; all equations, therefore, of the form

x3+2ax2+(a2+c)x + ac=0.............. (A),

is got

are easily solvible by this method: one root will always be a= -α. If b=0, or the equation be of the form x3+ax2+cx+ ac=0, where the final term is the product of the coefficients of the two preceding terms, the roots are xa, x=±√—c.